Solution: The equations that govern the motion are:

Problem 13.1 In Eample 13., suppose that the vehicle is dropped from a height h = 6m. What is the downward velocit 1 s after it is released? What is its downward velocit just before it reaches the ground? The equations that govern the motion are: a = g = 9.81 m/s v = gt s = 1 gt + h h v = gt = (9.81 m/s )(1 s) = 9.81 m/s. The downward velocit is 9.81 m/s. We need to first determine the time at which the vehicle hits the ground s = = 1 h gt + h t = g = (6 m) 9.81 m/s = 1.16 s Now we can solve for the velocit v = gt = (9.81 m/s )(1.16 s) = 1.8 m/s. The downward velocit is 1.8 m/s. Problem 13. The milling machine is programmed so that during the interval of time from t = tot = s, the position of its head (in inches) is given as a function of time b s = 4t t 3. What are the velocit (in in/s) and acceleration (in in/s ) of the head at t = 1s? The motion is governed b the equations s = (4 in/s)t ( in/s )t, s v = (4 in/s) ( in/s )t, a = ( in/s ). At t = 1 s, we have v =,a = 4 in/s. 9

Problem 13.3 In an eperiment to estimate the acceleration due to gravit, a student drops a ball at a distance of 1 m above the floor. His lab partner measures the time it takes to fall and obtains an estimate of.46 s. What do the estimate the acceleration due to gravit to be? Let s be the ball s position relative to the floor. Using the value of the acceleration due to gravit that the obtained, and assuming that the ball is released at t =, determine s (in m) as a function of time. s The governing equations are a = g v = gt s = 1 gt + h s When the ball hits the floor we have = 1 gt + h g = h (1 m) = = 9.45 m/s t (.46 s) g = 9.45 m/s The distance s is then given b s = 1 (9.45 m/s ) + 1m. s = (4.73 m/s )t + 1. m. Problem 13.4 The boat s position during the interval of time from t = stot = 1 s is given b s = 4t + 1.6t.8t 3 m. Determine the boat s velocit and acceleration at t = 4s. What is the boat s maimum velocit during this interval of time, and when does it occur? s = 4t + 1.6t.8t 3 v = ds a = dv = 4 + 3.t.4t = 3..48t a) v(4s) = 1.96 m/s a(4s) = 1.8 m/s b) a = 3..48t = t = 6.67s v(6.67s) = 14.67 m/s 1

Problem 13.5 The rocket starts from rest at t = and travels straight up. Its height above the ground as a function of time can be approimated b s = bt + ct 3, where b and c are constants. At t = 1 s, the rocket s velocit and acceleration are v = 9 m/s and a = 8. m/s. Determine the time at which the rocket reaches supersonic speed (35 m/s). What is its altitude when that occurs? The governing equations are s = bt + ct 3, s v = bt + 3ct, a = b + 6ct. Using the information that we have allows us to solve for the constants b and c. (9 m/s) = b(1 s) + 3c(1 s), (8. m/s ) = b + 6c(1 s). Solving these two equations, we find b = 8.8 m/s, c =.177 m/s3. When the rocket hits supersonic speed we have (35 m/s) = (8.8 m/s )t + 3(.177 m/s3 )t t = 13. s. The altitude at this time is s = (8.8 m/s )(13. s) + (.177 m/s3 )(13. s)3 s = 194 m. Problem 13.6 The position of a point during the interval of time from t = to t = 6 s is given b s = 1 t 3 + 6t + 4t m. What is the maimum velocit during this interval of time, and at what time does it occur? What is the acceleration when the velocit is a maimum? s = 1 t 3 + 6t + 4t m dv = (it could be a minimum) da = 3 so we have a maimum. This occurs at t = 4 s. At this point v = 3 t + 1t + 4 m/s Maimum velocit occurs where a = Ma velocit is at t = 4 s. where v = 8 m/s and a = m/s a = 3t + 1 m/s Please purchase PDF Split-Merge on www.verpdf.com to remove this watermark. 11

Problem 13.7 The position of a point during the interval of time from t = tot = 3 seconds is s = 1 + 5t t 3 m. What is the maimum velocit during this interval of time, and at what time does it occur? What is the acceleration when the velocit is a maimum? The velocit is ds = 1t 3t. The maimum occurs when dv = 1 6t =, from which This is indeed a maimum, since d v = 6 <. The maimum velocit is v = [ 1t 3t ] = 8.33 m/s t=1.667 t = 1 6 = 1.667 seconds. The acceleration is dv = when the velocit is a maimum. Problem 13.8 The rotating crank causes the position of point P as a function of time to be s =.4 sin (πt) m. P (c) Determine the velocit and acceleration of P at t =.375 s. What is the maimum magnitude of the velocit of P? When the magnitude of the velocit of P is a maimum, what is the acceleration of P? s s =.4 sin(πt) v = ds a = dv =.8π cos(πt) = 1.6π sin(πt) a) v(.375s) = 1.777 m/s a(.375) = 11. m/s b) v ma =.8π =.513 m/s c) v ma t =,nπ a = Problem 13.9 For the mechanism in Problem 13.8, draw graphs of the position s, velocit v, and acceleration a of point P as functions of time for t s. Using our graphs, confirm that the slope of the graph of s is zero at times for which v is zero, and the slope of the graph of v is zero at times for which a is zero. 1

Problem 13.1 A seismograph measures the horizontal motion of the ground during an earthquake. An engineer analzing the data determines that for a 1-s interval of time beginning at t =, the position is approimated b s = 1 cos(πt) mm. What are the maimum velocit and maimum acceleration of the ground during the 1-s interval? The velocit is ds = (π)1 sin(πt) mm/s =.π sin(πt) m/s. The velocit maima occur at dv =.4π cos(πt) =, from which πt = (n 1)π, or t = n = 1,, 3,...M, where (n 1), 4 (M 1) 4 1 seconds. These velocit maima have the absolute value ds = [.π] =.68 m/s. (n 1) t= 4 The acceleration is d s =.4π cos(πt). The acceleration maima occur at d 3 s 3 = d v =.8π 3 sin(πt) =, from which πt = nπ, ort = n,n=, 1,,...K, where K 1 seconds. These acceleration maima have the absolute value dv nπ =.4π = 3.95 m/s. t= Problem 13.11 In an assembl operation, the robot s arm moves along a straight horizontal line. During an interval of time from t = tot = 1 s, the position of the arm is given b s = 3t t 3 mm. Determine the maimum velocit during this interval of time. What are the position and acceleration when the velocit is a maimum? s s = 3t t 3 mm v = 6t 6t mm/s a = 6 1t mm/s da = 1 mm/s 3 Maimum velocit occurs when dv = a =. This occurs at = 6 1t or t = 1/ second. (since da/ <, we have a maimum). The velocit at this time is v = (6) v = 15 mm/s ( ) ( ) 1 1 6 mm/s 4 The position and acceleration at this time are s = 7.5.5 mm s = 5mm a = mm/s 13

Problem 13.1 In Active Eample 13.1, the acceleration (in m/s ) of point P relative to point O is given as a function of time b a = 3t. Suppose that at t = the position and velocit of P are s = 5 m and v = m/s. Determine the position and velocit of P at t = 4s. The governing equations are a = (3 m/s 4 )t v = 1 3 (3 m/s4 )t 3 + ( m/s) O s P s s = 1 1 (3 m/s4 )t 4 + ( m/s)t + (5 m) At t = 4 s, we have s = 77 m,v = 66 m/s. Problem 13.13 The Porsche starts from rest at time t =. During the first 1 seconds of its motion, its velocit in km/h is given as a function of time b v =.8t.88t, where t is in seconds. What is the car s maimum acceleration in m/s, and when does it occur? What distance in km does the car travel during the 1 seconds?.8 km hr.88 km hr First convert the numbers into meters and seconds )( ) 1hr = 6.33 m/s 36 s ( 1 m 1km ( )( ) 1 m 1hr =.44 m/s 1km 36 s The governing equations are then s = 1 (6.33 m/s)(t /s) 1 3 (.88 m/s)(t3 /s ), v = (6.33 m/s)(t/s) (.88 m/s)(t /s ), a = (6.33 m/s ) (.88 m/s)(t/s ), d a = (.88 m/s)(1/s ) = 1.76 m/s 3. The maimum acceleration occurs at t = (and decreases linearl from its initial value). a ma = 6.33 m/s @ t = In the first 1 seconds the car travels a distance [ ( 1 s =.8 km ) (1 s) 1 (.88 km ) (1 s) 3 ]( ) 1hr hr s 3 hr s 36 s s =.35 km. 14

Problem 13.14 The acceleration of a point is a = t m/s. When t =,s = 4 m and v = 1 m/s. What are the position and velocit at t = 3s? v = a+ C 1, The velocit is where C 1 is the constant of integration. Thus v = t + C 1 = 1t + C 1. At t =, v = 1 m/s, hence C 1 = 1 and the velocit is v = 1t 1 m/s. The position is s = v+ C, At t =, s = 4 m, thus C = 4. The position is s = ( ) 1 t 3 1t + 4 m. 3 At t = 3 seconds, [ ] 1 s = 3 t3 1t + 4 = 1 m. t=3 The velocit at t = 3 seconds is v = [ 1t 1 ] = 8 m/s. t=3 where C is the constant of integration. s = (1t 1) + C = ( ) 1 t 3 1t + C. 3 Problem 13.15 The acceleration of a point is a = 6t 36t m/s. When t =,s = and v = m/s. What are position and velocit as a function of time? The velocit is v = a+ C 1 = (6t 36t ) + C 1 = 3t 1t 3 + C 1. The position is s = v+ C = (3t 1t 3 + ) + C At t =, v = m/s, hence C 1 =, and the velocit as a function of time is v = 3t 1t 3 + m/s. = 1t 3 3t 4 + t + C. At t =, s =, hence C =, and the position is s = 1t 3 3t 4 + t m 15

Problem 13.16 As a first approimation, a bioengineer studing the mechanics of bird flight assumes that the snow petrel takes off with constant acceleration. Video measurements indicate that a bird requires a distance of 4.3 m to take off and is moving at 6.1 m/s when it does. What is its acceleration? The governing equations are a = constant, v = at, s = 1 at. Using the information given, we have 6.1 m/s = at, 4.3 m= 1 at. Solving these two equations, we find t = 1.41 s and a = 4.33 m/s. Problem 13.17 Progressivel developing a more realistic model, the bioengineer net models the acceleration of the snow petrel b an equation of the form a = C(1 + sin ωt), where C and ω are constants. From video measurements of a bird taking off, he estimates that ω = 18/s and determines that the bird requires 1.4 s to take off and is moving at 6.1 m/s when it does. What is the constant C? acceleration We find an epression for the velocit b integrating the a = C(1 + sin ωt), v = Ct + C ω (1 cos ωt) = C ( t + 1 ω 1 ω cos ωt ). Using the information given, we have 6.1 m/s = C (1.4 s + s 18 s ) 18 cos[18(1.4)] Solving this equation, we find C = 4.8 m/s. 16

Problem 13.18 Missiles designed for defense against ballistic missiles have attained accelerations in ecess of 1 g s, or 1 times the acceleration due to gravit. Suppose that the missile shown lifts off from the ground and has a constant acceleration of 1 g s. How long does it take to reach an altitude of 3 m? How fast is it going when it ranches that altitude? The governing equations are a = constant, v = at, s = 1 at Using the given information we have 3 m = 1 1(9.81 m/s )t t =.47 s The velocit at that time is v = 1(9.81 m/s )(.47 s) = 43 m/s. v = 43 m/s. Problem 13.19 Suppose that the missile shown lifts off from the ground and, because it becomes lighter as its fuel is epended, its acceleration (in g s) is given as a function of time in seconds b a = 1 1.t. What is the missile s velocit in kilometres per hour 1s after liftoff? We find an epression for the velocit b integrating the acceleration (valid onl for <t<5s). t t ( ) 1g 1gs v = a = = 1.(t/s). ln 1 1.[t/s] At time t = 1 s, we have v = 1(. ( 9 81 m/s )(1 s) ln. v = 394 k m/ h. 1 1. ) = 195 m/s. 17

Problem 13. The airplane releases its drag parachute at time t =. Its velocit is given as a function of time b 8 v = 1 +.3t m/s. What is the airplane s acceleration at t = 3s? v = 8 1 +.3t ; a = dv 5.6 = (1 +.3t) a(3 s) = 6.66 m/s Problem 13.1 How far does the airplane in Problem 13. travel during the interval of time from t = to t = 1 s? v = 8 1 s 1 +.3t ; s = 8 = 5 ln 1 +.3t ( 1 + 3. 1 ) = 359 m Problem 13. The velocit of a bobsled is v = 1t m/s. When t = s, the position is s = 5 m. What is its position at t = 1 s? The equation for straight line displacement under constant acceleration is s = a(t t ) + v(t )(t t ) + s(t ). Choose t =. At t =, the acceleration is [ ] dv(t) a = = 1 m/s, t= the velocit is v(t ) = 1() = m/s, and the initial displacement is s(t ) = 5 m. At t = 1 seconds, the displacement is s = 1 (1 ) + (1 ) + 5 = 55 m 18

Problem 13.3 In September, 3, Ton Schumacher started from rest and drove 4 km in 4.498 seconds in a National Hot Rod Association race. His speed as he crossed the finish line was 58 km/h. Assume that the car s acceleration can be epressed b a linear function of time a = b + ct. Determine the constants b and c. What was the car s speed s after the start of the race? a = b + ct, v = bt + ct, s = bt + ct3 6 Both constants of integration are zero. 58 km/h = b(4.498 s) + c (4.498 s) 4 km = b (4.498 s) + c (4.498 s)3 6 b = 54 m/s 3 c = 9.5 m/s v = b( s) + c (s) = 89 m/s Problem 13.4 The velocit of an object is v = t m/s. When t = 3 seconds, its position is s = 6 m. What are the position and acceleration of the object at t = 6s? dv(t) = 4t m/s. The acceleration is At t = 6 seconds, the acceleration is a = 4 m/s. Choose the initial conditions at t = 3 seconds. The position is obtained from the velocit: 6 [ s(t t ) = v(t) + s(t ) = t ] 6 3 3 t3 + 6 = 17 m. 3 Problem 13.5 An inertial navigation sstem measures the acceleration of a vehicle from t = tot = 6s and determines it to be a = +.1t m/s. At t =, the vehicle s position and velocit are s = 4 m, v = 4 m/s, respectivel. What are the vehicle s position and velocit at t = 6s? a = +.1t m/s v = 4 m/s s = 4 m Integrating v = v + t +.1t / s = v t + t +.1t 3 /6 + s Substituting the known values at t = 6s,weget v = 55.8 m/s s = 531.6 m 19

Problem 13.6 In Eample 13.3, suppose that the cheetah s acceleration is constant and it reaches its top speed of 1 km/h in 5 s. What distance can it cover in 1 s? The governing equations while accelerating are a = constant, v = at, s = 1 at. Using the information supplied, we have ( ) 1 1 m = a(5 s) a = 6. 6 7 m/s 36 s The distance that he travels in the first 1 s (5 seconds accelerating and then the last 5 seconds traveling at top speed) is s = 1 ( ) ( ) (5 s) 1 1 m 6.67 m/s + (1 s 5s) = 5 m. 36 s s = 5 m. Problem 13.7 The graph shows the airplane s acceleration during its takeoff. What is the airplane s velocit when it rotates (lifts off) at t = 3 s? a 9 m/ s Velocit = Area under the curve v = 1 (3 m/s + 9 m/s )(5 s) + (9 m/s )(5 s) = 55 m/s 3 m/s 5 s 3 s t Problem 13.8 Determine the distance traveled during its takeoff b the airplane in Problem 13.7. for t 5s ( ) ( ) a = 6 m/s t + (3 m/s 6 m/s ), v = 5s 5s t + (3 m/s )t ( ) 6 m/s t 3 s = 5s 6 + (3 m/s ) t v(5 s) = 3 m/s, s(5 s) = 6.5 m for 5 s t 3 s a = 9 m/s, v = (9 m/s )(t 5s) + 3 m/s, s = (9 m/s ) (t 5s) + (3 m/s)(t 5s) + 6.5 m s(3 s) = 365 m

Problem 13.9 The car is traveling at 48 km/h when the traffic light 9 m ahead turns ellow. The driver takes one second to react before he applies the brakes. 48 km/h After he applies the brakes, what constant rate of deceleration will cause the car to come to a stop just as it reaches the light? How long does it take the car to travel the 9 m? 9 m for t 1s a =, v= 48 km/h = 13.33 m/s,s= ( 13.33 m/s)t s(1 s) = 13.33 m for t>1s a = c(constant), v = ct + 13.33 m/s, s = c t + ( 13.33 m/s )t + 13.33 m At the stop we have 9 m = c t + ( 13.33 m/s )t + 13.33 m = ct + 13.33 m/s a) c = 1. 17 m/s b) t = 11.41 s Problem 13.3 The car is traveling at 48 km/h when the traffic light 9 m ahead turns ellow. The driver takes 1 s to react before he applies the accelerator. If the car has a constant acceleration of m/s and the light remains ellow for 5 s, will the car reach the light before it turns red? How fast is the car moving when it reaches the light? First, convert the initial speed into m/s. 48 km/h = 13.33 m/s. At the end of the 5 s, the car will have traveled a distance [ ] 1 d = ( 13.33 m/s )(1 s) + ( m/s )(5 s 1s) + ( 13.33 m/s )(5 s 1s) = 8.65 m. When the light turns red, the driver will still be 7.35 m from the light. No. To find the time at which the car does reach the light, we solve [ ] 1 9 m = ( 13.33 m/s)(1 s) + ( m/s )(t 1s) + ( 13.33 m/s )(t 1s) t = 5. 34 s. The speed at this time is v = 13.33 m/s + ( m/s ) (5. 34 s 1s) =.1 m/s. v = 79. k m/ h. 1

Problem 13.31 A high-speed rail transportation sstem has a top speed of 1 m/s. For the comfort of the passengers, the magnitude of the acceleration and deceleration is limited to m/s. Determine the time required for a trip of 1 km. Strateg: A graphical approach can help ou solve this problem. Recall that the change in the position from an initial time t to a time t is equal to the area defined b the graph of the velocit as a function of time from t to t. Divide the time of travel into three intervals: The time required to reach a top speed of 1 m/s, the time traveling at top speed, and the time required to decelerate from top speed to zero. From smmetr, the first and last time intervals are equal, and the distances traveled during these intervals are equal. The initial time is obtained from v(t 1 ) = at 1, from which t 1 = 1/ = 5 s. The distance traveled during this time is s(t 1 ) = at1 / from which s(t 1) = (5) / = 5 m. The third time interval is given b v(t 3 ) = at 3 + 1 =, from which t 3 = 1/ = 5 s. Check. The distance traveled is s(t 3 ) = a t 3 + 1t 3, from which s(t 3 ) = 5 m. Check. The distance traveled at top speed is s(t ) = 1 5 5 = 95 m = 95 km. The time of travel is obtained from the distance traveled at zero acceleration: s(t ) = 95 = 1t, from which t = 95. The total time of travel is t total = t 1 + t + t 3 = 5 + 95 + 5 = 15 s = 17.5 minutes. A plot of velocit versus time can be made and the area under the curve will be the distance traveled. The length of the constant speed section of the trip can be adjusted to force the length of the trip to be the required 1 km. Problem 13.3 The nearest star, Proima Centauri, is 4. light ears from the Earth. Ignoring relative motion between the solar sstem and Proima Centauri, suppose that a spacecraft accelerates from the vicinit of the Earth at.1 g (.1 times the acceleration due to gravit at sea level) until it reaches one-tenth the speed of light, coasts until it is time to decelerate, then decelerates at.1 g until it comes to rest in the vicinit of Proima Centauri. How long does the trip take? (Light travels at 3 1 8 m/s.) The distance to Proima Centauri is d = (4. light - ear)(3 1 8 m/s)(365.4 da) = 3.995 1 16 m. ( ) 864 s 1 da Divide the time of flight into the three intervals. The time required to reach.1 times the speed of light is t 1 = v a = 3 17 m/s.981 m/s = 3.581 18 seconds. The distance traveled is s(t 1 ) = a t 1 + v()t + s(), where v() = and s() = (from the conditions in the problem), from which s(t 1 ) = 4.587 1 15 m. From smmetr, t 3 = t 1, and s(t 1 ) = s(t 3 ). The length of the middle interval is s(t ) = d s(t 1 ) s(t 3 ) = 3.777 1 16 m. The time of flight at constant velocit is t = 3.777 116 m 3 1 7 = 1.6 1 9 seconds. The total time of flight is t total = t 1 + t + t 3 = 1.63751 1 9 seconds. In solar ears: t total = ( 1.63751 1 9 sec ) ( )( ) 1 solar ears 1 das 365.4 das 864 sec = 51.9 solar ears

Problem 13.33 A race car starts from rest and accelerates at a = 5 + t m/s for 1 seconds. The brakes are then applied, and the car has a constant acceleration a = 3 m/s until it comes to rest. Determine the maimum velocit, the total distance traveled; (c) the total time of travel. (c) the total time of travel is t = 15. The total distance traveled is For the first interval, the velocit is v(t) = (5 + t) + v() = 5t + t since v() =. The velocit is an increasing monotone function; hence the maimum occurs at the end of the interval, t = 1 s, from which s(t 1) = a (t 1) + v(1)(t 1) + s(1), from which s(5) = 3 5 + 15(5) + 583.33 = 958.33 m v ma = 15 m/s. The distance traveled in the first interval is s(1) = [ 1 5 (5t + t ) = t + 1 ] 1 3 t3 The time of travel in the second interval is = 583.33 m. v(t 1) = = a(t 1) + v(1), t 1 s, from which (t 1) = 15 3 = 5, and Problem 13.34 When t =, the position of a point is s = 6 m and its velocit is v = m/s. From t = tot = 6 s, the acceleration of the point is a = + t m/s. From t = 6 s until it comes to rest, its acceleration is a = 4 m/s. What is the total time of travel? What total distance does the point move? For the first interval the velocit is v(t) = ( + t ) + v() = [t + 3 ] t3 + m/s. The total time of travel is t total = 39.5 + 6 = 45.5 seconds. The velocit at the end of the interval is v(6) = 158 m/s. The displacement in the first interval is ( s(t) = t + ) [ 3 t3 + + 6 = t + 1 ] 6 t4 + t + 6. The distance traveled is s(t 6) = 4 (t 6) + v(6)(t 6) + s(6) = (39.5) + 158(39.5) + 7, The displacement at the end of the interval is s(6) = 7 m. For the second interval, the velocit is v(t 6) = a(t 6) + v(6) =,t 6, from which (t 6) = v(6) a = 158 4 = 39.5. from which the total distance is s total = 339 m 3

Problem 13.35 Zoologists studing the ecolog of the Serengeti Plain estimate that the average adult cheetah can run 1 km/h and that the average springbuck can run 65 km/h. If the animals run along the same straight line, start at the same time, and are each assumed to have constant acceleration and reach top speed in 4 s, how close must the a cheetah be when the chase begins to catch a springbuck in 15 s? The top speeds are V c = 1 km/h = 7.78 m/s for the cheetah, and V s = 65 km/h = 18.6 m/s. The acceleration is a c = V c 4 = 6.94 m/s for the cheetah, and a s = V s 4 = 4.513 m/s for the springbuck. Divide the intervals into the acceleration phase and the chase phase. For the cheetah, the distance traveled in the first is s c (t) = 6.94 (4) = 55.56 m. The total distance traveled at the end of the second phase is s total = V c (11) + 55.56 = 361.1 m. For the springbuck, the distance traveled during the acceleration phase is s s (t) = 4.513 (4) = 36.11 m. The distance traveled at the end of the second phase is s s (t) = 18.6(11) + 36.1 = 34.7 m. The permissible separation between the two at the beginning for a successful chase is d = s c (15) s s (15) = 361.1 34.7 = 16.4 m Problem 13.36 Suppose that a person unwisel drives 1 km/h in a 88 km/h zone and passes a police car going 88 km/h in the same direction. If the police officers begin constant acceleration at the instant the are passed and increase their speed to 19 km/h in 4 s, how long does it take them to be even with the pursued car? The conversion from mi/h to m/s is The distance traveled b the pursued car during this acceleration is km h = 1 m 36 s =.78 m/s The acceleration of the police car is a = ( 19 88 )(. 78 ) m/s 4s =.85 m/s. The distance traveled during acceleration is s(t 1 ) =. 85 (4) + 88(. 78)(4) = 11 m. s c (t 1 ) = 1 (. 78 ) t 1 = 33.36 (4) = 133.4 m. The separation between the two cars at 4 seconds is d = 133.4 11 = 1.4 m. This distance is traversed in the time 1.4 t = = 1. 9. ( 19 88 )(. 78 ) The total time is t total = 1. 9 + 4 = 5.9 seconds. Problem 13.37 If θ = 1 rad and dθ is the velocit of P relative to O? = 1 rad/s, what Strateg: You can write the position of P relative to O as s = ( m) cos θ + ( m) cos θ and then take the derivative of this epression with respect to time to determine the velocit. The distance s from point O is s = ( m) cos θ + ( m) cos θ. O m s m P The derivative is ds = 4 sin θ dθ. For θ = 1 radian and dθ = 1 radian/second, ds = v(t) = 4(sin(1 rad)) = 4(.841) = 3.37 m/s 4

Problem 13.38 In Problem 13.37, if θ = 1 rad, dθ/ = rad/s and d θ/ =, what are the velocit and acceleration of P relative to O? The velocit is ds = 4 sin θ dθ = 4(sin(1 rad))( ) = 6.73 m/s. The acceleration is d s = 4 cos θ ( ) dθ 4 sin θ ( d ) θ, from which d s = a = 4 cos(1 rad)(4) = 8.64 m/s Problem 13.39 If θ = 1 rad and dθ is the velocit of P relative to O? = 1 rad/s, what mm 4 mm The acute angle formed b the 4 mm arm with the horizontal is given b the sine law: sin α = 4 sin θ, O s P from which sin α = ( ) sin θ. 4 For θ = 1 radian, α =.4343 radians. The position relative to O is. s = cos θ + 4 cos α. The velocit is ds = v(t) = sin θ ( ) ( ) dθ dα 4 sin α. From the epression for the angle α, cos α ( ) dα =.5 cos θ ( ) dθ, from which the velocit is v(t) = ( sin θ tan α cos θ) ( ) dθ. Substitute: v(t) = 18.4 mm/s. 5

Problem 13.4 In Active Eample 13.4, determine the time required for the plane s velocit to decrease from 5 m/s to 1 m/s. From Active Eample 13.4 we know that the acceleration is given b a = (.4/m)v. We can find an epression for the velocit as a function of time b integrating a = dv 1 m/s 5 m/s = (.4/m)v dv v = (.4/m)t ( dv v = 1 ) 1 m/s ( = 1s v 5 m/s 1 m + 1s ) ( = 4s ) = (.4/m)t 5 m 5 m ( t = 4s )( ) 1m = s. 5 m.4 t = s. Problem 13.41 An engineer designing a sstem to control a router for a machining process models the sstem so that the router s acceleration (in cm/s ) during an interval of time is given b a =.4v, where v is the velocit of the router in cm/s. When t =, the position is s = and the velocit is v = cm/s. What is the position at t = 3s? We will first find the velocit at t = 3. a = dv ( ).4 = v, s s v dv 3 s ( ).4 cm/s v =, s ( ) v ln = cm/s (.4 s ) (3 s) = 1. v = ( cm/s)e 1. =.6 cm/s. Now we can find the position a = vdv ( ).4 = v, ds s.6 cm/s cm/s vdv v ( ).4 s = ds s ( ).4 (.6 cm/s) ( cm/s)= s s s = 1.398 cm/s = 3.49 cm.4/ s s = 3.49 cm 6

Problem 13.4 The boat is moving at 1 m/s when its engine is shut down. Due to hdrodnamic drag, its subsequent acceleration is a =.5v m/s, where v is the velocit of the boat in m/s. What is the boat s velocit 4 s after the engine is shut down? a = dv v 1 m/s = (.5 m 1 )v dv t v = (.5 m 1 ) 1 m/s v = 1 + (.5 s 1 )t 1 v v = (.5 m 1 )t 1 m/s v(4 s) = 3.33 m/s Problem 13.43 In Problem 13.4, what distance does the boat move in the 4 s following the shutdown of its engine? From Problem 13.4 we know v = ds 1 m/s 4s 1 m/s = 1 + (.5 s 1 s(4 s) = )t 1 + (.5 s 1 )t [ ] + (1 s 1 )(4 s) s(4 s) = ( m) ln = 1.97 m Problem 13.44 A steel ball is released from rest in a container of oil. Its downward acceleration is a =.4.6v cm/s, where v is the ball s velocit in cm/s.what is the ball s downward velocit s after it is released? a = dv v = (.4 cm/s) (.6 s 1 )v dv (.4 cm /s) (.6 s 1 )v = 5 3 ln ( v + 4 cm/s 4 cm/s t ) = t v = (4 cm/s) (1 ) e (.6 s 1 )t v( s) =.795 cm/s Problem 13.45 In Problem 13.44, what distance does the ball fall in the first s after its release? From 13.44 we know v = ds = (4 cm/s) (1 ) e (.6 s 1 )t t ( ) s( s) = (4 cm/s) 1 e (.6 s 1 )t = cm 3 s( s) = 3.34 cm. ( ) e (.6 s 1 )t 1 + (4 cm/s )t 7

Problem 13.46 The greatest ocean depth et discovered is the Marianas Trench in the western Pacific Ocean. A steel ball released at the surface requires 64 minutes to reach the bottom. The ball s downward acceleration is a =.9g cv, where g = 9.81 m/s and the constant c = 3. s 1. What is the depth of the Marianas Trench in kilometers? a = dv =.9g cv. Separating variables and integrating, v dv t.9g cv = = t. Integrating and solving for v, v = ds =.9g c (1 e ct ). Integrating, s t ds = We obtain s =.9g c.9g c (1 e ct ). (t + e ct 1 c c ). At t = (64)(6) = 384 s, we obtain s = 11,5 m. Problem 13.47 The acceleration of a regional airliner during its takeoff run is a = 14.3v m/s, where v is its velocit in m/s. How long does it take the airliner to reach its takeoff speed of m/s? a = dv = (14 m/s ) (.3 m 1 )v m/s dv t (14 m/s ) (.3 m 1 )v = t = 5.1 s Problem 13.48 In Problem 13.47, what distance does the airliner require to take off? a = v dv ds = (14 m/s ) (.3 m 1 )v m/s vdv (14 m/s ) (.3 m 1 )v = s ds s = 343 m 8

Problem 13.49 A sk diver jumps from a helicopter and is falling straight down at 3 m/s when her parachute opens. From then on, her downward acceleration is approimatel a = g cv, where g = 9.81 m/s and c is a constant. After an initial transient period she descends at a nearl constant velocit of 5 m/s. (c) What is the value of c, and what are its SI units? What maimum deceleration is the sk diver subjected to? What is her downward velocit when she has fallen meters from the point at which her parachute opens? Assume c>. After the initial transient, she falls at a constant velocit, so that the acceleration is zero and cv = g, from which Integrate: ( 1 ) ln g cv =s + C. c (c) c = g 9.81 m/s = v (5) m =.394 m 1 /s The maimum acceleration (in absolute value) occurs when the parachute first opens, when the velocit is highest: a ma = g cv = g c(3) =343.4 m/s Choose coordinates such that distance is measured positive downward. The velocit is related to position b the chain rule: dv = dv ds = v dv ds ds = a, from which vdv g cv = ds. When the parachute opens s = and v = 3 m/s, from which ( ) 1 C = ln g 9c = 7.4398. c The velocit as a function of distance is ln g cv = c(s + C). For s = m, v = 14.4 m/s Problem 13.5 The rocket sled starts from rest and accelerates at a = 3 + t m/s until its velocit is 4 m/s. It then hits a water brake and its acceleration is a =.3v m/s until its velocit decreases to 1 m/s. What total distance does the sled travel? a = 3 + t m/s v = 3t + t m/s s = 15t + t 3 /3m Acceleration Phase When v = 4 m/s, acceleration ends. At this point, t = 1 s and s = 1833 m. Deceleration Phase starts at s 1 = 1833 m, v 1 = 4 m/s. Let us start a new clock for the deceleration phase. v f = 1 m/s a = v dv ds =.3v sf s 1 ds = 1 (.3) vf v 1 vdv v s f 1833 m = 1 [ln(1) ln(4)] (.3) s f = 3 m 9

Problem 13.51 In Problem 13.5, what is the sled s total time of travel? From the solution to Problem 13.5, the acceleration takes 1 s. At t = 1 s, the velocit is 4 m/s. We need to find out how long it takes to decelerate from 4 m/s to 1 m/s and add this to the 1 s required for acceleration. The deceleration is given as a = dv td.3 = =.3v m/s 1 4 dv v.3t d = 1 1 v 4 t d =.5 s.3t d = 3 4 ( 1 = 1 1 ) 4 t = 1 + t d = 1.5 s Problem 13.5 A car s acceleration is related to its position b a =.1s m/s. When s = 1 m, the car is moving at 1 m/s. How fast is the car moving when s = 4 m? a = v dv =.1s m/s ds vf 1 4 vdv=.1 sds 1 [ v ] vf [ s ] 4 m =.1 1 m/s 1 m vf = 1 +.1 (4 1 ) v f = 4.5 m/s Problem 13.53 Engineers analzing the motion of a linkage determine that the velocit of an attachment point is given b v = A + 4s m/s, where A is a constant. When s = m, its acceleration is measured and determined to be a = 3 m/s. What is its velocit of the point when s = m? The velocit as a function of the distance is When s = m, a = 3 m/s, from which A = 4. v dv ds = a. Solve for a and carr out the differentiation. The velocit at s = mis v = 4 + 4( ) = m/s a = v dv ds = (A + 4s )(8s). 3

Problem 13.54 The acceleration of an object is given as a function of its position in feet b a = s ( m/s ). When s =, its velocit is v = 1 m/s. What is the velocit of the object when s = m? We are given a = vdv ( = ds ) s, m-s v 1m/s ( m vdv = ) s ds m-s v (1 m/s) v = 3.4 m/s. ( (m) 3 = m-s) 3 Problem 13.55 Gas guns are used to investigate the properties of materials subjected to high-velocit impacts. A projectile is accelerated through the barrel of the gun b gas at high pressure. Assume that the acceleration of the projectile is given b a = c/s, where s is the position of the projectile in the barrel in meters and c is a constant that depends on the initial gas pressure behind the projectile. The projectile starts from rest at s = 1.5 m and accelerates until it reaches the end of the barrel at s = 3 m. Determine the value of the constant c necessar for the projectile to leave the barrel with a velocit of m/s. s a = v dv ds = c s, ( m/s) m/s ( ) 3m = c ln 1.5 m 3m c vdv = 1.5m s ds c = 8.85 1 3 m /s Problem 13.56 If the propelling gas in the gas gun described in Problem 13.55 is air, a more accurate modeling of the acceleration of the projectile is obtained b assuming that the acceleration of the projectile is given b a = c/s γ, where γ = 1.4 is the ratio of specific heats for air. (This means that an isentropic epansion process is assumed instead of the isothermal process assumed in Problem 13.55.) Determine the value of the constant c necessar for the projectile to leave the barrel with a velocit of m/s. a = v dv ds = c s 1.4, ( m/s) m/s 3m c vdv = ds 1.5m s1.4 =.5c ((3m).4 (1.5m).4) c = 38.86 1 3 m.4 /s 31

Problem 13.57 A spring-mass oscillator consists of a mass and a spring connected as shown. The coordinate s measures the displacement of the mass relative to its position when the spring is unstretched. If the spring is linear, the mass is subjected to a deceleration proportional to s. Suppose that a = 4s m/s, and that ou give the mass a velocit v = 1 m/s in the position s =. s How far will the mass move to the right before the spring brings it to a stop? What will be the velocit of the mass when it has returned to the position s =? The velocit of the mass as a function of its position is given b vdv/ds = a. Substitute the given acceleration, separate variables, and integrate: vdv= 4sds, from which v / = s + C. The initial velocit v() = 1 m/s at s =, from which C = 1/. The velocit is v / = s + 1/. The velocit is zero at the position given b = (s 1 ) + 1, 1 from which s 1 =± 4 =±1 m. Since the displacement has the same sign as the velocit, s 1 =+1/ m is the distance traveled before the spring brings it to a stop. At the return to s =, the velocit is v =± =±1 m/s. From the phsical situation, the velocit on the first return is negative (opposite the sign of the initial displacement), v = 1 m/s. Problem 13.58 In Problem 13.57, suppose that at t = ou release the mass from rest in the position s = 1 m. Determine the velocit of the mass as a function of s as it moves from the initial position to s =. From the solution to Problem 13.57, the velocit as a function of position is given b v = s + C. At t =,v = and s = 1 m, from which C = (1) =. The velocit is given b v(s) =±( 4s + 4) 1 =± 1 s m/s. From the phsical situation, the velocit is negative (opposite the sign of the initial displacement): v = 1 s m/s. [Note: From the initial conditions, s 1 alwas.] 3

Problem 13.59 A spring-mass oscillator consists of a mass and a spring connected as shown. The coordinate s measures the displacement of the mass relative to its position when the spring is unstretched. Suppose that the nonlinear spring subjects the mass to an acceleration a = 4s s 3 m/s and that ou give the mass a velocit v = 1 m/s in the position s =. How far will the mass move to the right before the springs brings it to a stop? What will be the velocit of the mass when it has returned to the position s =? s Find the distance when the velocit is zero. a = vdv ( ) ( ) 4 = ds s s + m s s 3 d vdv = 1m/s (1 m/s) = [( ) ( ) ] 4 s s + m s s 3 ds ( ) 4 d ( ) d 4 s m s 4 Solving for d we find d =.486 m. When the cart returns to the position s =, we have a = vdv ( ) ( ) 4 = ds s s + m s s 3 v [( ) ( ) ] 4 vdv = 1m/s s s + m s s 3 ds = v (1 m/s) =, v =±1 m/s. The cart will be moving to the left, so we choose v = 1 m/s. Problem 13.6 The mass is released from rest with the springs unstretched. Its downward acceleration is a = 3. 5s m/s, where s is the position of the mass measured from the position in which it is released. How far does the mass fall? What is the maimum velocit of the mass as it falls? The acceleration is given b s a = dv = dv ds = v dv ds ds = 3. 5s m/s. Integrating, we get v s vdv= (3. 5s)ds or v = 3.s 5s. The mass falls until v =. Setting v =, we get = (3. 5s)s. Wefind v = ats = and at s = 1.88 m. Thus, the mass falls 1.88 m before coming to rest. From the integration of the equation of motion, we have v = (3.s 5s ). The maimum velocit occurs where dv ds =. From the original equation for acceleration, we have a = v dv ds = (3. 5s) m/s. Since we want maimum velocit, we can assume that v = at this point. Thus, = (3. 5s), ors = (3./5) m when v = v MAX. Substituting this value for s into the equation for v, weget ( (3.) vmax = 5 or v MAX = 4.55 m/s ) (5)(3.) 5, 33

Problem 13.61 Suppose that the mass in Problem 13.6 is in the position s = and is given a downward velocit of 1 m/s. How far does the mass fall? What is the maimum velocit of the mass as it falls? a = v dv ds = (3. m/s ) (5 s )s v s vdv = [(3. m/s ) (5 s )s]ds 1 m/s v (1 m/s) = (3. m/s ) s (5 s ) s v = (1 m/s) + (64.4 m/s )s (5 s )s The mass falls until v = = (1 m/s) + (64.4 m/s )s (5 s )s s =. m The maimum velocit occurs when a = = (3. m/s ) (5 s )s s =.644 m v = (1 m/s) + (64.4 m/s )(.644 m) (5 s )(.644 m) v = 1.99 m/s Problem 13.6 If a spacecraft is 161 km above the surface of the earth, what initial velocit v straight awa from the earth would be required for the vehicle to reach the moon s orbit 38, 94 km from the center of the earth? The radius of the earth is 637 km. Neglect the effect of the moon s gravit. (See Eample 13.5.) 161 km 38,94 km For computational convenience, convert the acceleration due to Earth s gravit into the units given in the problem, namel miles and hours: g = ( 9. 81 m 1 s )( )( 1 km 36 s ) 1h = 17137. 6 km/h. 1 m The velocit as a function of position is given b v dv ds = a = gr E s. Separate variables, vdv= gre ds s. Integrate: v = gr E ( 1 ) + C. Suppose that the velocit at the distance of the Moon s orbit is zero. Then ( 637 ) = ( 17137. 6 ) + C, 3894 from which C = 696164 km /h. At the 161 km altitude, the equation for the velocit is ( ) v = g RE + C. R E + 161 From which v = 1553351991= 39,413 km/h Converting: ( )( )( ) 39413 km 1 m 1h v = = 1, 948 m/s. 1h 1km 36 s Check: Use the result of Eample 13.5 v = gr E (where H >s alwas), and H = 38, 94, ( 1 1 ), s H from which v = 39,413 km/h. check. 34

Problem 13.63 The moon s radius is 1738 km. The magnitude of the acceleration due to gravit of the moon at a distance s from the center of the moon is 4.89 1 1 m/s. s Suppose that a spacecraft is launched straight up from the moon s surface with a velocit of m/s. Set G = 4.89 1 1 m 3 /s, r = 1.738 1 6 m, v = m/s a = v dv v r ds = G s G v vdv = ds v s v ( 1 = G r 1 ) r ( ) v = v r r + G rr r What will the magnitude of its velocit be when it is 1 km above the surface of the moon? What maimum height above the moon s surface will it reach? v(r + 1. 1 6 m) = 1395 m/s The maimum velocit occurs when v = r = G = 61 km G r v h = r r = 47 km Problem 13.64* The velocit of an object subjected onl to the earth s gravitational field is [ ( 1 v = v + gr E s 1 )] 1/, s where s is the object s position relative to the center of the earth, v is the velocit at position s, and R E is the earth s radius. Using this equation, show that the object s acceleration is given as a function of s b a = gre /s. ( 1 v = v + gr E s 1 ) 1/ s a = dv = v dv ds Rewrite the equation given as v = v + gr E s gr E s Take the derivative with respect to s. v dv ds = gr E s Thus a = v dv ds gr E s Problem 13.65 Suppose that a tunnel could be drilled straight through the earth from the North Pole to the South Pole and the air was evacuated. An object dropped from the surface would fall with the acceleration a = gs/r E, where g is the acceleration of gravit at sea level, R E is radius of the earth, and s is the distance of the object from the center of the earth. (The acceleration due to gravitation is equal to zero at the center of the earth and increases linearl with the distance from the center.) What is the magnitude of the velocit of the dropped object when it reaches the center of the earth? s N S R E Tunnel v dv ds = gs R E. The velocit as a function of position is given b Separate variables and integrate: ( ) g v = s + C. R E At s = R E, v =, from which C = gr E. Combine and reduce: ( ) v = gr E 1 s RE At the center of the earth s =, and the velocit is v = gr E 35

Problem 13.66 Determine the time in seconds required for the object in Problem 13.65 to fall from the surface of the earth to the center. The earth s radius is 637 km. From Problem 13.65, the acceleration is a = v dv ds = g s R E v s ( ) g vdu= sds R E R E ( ) g v = (RE R s ) E Recall that v = ds/ v = ds g =± RE R s E R E ds g =± RE s R E tf ( ) g s t f =±sin 1 =±sin 1 (1) R E R E R E RE π t f =± =±166 s =±1.1 min g Problem 13.67 In a second test, the coordinates of the position (in m) of the helicopter in Active Eample 13.6 are given as functions of time b = 4 + t, = 4 + 4t + t. What is the magnitude of the helicopter s velocit at t = 3s? What is the magnitude of the helicopter s acceleration at t = 3s? We have = (4 m) + ( m/s)t, = (4 m) + (4 m/s)t + (1 m/s )t, v = ( m/s), v = (4 m/s) + (1 m/s )t, a =, a = (1 m/s ). At t = 3 s, we have = 1 m, v = m/s, a =, = 5 m, v = 1 m/s, a = m/s. Thus v = v + v = 1. m/s v = 1. m/s. a = + ( m/s ) = m/s a = m/s. 36

Problem 13.68 In terms of a particular reference frame, the position of the center of mass of the F-14 at the time shown (t = ) is r = 1i + 6j + k (m). The velocit from t = tot = 4sisv = (5 + 6t)i + (1 + t )j (4 + t )k (m/s). What is the position of the center of mass of the plane at t = 4s? r = 1i + 6j + k m v = (5 + 6t)i + (1 + t )j (4 + t )k m/s 4 4 = v = 5t + 3t + 4 = (5)(4) + 3(4) + 1 m = 66. m 4 4 = v = 1t + t 3 /3 + 4 = 1(4) + (4) 3 /3 + 6m= 75.3 m 4 z 4 = v z = (4t + t 3 /3) + z z 4 = 4(4) (4) 3 /3 + = 36.7 m r t=4s = 66i + 75.3j 36.7k (m) Problem 13.69 In Eample 13.7, suppose that the angle between the horizontal and the slope on which the skier lands is 3 instead of 45. Determine the distance d to the point where he lands. 3 m d 45 The skier leaves the surface at 1 m/s. The equations are a =, a = 9.81 m/s, v = (1 m/s) cos, v = (9.81 m/s )t (1 m/s) sin s = (1 m/s) cos t, s = 1 (9.81 m/s )t (1 m/s) sin t When he hits the slope, we have s = d cos 3 = (1 m/s) cos t, s = ( 3 m) d sin 3 = 1 (9.81 m/s )t (1 m/s) sin t Solving these two equations together, we find t = 1.1 s, d = 11. m. 37

Problem 13.7 A projectile is launched from ground level with initial velocit v = m/s. Determine its range R if θ = 3 ; θ = 45 (c) θ = 6. Set g = 9.81 m/s,v = m/s a = g, v = gt + v sin θ, s = 1 gt + v sin θ t a =, v = v cos θ, s = v cos θ t u R When it hits the ground, we have = 1 gt + v sin θ t t = v sin θ g R = v cos θ t R = v sin θ g a) θ = 3 R = 35.3 m b) θ = 45 R = 4.8 m c) θ = 6 R = 35.3 m Problem 13.71 Immediatel after the bouncing golf ball leaves the floor, its components of velocit are v =.66 m/s and v = 3.66 m/s. Determine the horizontal distance from the point where the ball left the floor to the point where it hits the floor again. The ball leaves the floor at =, =. Determine the ball s coordinate as a function of. (The parabolic function ou obtain is shown superimposed on the photograph of the ball.) The governing equations are a =, a = g, v = v, v = gt + v, = v t, = 1 gt + v t When it hits the ground again = 1 gt + v t t = v g (.66 m/s)(3.66 m/s) = 9.81 m/s =.494 m. At an point of the flight we have t =, = 1 ( ) ( ) v g + v v v ( ) = 1 9.81 m/s [.66 m/s] 3.66 m/s +.66 m/s ( ) = v t = v = v v g g ( ) 11. = + 5.53. m 38

Problem 13.7 Suppose that ou are designing a mortar to launch a rescue line from coast guard vessel to ships in distress. The light line is attached to a weight fired b the mortar. Neglect aerodnamic drag and the weight of the line for our preliminar analsis. If ou want the line to be able to reach a ship 91 m awa when the mortar is fired at 45 above the horizontal, what muzzle velocit is required? 45 From 13.7 we know that R = v sin θ Rg v = g sin θ v = ( 91)( 9.81 m/s ) sin(9 ) = 9.9 m/s Problem 13.73 In Problem 13.7, what maimum height above the point where it was fired is reached b the weight? From Problem 13.7 we have v = gt + v sin θ, s = 1 gt + v sin θ t When we reach the maimum height, = gt + v sin θ t = v sin θ g h = 1 gt + v sin θ t h = 1 g ( v sin θ g ) ( ) v sin θ + v sin θ g h = v sin θ g Putting in the numbers we have h = ( 9.9 m/s ) sin (45 ) ( =.78 m 9.81 m/s ) 39

Problem 13.74 When the athlete releases the shot, it is 1.8 m above the ground and its initial velocit is v = 13.6 m/s. Determine the horizontal distance the shot travels from the point of release to the point where it hits the ground. 3 v The governing equations are a =, v = v cos 3, s = v cos 3 t, a = g v = gt + v sin 3 s = 1 gt + v sin 3 t + h When it hits the ground, we have s = (13.6 m/s) cos 3 t, s = = 1 (9.81 m/s )t + (13.6 m/s) sin 3 t + 1.8 m. Solving these two equations, we find t = 1.6 s, s = 19. m. Problem 13.75 A pilot wants to drop surve markers at remote locations in the Australian outback. If he flies at a constant velocit v = 4 m/s at altitude h = 3 m and the marker is released with zero velocit relative to the plane, at what horizontal d from the desired impact point should the marker be released? h We want to find the horizontal distance traveled b the marker before it strikes the ground ( goes to zero for t>.) a = a = g d v = v v = v gt = + v t = + v t gt / From the problem statement, =,v =,v = 4 m/s, and = 3 m The equation for becomes = 3 (9.81)t / Solving with =, we get t f =.47 s. Substituting this into the equation for, we get f = 4t f = 98.9 m 4 c 8 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copright laws as the currentl eist. No portion of this material ma be reproduced, in an form or b an means, without permission in writing from the publisher.

Problem 13.76 If the pitching wedge the golfer is using gives the ball an initial angle θ = 5, what range of velocities v will cause the ball to land within 3 m of the hole? (Assume the hole lies in the plane of the ball s trajector). v θ 3 m 3 m Set the coordinate origin at the point where the golfer strikes the ball. The motion in the horizontal () direction is given b a =, V = V cos θ, = (V cos θ )t. The motion in the vertical () direction is given b a = g, V = V sin θ gt, = (V sin θ )t gt. From the equation, we can find the time at which the ball reaches the required value of (7 or 33 metres). This time is t f = f /(V cos θ ). We can substitute this information the equation for Y with Y f = 3m and solve for V. The results are: For hitting (7,3) metre, V = 31. m/s. For hitting (33,3) metre, V = 34. m/s. Problem 13.77 A batter strikes a baseball.9 m above home plate and pops it up. The second baseman catches it 1.8 m above second base 3.68 s after it was hit. What was the ball s initial velocit, and what was the angle between the ball s initial velocit vector and the horizontal? Second base 7 m The equations of motion g = 9.81 m/s a = a = g v = v cos θ v = gt + v sin θ s = v cos θ t s = 1 gt + v sin θ t +.9m When the second baseman catches the ball we have 38. m = v cos θ (3.68 s) Home plate 7 m 1.8m= 1 ( 9.81 m/s ) (3.68 s) + v sin θ (3.68 s) Solving simultaneousl we find v = 1 m/s, θ = 6.4 41

Problem 13.78 A baseball pitcher releases a fastball with an initial velocit v = 144.8 km/h. Let θ be the initial angle of the ball s velocit vector above the horizontal. When it is released, the ball is 1.83 m above the ground and 17.7 m from the batter s plate. The batter s strike zone etends from.56 m above the ground to 1.37 m above the ground. Neglecting aerodnamic effects, determine whether the ball will hit the strike zone if θ = 1 ; if θ =. 1.37 m.56 m 17.7m The initial velocit is v = 144.8 km/h = 4.3 m/s. The velocit equations are (1) () (3) dv dv =, from which v = v cos θ. = g, from which v = gt + v sin θ. d = v cos θ, from which (t) = v cos θt, since the initial position is zero. Substitute: (t p ) = h = g ( ) d + d tan θ + 1.83. v cos θ For θ = 1, h = 1. m,yes, the pitcher hits the strike zone. For θ =, h = 1. 5 m No, the pitcher misses the strike zone. (4) d = gt + v sin θ, from which (t) = g t + v sin θt + 1.83, since the initial position is () = 1.83 m. At a distance d = 17.7 m, the height is h. The time of passage across the home plate is (t p ) = d = v cos θt p, from which t p = d v cos θ. Problem 13.79 In Problem 13.78, assume that the pitcher releases the ball at an angle θ = 1 above the horizontal, and determine the range of velocities v (in m/s) within which he must release the ball to hit the strike zone. From the solution to Problem 13.78, h = g ( ) d + d tan θ + 1.83, v cos θ where d = 17.7 m, and 1.37 h.56 m. Solve for the initial velocit: gd v = cos θ(d tan θ + 1.83 h). For h = 1. 37, v = 44.74 m/s. For h =. 56 m, v = 31.15 m/s. The pitcher will hit the strike zone for velocities of release of 31. 15 v 44. 74 m/s, and a release angle of θ = 1. Check: The range of velocities in miles per hour is 11. 1km/h v 161.1 km/h, which is within the range of major league pitchers, although the 16.9 km upper value is achievable onl b a talented few (Nolan Ran, while with the Houston Astros, would occasionall in a game throw a 168.9 km fast ball, as measured b hand held radar from behind the plate). 4

Problem 13.8 A zoolog student is provided with a bow and an arrow tipped with a sringe of sedative and is assigned to measure the temperature of a black rhinoceros (Diceros bicornis). The range of his bow when it is full drawn and aimed 45 above the horizontal is 1 m. A truculent rhino charges straight toward him at 3 km/h. If he full draws his bow and aims above the horizontal, how far awa should the rhino be when the student releases the arrow? The strateg is to determine the range and flight time of the arrow when aimed above the horizontal, to determine the distance traveled b the rhino during this flight time, and then (c) to add this distance to the range of the arrow. Neglect aerodnamic drag on the arrow. The equations for the trajector are: Denote the constants of integration b V,V,C,C, and the velocit of the arrow b V A. (1) dv =, from which v = V.Att =,V = V A cos θ. () (3) (4) dv = g, from which v = gt + V. At t =,V = V A sin θ. d = v = V A cos θ, from which (t) = V A cos θt + C. At t =, () =, from which C =. d = v = gt + V A sin θ, from which = g t + V A sin θt + C. At t =, =, from which C =. The time of flight is given b ( (t flight ) = = g ) t flight + V A sin θ t flight, from which t flight = V A sin θ. g The range is given b (t flight ) = R = V A cos θt flight = V A cos θ sin θ. g The maimum range (1 meters) occurs when the arrow is aimed 45 above the horizon. Solve for the arrow velocit: V A = grma = 31.3 m/s. The time of flight when the angle is is t flight = V A sin θ g =.18 s, and the range is R = V A cos θt flight = 64.3 m. The speed of the rhino is 3 km/h = 8.33 m/s. The rhino travels a distance d = 8.33(.18) = 18. m. The required range when the arrow is released is d + R = 8.5 m. 43

Problem 13.81 The crossbar of the goalposts in American football is c = 3.5 m above the ground. To kick a field goal, the ball must make the ball go between the two uprights supporting the crossbar and be above the crossbar when it does so. Suppose that the kicker attempts a 36.58 m field goal, and kicks the ball with an initial velocit v = 1.3 m/s and θ = 4. B what vertical distance does the ball clear the crossbar? v θ c c Set the coordinate origin at the point where the ball is kicked. The (horizontal) motion of the ball is given b a =, V = V cos θ, = (V cos θ )t. The motion is given b a = g, V = V sin θ gt, = (V sin θ )t gt. Set = c = 36.58 m and find the time t c at which the ball crossed the plane of the goal posts. Substitute this time into the equation to find the coordinate Y B of the ball as it passes over the crossbar. Substituting in the numbers (g = 9.81 m/s ),wegett c =.4 s and B = 6. 11 m. Thus, the ball clears the crossbar b 3.7 m. Problem 13.8 An American football quarterback stands at A. At the instant the quarterback throws the football, the receiver is at B running at 6.1 m/s toward C, where he catches the ball. The ball is thrown at an angle of 45 above the horizontal, and it is thrown and caught at the same height above the ground. Determine the magnitude of the ball s initial velocit and the length of time it is in the air. Set as the horizontal motion of the football, as the vertical motion of the football and z as the horizontal motion of the receiver. Set g = 9.81 m/s, θ = 45. We have C Receiver s path Path of the ball a z =, v z = 6.1 m/s,s z = ( 6.1 m/s)t a = g, v = gt + v sin θ, s = 1 gt + v sin θ t 9 a =, v = v cos θ, s = v cos θ t When the ball is caught we have s z = ( 6.1 m/s)t B 9.1 m A = 1 gt + v sin θ t s = v cos θ t s = s z + ( 9.1 m) We can solve these four equations for the four unknowns s,s z,v,t We find t = 1.67 s, v = 11.6 m/s 44

Problem 13.83 The cliff divers of Acapulco, Meico must time their dives that the enter the water at the crest (high point) of a wave. The crests of the waves are 1 m above the mean water depth h = 4 m. The horizontal velocit of the waves is equal to gh. The diver s aiming point is m out from the base of the cliff. Assume that his velocit is horizontal when he begins the dive. What is the magnitude of the driver s velocit when he enters the water? How far from his aiming point must a wave crest be when he dives in order for him to enter the water at the crest? 6 m 1 m h 6.4 m m t =,v =, = 7 m, = a = g = 9.81 m/s V = V gt 7 m = gt / = 1matt IMPACT for an ideal dive to hit the crest of the wave t 1 = t IMPACT =.3 s 8.4 m V (t 1 ) =.59 m/s a = V = V X I = V t 1 + X At impact X I = 8.4 m. For impact to occur as planned, then V = 8.4/t 1 = 3.65 m/s = constant The velocit at impact is V = (V ) + [V (t 1 )] =.9 m/s The wave moves at gh = 6.6 m/s. The wave crest travels.3 seconds while the diver is in their s = ght1 = 14.4 m. 45

Problem 13.84 A projectile is launched at 1 m/s from a sloping surface. The angle α = 8. Determine the range R. 3 a 1 m/s R Set g = 9.81 m/s,v = 1 m/s. The equations of motion are a =, v = v cos(8 3 ), s = v cos 5 t a = g, v = gt + v sin(8 3 )t, s = 1 gt + v sin 5 t When the projectile hits we have R cos 3 = v cos 5 t R sin 3 = 1 gt + v sin 5 t t =.3 s, R = 17.1 m Problem 13.85 A projectile is launched at 1 m/s at 6 above the horizontal. The surface on which it lands is described b the equation shown. Determine the point of impact. 1 m/s 6 =.1 The motion in the direction is a =, v = V cos θ, = (V cos θ )t, and the motion in the direction is given b a = g, v = (V sin θ ) gt, = (V sin θ )t gt /. We know that V = 1 m/s and θ = 6. The equation of the surface upon which the projectile impacts is =.1. Thus, the time of impact, t I, can be determined b substituting the values of and from the motion equations into the equation for the surface. Hence, we get (V sin θ )t I g t I =.1(V cos θ ) ti. Evaluating with the known values, we get t I = 6.37 s Substituting this value into the motion equations reveals that impact occurs at (, ) = (318.4, 11.4) m. 46

Problem 13.86 At t =, a steel ball in a tank of oil is given a horizontal velocit v = i (m/s). The components of the ball s acceleration in m/s are a = 1.v, a = 8 1.v, a z = 1.v z. What is the velocit of the ball at t = 1s? Assume that the effect of gravit is included in the given accelerations. The equations for the path are obtained from: (1) dv = a = 1.v. Separate variables and integrate: dv v = 1., from which ln(v ) = 1.t + V. At t =, v () =, from which ( v ) ln = 1.t. Inverting: v (t) = e 1.t. () dv = a = 8 1.v. Separate variables and integrate: dv = 1., 8 1. + v from which ( ) 8 ln 1. + v = 1.t + V. At t =, v () =, from ( ln 1 + 1. ) 8 v = 1.t. (3) Inverting: v (t) = 8 1. (e 1.t 1). dv z = a z = 1.v z, from which ln(v z ) = 1.t + V z. Invert to obtain v z (t) = V z e 1.t.Att =, v z () =, hence V z = and v z (t) =. At t = 1 second, v (1) = e 1. =.64 m/s, and ( ) 8 v (1) = (1 e 1. ) = 4.66 m/s,or 1 v =.6i 4.66j (m/s). 47

Problem 13.87 In Problem 13.86, what is the position of the ball at t = 1 s relative to its position at t =? Use the solution for the velocit components from Problem 13.86. The equations for the coordinates: (1) () d = v = e 1.t, from which ( ) (t) = e 1.t + C. 1. At t =, () =, from which ( ) (t) = (1 e 1.t ). 1. d ( ) 8 = (e 1.t 1), from which 1. At t =, () =, from which ( )( 8 e 1.t (t) = 1. 1. + t 1 ). 1. (3) Since v z () = and z() =, then z(t) =. At t = 1, ( ) (1) = (1 e 1. ) = 1.165 m. 1. ( )( 8 e 1. (1) = 1. 1. + 1 1 ) =.784 m, 1. r = 1.165i.784j (m). or ( )( 8 e 1.t ) (t) = 1. 1. + t + C. Problem 13.88 The point P moves along a circular path with radius R. Show that the magnitude of its velocit is v =R dθ/. Strateg: Use Eqs. (13.3). = R cos θ θ P = R sin θ v = R sin θ ( ) dθ v = R cos θ ( ) dθ V = V + V V = R sin θ ( ) dθ + R cos θ ( ) dθ V = R ( dθ ) (sin θ + cos θ) V =R dθ 48

Problem 13.89 If = 15 mm, d = 3 mm/s, and d =, what are the magnitudes of the velocit and acceleration of point P? The equation for the location of the point P is R = +, from which = (R ) 1 =.598 m, and d d ( ) ( ) d = =.173 m/s, = 1 ( ) d + ( d )( ) d ( ) ( d ) 3 mm P =.4619 m/s. The magnitudes are: v P = ( ) d ( ) d + =.3464 m/s a p = ( d ) ( d ) + =.4619 m/s Problem 13.9 A car travels at a constant speed of 1 km/h on a straight road of increasing grade whose vertical profile can be approimated b the equation shown. When the car s horizontal coordinate is = 4 m, what is the car s acceleration? =.3 Denote C =.3 and V = 1 km/h = 7.78 m/s. The magnitude of the constant velocit is V = ( ) d ( ) d + The equation for the road is = C from which d = C ( ) d. Substitute and solve: d = V = 7.1 m/s. (C) + 1 d d d is positive (car is moving to right in sketch). The acceleration is ( ) = d V 4C ( ) V d = (C) + 1 ((C) + 1) 3. =.993 m/s = C ( ) d ( d ) + C =.4139 m/s, or a =.99i +.414j(m/s ) 49

Problem 13.91 Suppose that a projectile has the initial conditions shown in Fig. 13.1. Show that in terms of the coordinate sstem with its origin at the highest point of the trajector, the equation describing the trajector is g = v ( ). cos θ The initial conditions are t =, () =, () =, v () = v cos θ, and v () = v sin θ. The accelerations are a (t) =,a (t) = g. The path of the projectile in the, sstem is obtained b solving the differential equations subject to the initial conditions: (t) = (v cos θ )t, (t) = g t + (v sin θ )t Eliminate t from the equations b substituting t = v cos θ Substitute p = v cos θ sin θ, g = g( ) v v sin θ tan θ + tan θ + v sin θ cos θ g g v sin θ. g g = v ( ) cos θ to obtain g () = v + tan θ. cos θ At the peak, d d =, peak from which p = v cos θ sin θ, g and p = v sin θ. g The primed coordinates: = p reduce: = p. Substitute and = g( + p ) v cos θ + ( + p ) tan θ p. g = v (( ) + cos p θ + p ) + ( + p ) tan θ v sin θ. g 5

Problem 13.9 The acceleration components of a point are a = 4 cos t, a = 4 sin t, a z =. At t =, its position and velocit are r = i, v = j. Show that the magnitude of the velocit is constant; the velocit and acceleration vectors are perpendicular; (c) the magnitude of the acceleration is constant and points toward the origin; (d) the trajector of a point is a circle with its center at the origin. (1) () The equations for the path are dv = a = 4 cos(t), from which v (t) = sin(t) + V. At t =,v () =, from which V =. d = v = sin(t), from which (t) = cos(t) + C.Att =,() = 1, from which C =. dv = a = 4 sin(t), from which v (t) = cos(t) + V.At d t =,v () =, from which V =. = v = cos(t), from which (t) = sin(t) + C.Att =,() =, from which C =. (3) For a z = and zero initial conditions, it follows that v z (t) = and z(t) =. The magnitude of the velocit is v = ( sin(t)) + ( cos(t)) = = const. (c) (d) The velocit is v(t) = i sin(t) + j cos(t). The acceleration is a(t) = i4 cos(t) j4 sin(t). If the two are perpendicular, the dot product should vanish: a(t) v(t) = ( sin(t))( 4 cos(t))+ ( cos(t))( 4 sin(t)) =, and it does The magnitude of the acceleration: a = ( 4 cos(t)) + ( 4 sin(t)) = 4 = const. The unit vector parallel to the acceleration is e = a = i cos(t) j sin(t), a which alwas points to the origin. The trajector path is (t) = cos(t) and (t) = sin(t). These satisf the condition for a circle of radius 1: 1 = + Problem 13.93 When an airplane touches down at t =, a stationar wheel is subjected to a constant angular acceleration α = 11 rad/s until t = 1s. What is the wheel s angular velocit at t = 1s? At t =, the angle θ =. Determine θ in radians and in revolutions at t = 1s. θ α = 11 rad/s ω = αt + ω θ = ( 1 αt ) + ω t + θ From the problem statement, ω = θ = At t = 1s, ω = (11)(1) + = 11 rad/s At t = 1s, θ = 11(1) / = 55 radians (8.75 revolutions) 51

Problem 13.94 Let L be a line from the center of the earth to a fied point on the equator, and let L be a fied reference direction. The figure views the earth from above the north pole. Is dθ/ positive or negative? (Remember that the sun rises in the east.) Determine the approimate value of dθ/ in rad/s and use it to calculate the angle through which the earth rotates in one hour. u L L dθ >. dθ π rad 4(36) s = 7.7 1 5 rad/s. ( ) 36 s In one hour θ (7.7 1 5 rad/s)(1 hr) 1hr =.6 rad = 15 dθ 7.7 1 5 rad/s, θ 15. Problem 13.95 The angular acceleration of the line L relative to the line L is given as a function of time b α =.5 1.t rad/s.att =,θ = and the angular velocit of L relative to L is ω = 5 rad/s. Determine θ and ω at t = 3s. u L L α =.5 1.t ω =.5t.6t + 5 θ = 1.5t.t 3 + 5t θ(3) = 1.5(3).(3) 3 + 5(3) =.85 rad ω(3) =.5(3).6(3) + 5 = 7.1 rad/s 5

Problem 13.96 In Active Eample 13.8, suppose that the angular acceleration of the rotor is α =.ω, where ω is the angular velocit of the rotor in rad/s. How long does it take the rotor to slow from 1, rpm to 1 rpm? L Then Let α = kω, where k =.. a = dω = kω, u ω ω 1 dω ω = t k, ( 1 ) 1 + 1 = kt t = 1 1 ω ω 1 k ω 1 ω Convert the numbers to rad/s ( )( π rad 1 min ω 1 = 1 rpm rev 6 sec ( πrad ω = 1 rpm rev Using the numbers in the problem ( 1 1 t =. 147 rad/s 1 14.7 rad/s ) = 147 rad/s )( ) 1 min = 14.7 rad/s 6 sec ) L t = 43 s. Problem 13.97 The astronaut is not rotating. He has an orientation control sstem that can subject him to a constant angular acceleration of.1 rad/s about the vertical ais is either direction. If he wants to rotate 18 about the vertical ais (that is, rotate so that he is facing toward the left) and not be rotating in his new orientation, what is the minimum time in which he could achieve the new orientation? He could achieve the rotation in minimum time b accelerating until he has turned 9 and then decelerating for the same time while he rotates the final 9. Thus the time needed to turn 9 (π/ rad) is α =.1 rad/s, ω = αt θ = 1 αt t = θ α = (π/) = 5.65 s..1 rad/s The total maneuver time is t. t = 11. s. 53

Problem 13.98 The astronaut is not rotating. He has an orientation control sstem that can subject him to a constant angular acceleration of.1 rad/s about the vertical ais in either direction. Refer to problem 13.97. For safet, the control sstem will not allow his angular velocit to eceed 15 per second. If he wants to rotate 18 about the vertical ais (that is, rotate so that he is facing toward the left) and not be rotating in his new orientation, what is the minimum time in which he could achieve the new orientation? He could achieve the rotation in minimum time b accelerating until he has reached the angular velocit limit, then coasting until he has turned 9. He would then continue to coast until he needed to decelerate to a stop at the 18 position. The maneuver is smmetric in the spin up and the spin down phases. We will first find the time needed to reach the angular velocit limit and also find the angle through which he has rotated in this time. α =.1 rad/s ω = αt 1 t 1 = ω α = (15/18)π rad/s.1 rad/s =.6 s. θ 1 = 1 αt = 1 (.1 rad/s )(.6 s) =.343 rad. Now we need to find the coast time (constant angular velocit) ( π 15 ) rad.343 rad = 18 π rad/s t t = 4.69 s Thus the total time to complete a 9 turn is t = t 1 + t =.6 s + 4.69 s = 7.31 s. The time for the full 18 turn is t t = 14.6 s. Problem 13.99 The rotor of an electric generator is rotating at rpm when the motor is turned off. Due to frictional effects, the angular acceleration of the rotor after the motor is turned off is α =.1ω rad/s, where ω is the angular velocit in rad/s. What is the rotor s angular velocit one minute after the motor is turned off? After the motor is turned off, how man revolutions does the rotor turn before it comes to rest? Strateg: To do part, use the chain rule to write the angular acceleration as α = dω = dω dθ dθ = dω dθ ω. Let α = kω, where k =.1 s 1. Note that rpm =.9 rad/s. One minute after the motor is turned off α = dω ω dω t ( ) ω = kω ω ω = k ln = kt ω ω = ω e kt = (.9 rad/s)e (.1/s)(6 s) = 11.5 rad/s. ω = 11.5 rad/s (11 rpm). When the rotor comes to rest α = ω dω θ dθ = kω dω = kdθ ω = kθ ω θ = 1 k ( ω 1 ) = (.9 rad/s) 1.1 s ( ) 1 rev θ = 94 rad = 333 rev. π rad θ = 333 rev. 54

Problem 13.1 The needle of a measuring instrument is connected to a torsional spring that gives it an angular acceleration α = 4θ rad/s, where θ is the needle s angular position in radians relative to a reference direction. The needle is given an angular velocit ω = rad/s in the position θ =. θ What is the magnitude of the needle s angular velocit when θ = 3? What maimum angle θ does the needle reach before it rebounds? α = ω dω dθ = 4θ ω = 1 θ ω θ ωdω = 4 θdθ ω = θ ω = 1 (π/6) = 1.74 rad/s Maimum angle means ω =. θ = 1 rad = 57.3 Problem 13.11 The angle θ measures the direction of the unit vector e relative to the ais. The angular velocit of e is ω = dθ/ = rad/s, constant. Determine the derivative de/ when θ = 9 in two was: e Use Eq. (13.33). Epress the vector e in terms of its and components and take the time derivative of e. θ de = dθ n = ωn n n when θ = 9,n= i de = i rad/s when θ = 9 e θ ω = dθ = rad/s e θ = 9 de = sin θ e = (1) cos θi + (1) sin θj ( ) dθ i + cos θ ( ) dθ j de Evaluating at θ = 9 = dθ i = i rad/s 55

Problem 13.1 The angle θ measures the direction of the unit vector e relative to the ais. The angle θ is given as a function of time b θ = t rad. What is the vector de/ at t = 4s? de = ( ) dθ n, B definition: where ( n = i cos θ + π ) ( + j sin θ + π ) The angle is θ = [mod(t, π)] t=4 = mod(3, π) =.5841 rad, where mod(,) ( modulus ) is a standard function that returns the remainder of division of the first argument b the second. From which, [ ] de ( ( = 16 i cos.5841 + π ) ( + j sin.5841 + π )) t=4 = 8.83i + 13.35j is a unit vector in the direction of positive θ. The angular rate of change is [ ] dθ = [4t] t=4 = 16 rad/s. t=4 Problem 13.13 The line OP is of constant length R. The angle θ = ω t, where ω is a constant. Use the relations v = d and v = d to determine the velocit of P relative to O. Use Eq. (13.33) to determine the velocit of P relative to O, and confirm that our result agrees with the result of. O R θ P Strateg: In part, write the position vector of P relative to O as r = Re where e is a unit vector that points from O toward P. The point P is described b P = i + j. Take the derivative: dp = i ( ) ( ) d d + j. The coordinates are related to the angle θ b = R cos θ, = R sin θ. Take the derivative and note that R is a constant and θ = ω t, so that dθ d = ω : d ( ) dθ = R cos θ. = R sin θ ( ) dθ, Substitute into the derivative of the vector P, dp ( ) dθ = R ( i sin θ + j cos θ) = Rω ( i sin(ω t) + j cos(ω t)) Note that P = Re, and dp definition (Eq. (13.33)), de = ( ) dθ n, = R de when R is constant. Use the where n is a unit vector in the direction of positive θ, (i.e., perpendicular to e). Thus ( n = i cos θ + π ) ( + j sin θ + π ). Use the trigonometric sum-of-angles identities to obtain: n = i sin θ + j cos θ. Substitute, dp = Rω ( i sin(ω t) + j cos(ω t)) The results are the same. which is the velocit of the point P relative to the origin O. 56

Problem 13.14 In Active Eample 13.9, determine the motorccle s velocit and acceleration in terms of normal and tangential components at t = 5s. We are given the tangential acceleration a t = ( m/s ) + (. m/s 3 )t We can find the velocit at t = 5s. 5s v = ([ m/s ] + [. m/s 3 ]t) = [ m/s ][5 s] + 1 [. m/s3 ][5 s] = 1. m/s. s O The velocit is then v = (1.5 m/s) e t The accelerations are a t = ( m/s ) + (. m/s 3 )(5 s) = 3 m/s P e n 4 m a n = v ρ = (1.5 m/s) 4 m =.391 m/s. e t a = (3 m/s )e t + (.391 m/s )e n Problem 13.15 The armature starts from rest at t = and has constant angular acceleration α = rad/s.at t = 4 s, what are the velocit and acceleration of point P relative to point O in terms of normal and tangential components? O P We can find the angular velocit at t = 4s. α = rad/s 8 mm ω = ( rad/s )(4 s) = 8 rad/s. Then v = rω = (.8 m) (8 rad/s) =.64 m/s, a t = rα = (.8 m) ( rad/s ) =.16 m/s, a n = v r (.64 m/s) = = 5.1 m/s.8 m v = (.64 m/s)e t, a = (.16 m/s )e t + (5.1 m/s )e n. 57

Problem 13.16 Suppose ou want to design a medical centrifuge to subject samples to normal accelerations of 1 g s. If the distance from the center of the centrifuge to the sample is 3 mm, what speed of rotation in rpm is necessar? If ou want the centrifuge to reach its design rpm in 1 min, what constant angular acceleration is necessar? 3 mm The normal acceleration at a constant rotation rate is a n = Rω, giving an (1)9.81 ω = R = = 18.83 rad/s..3 The speed in rpm is ( )( )( ) rad 1 rev 6 s N = ω = 173 rpm. s π rad 1 min The angular acceleration is α = ω t = 18.83 6 = 3.1 rad/s Problem 13.17 The medical centrifuge shown in Problem 13.16 starts from rest at t = and is subjected to a constant angular acceleration α = 3 rad/s. What is the magnitude of the total acceleration to which the samples are subjected at t = 1s? α = 3, ω = 3t, θ = 1.5t a t = (.3 m)(3 rad/s ) =.9 m/s a n = (.3 m)(3 rad/s) =.7 m/s a = (.9) + (.7) m/s =.85 m/s 58

Problem 13.18 A centrifuge used to subject engineering components to high acceleration has a radius of 8 m. It starts from rest at t =, and during its two-minute acceleration phase it is programmed so that its angular acceleration is given as a function of time in seconds b α =.19.16t rad/s.att = 1 s, what is the magnitude of the acceleration a component is subjected to? We will first calculate the angular velocit 1 s ω = ([.19 rad/s ] [.16 rad/s 3 ]t) = [.19 rad/s][1 s] 1 [.16 rad/s3 ][1 s] = 11.5 rad/s The normal and tangential components of acceleration are a t = rα = (8 m)([.19 rad/s ] [.16 rad/s 3 ][1 s]) = a n = rω = (8 m)(11.5 rad/s) = 16 m/s Since the tangential component is zero, then the total acceleration is the same as the normal acceleration a = 16 m/s (18g s). Problem 13.19 A powerboat being tested for maneuverabilit is started from rest at t = and driven in a circular path 1 m in radius. The tangential component of the boat s acceleration as a function of time is a t =.4t m/s. What are the boat s velocit and acceleration in terms of normal and tangential components at t = 4s? What distance does the boat move along its circular path from t = tot = 4s? a t =.4t m/s a n =+v /r v =.t m/s At t = 4s, a =.4te t + v /re n a = 1.6e t +.853e n v = 3.e t m/s s =.t 3 /3 s 4s = 4.7 m 59

Problem 13.11 The angle θ = t rad. P What are the velocit and acceleration of point P in terms of normal and tangential components at t = 1s? What distance along the circular path does point P move from t = tot = 1s? θ 4 m O dθ θ = t = 4t = ω s = Rθ = 8t = 8(1) = 8m e t d θ = 4 rad s = α P s = rθ = 4θ = 8t v t = 16t m/s e N 4 m θ O v = rω = 4(4t) = 16t a t = dv = 16 m/s v = 16(1)e t m/s = 16 e t (m/s) a = Rαe t + Rω e N a = (4)(4)e t + (4)(4 )e N (m/s ) a = 16e t + 64e N (m/s ) Problem 13.111 The angle θ = t rad. What are the velocit and acceleration of point P in terms of normal and tangential components when P has gone one revolution around the circular path starting at t =? From the solution to Problem 13.11, θ = t rad ω = 4t rad/s α = 4 rad/s s = 8t m v t = 16t m/s a t = 16 m/s We want to know v and a when θ = π. Substituting into the first eqn, we find that θ = π when t = t 1 = 1.77 seconds. From the solution to Problem 13.11, v t = 16te t and a = Rαe t + Rω e N Substituting in the time t 1,weget v t = 8.4e t (m/s) a = 16e t + 1.1e N (m/s ) 6

Problem 13.11 At the instant shown, the crank AB is rotating with a constant counterclockwise angular velocit of 5 rpm. Determine the velocit of point B in terms of normal and tangential components; in terms of cartesian components. C ( )( ) π rad min ω = (5 rpm) = 54 rad/s rev 6 sec V B =.5 m (54 rad/s)e t = ( 6. e t ) m/s V B = ( 6. m/s)( cos 45 i sin 45 j) = ( 18. 53i 18. 53j) m/s 5 mm 45 B A Problem 13.113 The crank AB in Problem 13.11 is rotating with a constant counterclockwise angular velocit of 5 rpm. Determine the acceleration of point B in terms of normal and tangential components; in terms of cartesian components. ( )( ) π rad min ω = (5 rpm) = 54 rad/s rev 6 sec a t =, a n =. 5 m (54 rad/s) = 1378.8 1379 a P = ( 1379e n ) m/s a p = ( 1379 m/s )(cos 45 i sin 45 j) = ( 978i 978j) m/s Problem 13.114 Suppose that a circular tunnel of radius R could be dug beneath the equator. In principle, a satellite could be placed in orbit about the center of the earth within the tunnel. The acceleration due to gravit in the tunnel would be gr/r E, where g is the acceleration due to gravit at sea level and R E is the earth s radius. Determine the velocit of the satellite and show that the time required to complete one orbit is independent of the radius R. (See Eample 13.1.) R E R Equator Tunnel To be in orbit we must have a n = v R = gr R E g v = R R E The velocit of the satellite is given b the distance it travels in one orbit divided b the time needed to complete that orbit. g v = πr t = ν R t πr = R E g =. R R E Notice that the time does not depend on the radius R. 61

Problem 13.115 At the instant shown, the magnitude of the airplane s velocit is 13 m/s, its tangential component of acceleration is a t = 4 m/s, and the rate of change of its path angle is dθ/ = 5 /s. ω = (5 /s) ( ) π rad ( π ) 18 = rad/s 36 What are the airplane s velocit and acceleration in terms of normal and tangential components? What is the instantaneous radius of curvature of the airplane s path? a Pt = 4 m/s,a n = (13 m/s)ω = 11.34 m/s v p = (13e t ) m/s a p = ( 4e t + 11.34e n ) m/s θ ρ = v (13 m/s) = a n 11.34 m/s = 149 m Problem 13.116 In the preliminar design of a sunpowered car, a group of engineering students estimates that the car s acceleration will be.6 m/s. Suppose that the car starts from rest at A and the tangential component of its acceleration is a t =.6 m/s. What are the car s velocit and acceleration in terms of normal and tangential components when it reaches B? A 5 m B a t = v dv ds =.6 m/s v = (.6 m/s )s At point B ( S B = Thus + 5π v s vdv = (.6 m/s )ds ) m v B = 18.8 m/s, a Bn = v B = 6.68 m/s 5 m v B = (18.8e t ) m/s a B = (.6e t + 6.68e n ) m/s m Problem 13.117 After subjecting the car design described in Problem 13.116 to wind tunnel testing, the students estimate that the tangential component of the car s acceleration will be a t =.6.v m/s, where v is the car s velocit in m/s. If the car starts from rest at A, what are its velocit and acceleration in terms of normal and tangential components when it reaches B? ( At point B S B = + 5π a t = v dv vb ds =.6.v v B = 14. m/s, a Bn = v B = 4.3 m/s 5 m a t =.6.(14. m/s) =.197 m/s ) m vdv.6.v = sb ds Thus v B = (14.e t ) m/s a B = (.197e t + 4.3e n ) m/s 6

Problem 13.118 Suppose that the tangential component of acceleration of the car described in Problem 13.117 is given in terms of the car s position b a t =.4.1s m/s, where s is the distance the car travels along the track from point A. What are the car s velocit and acceleration in terms of normal and tangential components at point B? a t = dv v dv =.4.1s ds = dv ds =.4.1s m/s ds v SB vdv= (.4.1s)ds v = ] [.4s.1s S B From Fig. P13.116, S B = + πρ/4 where ρ = 5 m, so S B = 78.5 m Solving for v, v = 1.5 m/s v = 1.5e t (m/s) a = (.4.1s B )e t + v /ρe N (m/s ) Solving, a =.11e t +.95e N (m/s ) Problem 13.119 A car increases its speed at a constant rate from 64 km/h at A to 96 km/h at B. What is the magnitude of its acceleration s after the car passes point A? 36 m 3 B A 4 m 3 m 3 4 m v dv ds = a, Use the chain rule to obtain where a is constant. Separate variables and integrate: v = as + C. At ( ) s =, v() = 64 1 = 17. 78 m/s, 36 from which C = 316. 13. The acceleration is a = v C s The distance traveled from A to B is ( π ) s = ( 4) + (3) ( 36+ 3) = 8.6 m, 18 and the speed in ( ) [v(s)] s= 8.6 = 96 1 = 6.67 m/s, 36 The velocit is as a function of time is v(t) = v() + at = 17.78 +. 39t m/s. The distance from A is.39 s(t) = 17. 78t + t. At a point seconds past A, the distance is s() = 4.34 m, and the velocit is v() =.56 m/s. The first part of the hill ends at 43, so that at this point the car is still in the first part of the hill. The tangential acceleration is a t =.39 m/s. The normal acceleration is a n = v R = (.56 ) = 14.1 m/s. 36 The magnitude of the acceleration is a = +..39 14 1 = 14. 3 m/s Note: This is a large acceleration the driver (and passengers) would no doubt be uncomfortable. from the constant acceleration is a = ( 6.67 ) 316.13 ( 8.6) =. 39 m/s. 63

Problem 13.1 The car increases its speed at a constant rate from 64 km/h at A to 96 km/h at B. Determine the magnitude of its acceleration when it has traveled along the road a distance 36 m from A and 48 m from A. Use the solution in Problem 13.119. The velocit at distance 48 m from A is The velocit at a distance 36 m from A is v( 36) = as + C = ()(. 39)( 36) + 316. 13 =.1 m/s. At 36 m the car is in the first part of the hill. The tangential acceleration is a t =. 39 m/s from Problem 13.119. The normal acceleration is (v( 36)) a n = = (. 1 ) = 13.6 m/s. R 36 The magnitude of the acceleration is a =. + 39 13.6 = 13.81 m/s v( 48) = (. 39)( 48) + C = 3.4 m/s. At 48 m the car is on the second part of the hill. The tangential acceleration is unchanged: a t =.39 m/s. The normal acceleration is (v( 48)) 3.4 a n = = = 18.3 m/s. R 3 The magnitude of the acceleration is a = +.39 18.3 = 18.5 m/s [Note: The car will lift off from the road.] Problem 13.11 Astronaut candidates are to be tested in a centrifuge with 1-m radius that rotates in the horizontal plane. Test engineers want to subject the candidates to an acceleration of 5 g s, or five times the acceleration due to gravit. Earth s gravit effectivel eerts an acceleration of 1 g in the vertical direction. Determine the angular velocit of the centrifuge in revolutions per second so that the magnitude of the total acceleration is 5 g s. 1 m a n + g = (5g) a n = 4g a n = rω ω = a n /r 4(9.81 m/s ) ω = 1m =.19 rad/s Problem 13.1 In Eample 13.11, what is the helicopter s velocit in turns of normal and tangential components at t = 4s? In Eample 13.11 we find the and components of acceleration and velocit at t = 4s. a =.4 m/s,a =.36 m/s, v = 4.8 m/s,v = 4.3 m/s The total velocit of the helicopter is v = v + v = (4.8 m/s) + (4.3 m/s) = 6.46 m/s. B definition, the velocit is in the tangential direction, therefore v = 6.46 m/s)e t. 64

Problem 13.13 The athlete releases the shot with velocit v = 16 m/s at above the horizontal. What are the velocit and acceleration of the shot in terms of normal and tangential components when it is at the highest point of its trajector? What is the instantaneous radius of curvature of the shot s path when it is at the highest point of its trajector? a = v = v = 16 cos,v = 16 sin a = 9.81 m/s v = v 9.81t = 5.47 9.81t At highest point, v = (c) a N =v /ρ, ρ = v / a n ρ = (15) /9.81 = 3. m e t v = 16 cos e t = 15.e t (m/s) e N a = 9.81e n (m/s ) Problem 13.14 At t =, the athlete releases the shot with velocit v = 16 m/s. What are the velocit and acceleration of the shot in terms of normal and tangential components at t =.3 s? Use the relation a n = v /ρ to determine the instantaneous radius of curvature of the shot s path at t =.3 s. From the solution to Problem 13.13, a n =v /ρ v = 15. m/s v = v + v v = 5.47 9.81t m/s ρ = v / a n =(15.) /9.67 ρ = 4. m At t =.3 s, v = 15. m/s V v = 15.e t (m/s) We have the following geometr From the diagram tan r = v /v r = 9.55 e n m 9.81 s e t a n V r V a n =9.81 cos r = 9.67 m/s a t =9.81 sin r = 1.63 m/s a t a = 1,63e t + 9.67e n (m/s ) r 65

Problem 13.15 At t =, the athlete releases the shot with velocit v = 16 m/s. Use Eq. (13.4) to determine the instantaneous radius of curvature of the shot s path at t =.3 s. [ ( ) ] d 3/ 1 + d ρ = d d From the solution to 13.13, v = v 9.81t, hence = + v t 9.8/(t /), Also, v = v = + v t Hence t = /v and = v ( ) 9.81 ( ) v v We now have () d d = v ( ) 9.81 v v d d = 9.81/v We also know v = 16 m/s and v = v sin = 5.47 m/s v = v cos = 15.4 m/s At t =.3 s, = 4.5 m, d d =.168 d d =.434 and ρ = 4. m Problem 13.16 The cartesian coordinates of a point moving in the -plane are = + 4t m, = 1 t 3 m. What is the instantaneous radius of curvature of the path of the point at t = 3s? The components of the velocit: v = 8ti (3t )j. At t = 3 seconds, the magnitude of the velocit is v t=3 = (8t) + ( 3t ) = 36.1 m/s. The components of the acceleration are a = 8i (6t)j. The instantaneous path angle is tan β = v = 3t. v 8t At t = 3 seconds, β =.844 rad. The unit vector parallel to the path is e t = i cos β + j sin β. The unit vector normal to the path pointing toward the instantaneous radial center is ( e n = i cos β π ) ( + j sin β π ) = i sin β j cos β. The normal acceleration is the component of acceleration in the direction of e n. Thus, a n = e n a or a n = 8 sin β + (6t)cos β. Att = 3 seconds, a n = 5.98 m/s. The radius of curvature at t = 3 seconds is ρ = v a n = 18 m 66

Problem 13.17 The helicopter starts from rest at t =. The cartesian components of its acceleration are a =.6t m/s and a = 1.8.36t m/s. Determine the tangential and normal components of its acceleration at t = 6s. The solution will follow that of Eample 13.11, with the time changed to t = 6 s. The helicopter starts from rest (v,v ) = (, ) at t =. Assume that motion starts at the origin (, ). The equations for the motion in the direction are a =.6t m/s, v =.3t m/s, =.1t 3 m, and the equations for motion in the direction are a = 1.8.36t m/s, v = 1.8t.18t m/s, and =.9t.6t 3 m. At t = 6 s, the variables have the values a = 3.6 m/s, a =.36 m/s, v = 1.8 m/s, v = 4.3 m/s, = 1.6 m, and = 19.44 m. The magnitude of the velocit is given b v = v + v = 11.63 m/s. The tangential acceleration component is given b a T = a e T =.98a +.371a = 3.1 m/s. The magnitude of the acceleration is given b a = a + a = 3.6 m/s. The normal acceleration component is given b a N = a at = 1.67 m/s The unit vector in the tangential direction is given b e T = v v = v i + v j =.98i +.371j. v Problem 13.18 Suppose that when the centrifuge in Eample 13.1 is turned on, its motor and control sstem give it an angular acceleration (in rad/s ) α = 1.ω, where ω is the centrifuge s angular velocit. Determine the tangential and normal components of the acceleration of the sample at t =. s. 3 mm t =. s. We will first integrate to find the angular velocit at α = dω ω = ([1 rad/s ] [. s 1 ]ω) dω [1 rad/s ] [. s 1 ]ω =. s ( ) 1 s. ln [1 rad/s ] [. s 1 ]ω [1 rad/s =. s ] At this time, the angular acceleration is α = (1 rad/s ) (. s 1 )(.4 rad/s) = 11.95 rad/s. The components of acceleration are a t = rα = (.3 m)(11.95 rad/s ) = 3.59 m/s α n = rω = (.3 m)(.4 rad/s) = 1.7 m/s. a = (3.59e t + 1.7e n ) m/s. [1 rad/s ] [. s 1 ]ω = [1 rad/s ]e.4 ω = 1s. (1 rad/s )(1 e.4 ) =.4 rad/s 67

Problem 13.19* For astronaut training, the airplane shown is to achieve weightlessness for a short period of time b fling along a path such that its acceleration is a = and a = g. If the velocit of the plane at O at time t = isv = v i, show that the autopilot must fl the airplane so that its tangential component of the acceleration as a function of time is ( ) gt v a t = g ( ). gt 1 + v O angle is The velocit of the path is v(t) = v i gtj. The path β : tan β = v v = gt v, sin β = gt v + (gt). The unit vector parallel to the velocit vector is e = i cos β + j sin β. The acceleration vector is a = jg. The component of the acceleration tangent to the flight path is a t = g sin β., from which gt a t = g. v + (gt) Divide b v, [ ( ) ] 1 gt ( ) gt a t = g 1 + v v Problem 13.13* In Problem 13.19, what is the airplane s normal component of acceleration as a function of time? From Problem 13.19, the velocit is v(t) = v i gtj. The flight path angle is β, from which cos β = v v + (gt). The unit vector parallel to the flight path is e = i cos β + j sin β. The unit vector normal to e is ( e n = i cos β π ) ( + j sin β π ) = i sin β j cos β, pointing toward the instantaneous radial center of the path. The acceleration is a = jg. The component parallel to the normal component is a n = g cos β, from which [ ( ) ] 1 v gt a n = g = g 1 + v + (gt) v 68

d Problem 13.131 If = 1 mm, = mm/s, and d =, what are the velocit and acceleration of P in terms of normal and tangential components? P The equation for the circular guide is R = +, from which = R =.83 m, and d ( ) d = = v =.77 m/s. The velocit of point P is v p = iv + jv, from which the velocit is v = v + v =.1 m/s. The angular velocit 3 mm ω = v R The angle is =.771 rad/s. ( β = tan 1 ) = 19.5 a = dv = 1 = d ( ) d ( ) d + =.1591 m/s ( d )( ) d ( ) ( d ) The unit vector tangent to the path (normal to the radius vector for a circle) ise p = i sin β + j cos β, from which a t = a sin β = 53. mm/s since a = a n = Rω =.15 m/s. Check: a n = a cos β =.15 m/s check. 69

Problem 13.13* Suppose that the point P in Problem 13.131 moves upward in the slot with velocit v = 3e t (mm/s). When = 15 mm, what are d and d? The position in the guide slot is = R sin θ, from which ( θ = sin 1 ) = sin 1 (.5) = 3. R = R cos θ = 59.8 mm. From the solution to Problem 13.131, ( ) d ( ) v = = v. ( ) The velocit is v =3 = v + v = v + 1, from which and v = 15 mm/s (Since the point is moving upward in the slot, v is positive.). The velocit along the path in the guide slot is assumed constant, hence a t =. The normal acceleration is a n = v R = 3 mm/s directed toward the radius center, from which d = a n sin θ = 15 mm/s v = 3 ( ( ) ) 1 + 1 = 59.8 mm/s Problem 13.133* A car travels at 1 km/h on a straight road of increasing grade whose vertical profile can be approimated b the equation shown. When = 4 m, what are the tangential and normal components of the car s acceleration? =.3 The strateg is to use the acceleration in cartesian coordinates found in the solution to Problem 13.9, find the angle with respect to the -ais, ( ) d θ = tan 1, d and use this angle to transform the accelerations to tangential and normal components. From the solution to Problem 13.9 the accelerations are a =.993i +.4139j (m/s ). The angle at ( ) d θ = tan 1 d C = tan 1 (6 1 4 ) =4 = 13.5. =4 From trigonometr (see figure) the transformation is a t = a cos θ + a sin θ, a n = a sin θ + a cos θ, from which a t =.35...=, a n =.456 m/s Check: The velocit is constant along the path, so the tangential component of the acceleration is zero, a t = dv =, check. 7

Problem 13.134 A bo rides a skateboard on the concrete surface of an empt drainage canal described b the equation shown. He starts at = m, and the magnitude of his velocit is approimated b v = ( 9.81 )( ) m/s. =.3 Use Equation (13.4) to determine the instantaneous radius of curvature of the bo s path when he reaches the bottom. What is the normal component of his acceleration when he reaches the bottom? =.3, so d d =.6 and d d =.6. From Eq (13.4), ρ = [1 + (.6) ] 3/.6 m. At =, ρ = 16.7 m. The magnitude of the velocit is ( ) d ( ) d + = v = K( ) 1 = K( C ) 1, where K = 8.5, C =.3. From = C, d d = C = C Substitute: ( ) d, ( ) d ( d ) + C. At the bottom of the canal the values are ( ) d = K = 35.89 m/s. = ( ) ( d d ) =, = =, = ( d ) ( ) d = C = 77.8 m/s. = = The angle with respect to the ais at the bottom of the canal is ( ) d θ = tan 1 =. d = From the solution to Problem.133, the tangential and normal accelerations are a t = a cos θ + a sin θ, a n = a sin θ + a cos θ, from which a t =, and a n = 77.8 m/s. Check: The velocit is constant along the path, so the tangential component of the acceleration is zero, a t = dv =. check. B inspection, the normal acceleration at the bottom of the canal is identical to the component of the acceleration. check. d = K( C ) 1 (4C + 1) 1. Since the bo is moving the right, d >, and The acceleration is d = d. d = KC ( C ) 1 (4C + 1) 1 ( ) d K(4C )( C ) 1 (4C + 1) 3 ( ) d. 71

Problem 13.135 In Problem 13.134, what is the normal component of the bo s acceleration when he has passed the bottom and reached = 1 m? Use the results of the solutions to Problems 13.133 and 13.134. From the solution to Problem 13.134, at = 1 m, = ( ) = 18.57 m, from which C ( ) ( ) d = K( C ) 1 (4C + 1) 1 = 17.11 m/s. =1 =1 ( d ) ( ) d = K C =1 =1 ( C ) 1 (4C + 1) 1 + (4C )( C ) 1 (4C + 1) 3 =1 From the solution to Problem 13.133, a t = a cos θ + a sin θ,a n = a sin θ + a cos θ, from which a t = 3.78 m/s, a n = 11.84 m/s = 4.78 m/s. ( d ) ( = C =1 ( ) d + C ( ) d The angle is θ = tan 1 = 47.61. d =1 ( d ) ) = 9.58 m/s. =1 Problem 13.136* Using Eqs (13.41): Show that the relations between the cartesian unit vectors and the unit vectors e t and e n are i = cos θe t sin θe n and j = sin θe t + cos θe n Show that de t / = dθ/e n and de n / = dθ/e t. Equations (13.41) are e t = cos θi + sin θj and e n = sin θi + cos θj. Multipling the equation for e t b cos θ and the equation for e n b ( sin θ) and adding the two equations, we get i = cos θe t sin θe n. Similarl, b multipling the equation for e t b sin θ and the equation for e n b cos θ and adding, we get j = sin θe t + cos θe n. Taking the derivative of e t = cos θi + sin θj, we get = ( sin θi + cos θj) dθ dθ = e n. Similarl, taking the derivative of e n = sin θi + cos θj, we get de n / = (dθ/)e t de t 7

Problem 13.137 The polar coordinates of the collar A are given as functions of time in seconds b r = 1 +.t m and θ = t rad. What are the magnitudes of the velocit and acceleration of the collar at t = s? r u A We have r = (1 ft) + (. m/s )t, θ = ( rad/s)t, dr = (.4 ft/s )t, dθ = rad/s, d r =.4 m/s, At time t = s, we have r = 1.8 ft, dr d θ = =.8 m/s, d r =.4 m/s, dθ d θ θ = 8 rad, = rad/s, = The components of the velocit and acceleration are v r = dr a r = d r r =.8 m/s,v θ = r dθ = (1.8 m)( rad/s) = 3.6 m/s, ( ) dθ = (.4 m/s ) (1.8 m)( rad/s ) = 6.8 m/s, a θ = r d θ + dr dθ = + (.8 m/s)( rad/s) = 3. m/s. The magnitudes are v = v r + vθ = (.8 m/s) + (3.6 m/s) = 3.69 m/s, a = a r + aθ = ( 6.8 m/s ) + (3. m/s ) = 7.5 m/s. v = 3.69 m/s,a = 7.5 m/s. 73

Problem 13.138 In Active Eample 13.13, suppose that the robot arm is reprogrammed so that the point P traverses the path described b r = 1.5 sin πt m, θ =.5. cos πt rad. What is the velocit of P in terms of polar coordinates at t =.8 s? r P u r = [ 1.5 sin We have ( π t )] s [ ( m, θ =.5. cos π t )] rad, s dr ( = π cos π t ) m/s, s dθ ( =.4π sin π t ) rad/s. s At time t =.8 s, r = 1.48 m, dr The velocit is =.971 m/s, θ =.438 rad, dθ = 1. rad/s. v r = dr =.971 m/s,v θ = r dθ = (1.48 m)( 1. rad/s) = 1.76 m/s. v = (.971e r 1.76e θ ) m/s. Problem 13.139 At the instant shown, r = 3 m and θ = 3. The cartesian components of the velocit of point A are v = m/s and v = 8 m/s. A Determine the velocit of point A in terms of polar coordinates. What is the angular velocit dθ/ of the crane at the instant shown? r To transform to polar coordinates we have v r = v cos θ + v sin θ = ( m/s) cos 3 + (8 m/s) sin 3 = 5.73 m/s. u v θ = v sin θ + v cos θ = ( m/s) sin 3 + (8 m/s) cos 3 = 5.93 m/s. v A = (5.73e r + 5.93e θ )m/s. The angular velocit is found v θ = r dθ dθ = v θ r = 5.93 m/s 3m = 1.98 rad/s. dθ = 1.98 rad/s. 74

Problem 13.14 The polar coordinates of point A of the crane are given as functions of time in seconds b r = 3 +.t m and θ =.t rad. Determine the acceleration of point A in terms of polar coordinates at t = 3s. r A We have r = (3 m) + (. m/s )t, dr = (.4 m/s )t, d r =.4 m/s, u θ = (.)t rad/s. At t = 3s, r = 4.8 m, dr dθ d θ = 1. m/s, = (.4)t rad/s, =.4 rad/s d r =.4 m/s, θ =.18 rad, dθ =.1 rad/s, The acceleration components are d θ =.4 rad/s. ( ) a r = d r dθ r = (.4 m/s ) (4.8 m)(.1 rad/s), a θ = r d θ + dr dθ = (4.8 m)(.4 rad/s ) + (1. m/s)(.1 rad/s) a A = (.331e r +.48e θ ) m/s. 75

Problem 13.141 The radial line rotates with a constant angular velocit of rad/s. Point P moves along the line at a constant speed of 4 m/s. Determine the magnitude of the velocit and acceleration of P when r = m. (See Eample 13.14.) 4 m/s P rad/s r O The angular velocit of the line is dθ = ω = rad/s, The magnitude is v = 4 + 4 = 5.66 m/s. from which d θ =. The radial velocit of the point is dr = 4 m/s, The acceleration is a = [ (4)]e r + [(4)()]e θ = 8e r + 16e θ (m/s ). The magnitude is a = 8 + 16 = 17.89 m/s from which d r =. The vector velocit is v = ( ) dr e r + r ( ) dθ e θ = 4e r + 4e θ (m/s). 76

Problem 13.14 At the instant shown, the coordinates of the collar A are =.3 m, = 1.9 m. The collar is sliding on the bar from B toward C at a constant speed of 4 m/s. What is the velocit of the collar in terms of polar coordinates? Use the answer to part to determine the angular velocit of the radial line from the origin to the collar A at the instant shown. A C B 6 We will write the velocit in terms of cartesian coordi- nates. v = (4 m/s) cos 6 = m/s, v = (4 m/s) sin 6 = 3.46 m/s. The angle θ is ( θ = tan 1 ) ( ) 1.9 m = tan 1 = 39.6..3 m Now we can convert to polar coordinates v r = v cos θ + v sin θ = ( m/s) cos 39.6 + (3.46 m/s) sin 39.6, v θ = v sin θ + v cos θ = ( m/s) sin 39.6 + (3.46 m/s) cos 39.6. v = (3.75e r + 1.4e θ ) m/s. dθ = v θ r = 1.4 m/s dθ =.468 rad/s. (.3 m) + (1.9 m) =.468 rad/s. 77

Problem 13.143 At the instant shown, the coordinates of the collar A are =.3 m, = 1.9 m. The collar is sliding on the bar from B toward C at a constant speed of 4 m/s. What is the acceleration of the collar in terms of polar coordinates? Use the answer to part to determine the angular acceleration of the radial line from the origin to the collar A at the instant shown. A C B 6 The velocit is constant, so the acceleration is zero. a A =. Form Problem 13.141 we know that dr r = (.3 m) + (1.9 m) =.98 m, = v r = 3.75 m/s, dθ =.468 rad/s. Using the θ component of the acceleration we can solve for the angular acceleration. a θ = r d θ + dr dθ = d θ = r dr dθ = (3.75 m/s)(.486 rad/s).98 m d θ = 1.18 rad/s. 78

Problem 13.144 A boat searching for underwater archaeological sites in the Aegean Sea moves at.6 m/s and follows the path r = 1θ m, where θ is in radians. When θ = π rad, determine the boat s velocit in terms of polar coordinates and in terms of cartesian coordinates. The velocit along the path is v =.6 m/s. The path is r = 1θ. The velocit v r = dr = d dθ (1θ) = 1 m/s. The velocit along the path is related to the components b v = v r + v θ = ( dr ) ( ) dθ + r =.6. At θ = π, r = 1(π) = 6.8 m. Substitute: (.6 = 1 dθ ) ( ) dθ ( dθ + r = (1 + 6.8 ) ), from which dθ =.33 rad/s, v r = 1 dθ =.33 m/s, v θ = r dθ =.3 m/s From geometr, the cartesian components are v = v r cos θ + v θ sin θ, and v = v r sin θ + v θ cos θ. Atθ = π, v = v r, and v = v θ Problem 13.145 The collar A slides on the circular bar. The radial position of A (in meters) is given as a function of θ b r = cos θ. At the instant shown, θ = 5 and dθ/ = 4 rad/s. Determine the velocit of A in terms of polar coordinates. A r r = cos θ, ṙ = sin θ θ, r = sin θ θ cos θ θ Using the given data we have θ = 5, θ = 4, θ = u r = 1.813, ṙ = 3.381, r = 9. v =ṙe r + r θe θ = ( 3.381e r + 7.5e θ ) m/s 79

Problem 13.146 In Problem 13.145, d θ/ = at the instant shown. Determine the acceleration of A in terms of polar coordinates. See Problem 13.145 a = ( r r θ )e r + (r θ + ṙ θ)e θ = ( 58.e r 7.e θ ) m/s Problem 13.147 The radial coordinate of the earth satellite is related to its angular position θ b Satellite r = 1.91 17 1 +.5 cos θ m. The product of the radial position and the transverse component of the velocit is rv θ = 8.7 1 1 m /s. What is the satellite s velocit in terms of polar coordinates when θ = 9? r θ At θ = 9, r = 1.91 1 7 m = p r = ṙ = We also know that p 1 +.5 cos θ, ( p)(.5)( sin θ) θ (1 +.5 cos θ) rv θ = 8.7 1 1 m /s However v θ = r θ, hence r θ = 8.7 1 1 m /s Solving for θ, we get θ =.39 rad/s and ṙ = 83 m/s and from above v θ = 4565 m/s v = 83e r + 4565e θ (m/s) 8

Problem 13.148* In Problem 13.147, what is the satellite s acceleration in terms of polar coordinates when θ = 9? Set A = 1.91 1 7 m,b= 8.7 1 1 m /s r = A 1 +.5 cos θ, rv θ = r(r θ) = B θ = B r = ( B A ) (1 +.5 cos θ) ( ) B θ = A (1 +.5 cos θ)sin θ θ = ṙ = r =.5A sin θ (1 +.5 cos θ) θ.5b sin θ = A.5B cos θ A ( B θ =.5 A 3 When θ = 9 we have ( B A 4 ) cos θ(1 +.5 cos θ) ) (1 +.5 cos θ) 3 sin θ Thus r = A, ṙ = B A, r =, θ = B A, θ = B A 4 a = ( r r θ )e r + (r θ + ṙ θ)e θ = ( 1.91e r ) m/s 81

Problem 13.149 A bead slides along a wire that rotates in the -plane with constant angular velocit ω. The radial component of the bead s acceleration is zero. The radial component of its velocit is v when r = r. Determine the polar components of the bead s velocit as a function of r. Strateg: The radial component of the bead s velocit is v r = dr, and the radial component of its acceleration is a r = d r r ( ) dθ = B using the chain rule, dv r = dv r dr dr = dv r dr v r. ( ) dvr rω. ou can epress the radial component of the acceleration in the form a r = dv r dr v r rω. r ω From the strateg: a r = = v r dv r dr ω r. Separate variables and integrate: v r dv r = ω rdr, from which vr = r ω + C. The transverse component is v θ = r ( ) dθ = rω, from which v = v + ω (r r )e r + rω e θ At r = r,v r = v, from which C = v ω r, and v r = v + ω (r r ). Problem 13.15 If the motion of a point in the plane is such that its transverse component of acceleration a θ is zero, show that the product of its radial position and its transverse velocit is constant: rv θ = constant. We are given that a θ = rα + v r ω =. Multipl the entire relationship b r. We get ( = (r α + rv r ω) = r ( dω ) + r ( ) ) dr ω = d r (r ω). d Note that if (r ω) =, then r ω = constant. Now note that v θ = rω. We have r ω = r(rω) = rv θ = constant. This was what we needed to prove. 8

Problem 13.151* From astronomical data, Kepler deduced that the line from the sun to a planet traces out equal areas in equal times (Fig. a). Show that this result follows from the fact that the transverse component a θ of the planet s acceleration is zero. [When r changes b an amount dr and θ changes b an amount dθ (Fig. b), the resulting differential element of area is da = 1 r(rdθ)]. t + t A t A t 1 + t t 1 that r ω = r dθ From the solution to Problem 13.15, a θ = implies = constant. The element of area is or da = 1 r(rdθ), da Thus, if da = 1 ( r r dθ ) = 1 r ω = constant. = constant, then equal areas are swept out in equal times. dθ r + dr r θ da Problem 13.15 The bar rotates in the plane with constant angular velocit ω = 1 rad/s. The radial component of acceleration of the collar C (in m/s ) is given as a function of the radial position in meters b a r = 8r. When r = 1 m, the radial component of velocit of C is v r = m/s. Determine the velocit of C in terms of polar coordinates when r = 1.5 m. Strateg: Use the chain rule to write the first term in the radial component of the acceleration as C v d r = dv r = dv r dr dr = dv r dr v r r We have ( ) a r = d r dθ r = (8 rad/s )r, d r = ( [ ] dθ (8 rad/s )) r = ([1 rad/s] [8 rad/s ])r = ([136 rad/s )r Using the supplied strateg we can solve for the radial velocit d r = v dv r r dr = (136 rad/s )r vr 1.5 m v r dv r = (136 rad/s ) rdr m/s 1m ν r ( m/s) = (136 rad/s ) Solving we find v r = 13. m/s. We also have v θ = r dθ ( [1.5 m] = (1.5 m)(1 rad/s) = 18 m/s. ) [1 m] Thus v = (13.e r + 18e θ ) m/s. 83

Problem 13.153 The hdraulic actuator moves the pin P upward with constant velocit v = j (m/s). Determine the velocit of the pin in terms of polar coordinates and the angular velocit of the slotted bar when θ = 35. P v P = j (m/s) r = re r v =ṙe r + r θe θ θ m Also, r = i + j (m) v =ẏj (m/s) = j (m/s) Hence V =ṙe r + r θe θ ṙ =ẏ sin θ tan θ = V = 1.15e r + 1.64e θ (m/s) r θ =ẏ cos θ θ r = + ẏ θ = 35, Solving, we get = 1.4 m, ṙ = 1.15 m/s, r =.44 m, θ =.671 rad s e θ θ e r Problem 13.154 The hdraulic actuator moves the pin P upward with constant velocit v = j (m/s). Determine the acceleration of the pin in terms of polar coordinates and the angular acceleration of the slotted bar when θ = 35. From Problem 13.153 V = j m/s, constant ÿ = ( sec θ)(sec θ tan θ) θ + sec θ θ a = dv θ = ω =.671 rad/s θ = 35 θ = [ sec θ tan θ]( θ) sec θ θ = ω ẏ = m/s θ =.631 rad/s θ = α ÿ = tan θ = = = tan θ ẏ = sec θ θ θ = m 84

Problem 13.155 In Eample 13.15, determine the velocit of the cam follower when θ = 135 in terms of polar coordinates and in terms of cartesian coordinates. r Follower θ = 135, ω = dθ/ = 4rad/s, and α =. dr r =.15(1 +.5 cos θ) 1 =.3 m. =.75 dθ sin θ(1 +.5 cos θ) =.58 m/s. v = dr e r + r dθ e θ =.58e r +.98e θ (m/s). u Cam v = v r cos θ v θ sin θ = 1.15 m/s. v = v r sin θ + v θ cos θ =.97 m/s. Problem 13.156* In Eample 13.15, determine the acceleration of the cam follower when θ = 135 in terms of polar coordinates and in terms of cartesian coordinates. See the solution of Problem 13.155. d ( ) r dθ =.75 cos θ(1 +.5 cos θ) +.75 ( ) dθ sin θ(1 +.5 cos θ) 3 =.195 m/s. a = [ d r r ( ) ] dθ [ e r + r d θ + dr ] dθ e θ = 3.5e r + 4.6e θ (m/s ). a = a r cos θ a θ sin θ =.381 m/s a = a r sin θ + a θ cos θ = 5.36 m/s. 85

Problem 13.157 In the cam-follower mechanism, the slotted bar rotates with constant angular velocit ω = 1 rad/s and the radial position of the follower A is determined b the profile of the stationar cam. The path of the follower is described b the polar equation r = 1 +.5 cos(θ) m. r A Determine the velocit of the cam follower when θ = 3 in terms of polar coordinates and in terms of cartesian coordinates. θ θ = 3, ω = dθ/ = 1 rad/s, and α =. r = 1 +.5 cos θ dr = 1.5 m. = dθ sin θ = 8.66 m/s. v = dr e r + r dθ e θ = 8.66e r + 1.5e θ ( m/s). v = v r cos θ v θ sin θ = 13.75 m/s, v = v r sin θ + v θ cos θ = 6.5 m/s. Problem 13.158* In Problem 13.157, determine the acceleration of the cam follower when θ = 3 in terms of polar coordinates and in terms of cartesian coordinates. See the solution of Problem 13.157. d r = θ cos θ = 1 m/s. a = [ d r r ( ) ] dθ [ e r + r d θ + dr ] dθ e θ. = 5e r 173e θ ( m/s ). a = a r cos θ a θ sin θ = 18 m/s, a = a r sin θ + a θ cos θ = 63 m/s. 86

Problem 13.159 The cartesian coordinates of a point P in the plane are related to its polar coordinates of the point b the equations = r cos θ and = r sin θ. (c) Show that the unit vectors i, j are related to the unit vectors e r, e θ b i = e r cos θ e θ sin θ and j = e r sin θ + e θ cos θ. Beginning with the epression for the position vector of P in terms of cartesian coordinates, r = i + j, derive Eq. (13.5) for the position vector in terms of polar coordinates. B taking the time derivative of the position vector of point P epressed in terms of cartesian coordinates, derive Eq. (13.47) for the velocit in terms of polar coordinates. θ r e θ P e r From geometr (see Figure), the radial unit vector is e r = i cos θ + j sin θ, and since the transverse unit vector is at right angles: ( e θ = i cos θ + π ) ( + j sin θ + π ) = i sin θ + j cos θ. Solve for i b multipling e r b cos θ, e θ b sin θ, and subtracting the resulting equations: i = e r cos θ e θ sin θ. Solve for j b multipling e r b sin θ, and e θ b cos θ, and the results: j = e r sin θ + e θ cos θ The position vector is r = i + j = (r cos θ)i + (r sin θ)j = r(i cos θ + j sin θ). Use the results of Part epressing i, j in terms of e r, e θ : r = r(e r cos θ e θ cos θ sin θ + e r sin θ + e θ sin θ cos θ) = re r (c) The time derivatives are: ( dr dr = v = i ( dr + j ) dθ cos θ r sin θ ) dθ sin θ + r cos θ, from which v = dr dθ (i cos θ + j sin θ) + r ( i sin θ + j cos θ). Substitute the results of Part v = dr e r + r dθ e θ = dr e r + rωe θ 87

Problem 13.16 The airplane flies in a straight line at 643.6 km/h.the radius of its propellor is 1..54 m, and the propeller turns at rpm in the counterclockwise direction when seen from the front of the airplane. Determine the velocit and acceleration of a point on the tip of the propeller in terms of clindrical coordinates. (Let the z-ais be oriented as shown in the figure.) 1.54 m z The speed is The radial acceleration is v = 643.6 1/36 = 178.8 m/s The angular velocit is ω = + ( π rad 1 rev )( ) 1 min = 9.4 rad/s. 6 s The radial velocit at the propeller tip is zero. The transverse velocit is v θ = ωr = 319. m/s. The velocit vector in clindrical coordinates is a r = rω = 1.54 (9.4) = 6685 m/s. The transverse acceleration is a θ = r d θ + ( dr )( ) dθ =, since the propeller rotates at a constant angular velocit. The acceleration a z =, since the airplane travels at constant speed. Thus a = 6685e r ( m/s ) v = 319.e θ + 178.8ez ( m/s). Problem 13.161 A charged particle P in a magnetic field moves along the spiral path described b r = 1m, θ = z rad, where z is in meters. The particle moves along the path in the direction shown with constant speed v =1 km/s. What is the velocit of the particle in terms of clindrical coordinates? P z 1 km/s The radial velocit is zero, since the path has a constant radius. The magnitude of the velocit is v = r ( dθ ) + The angular velocit is Substitute: v = ( ) dz = 1 m/s. r ( dθ dθ = dz. ) + 1 4 ( ) dθ The velocit along the clindrical ais is dz = 1 ( ) dθ = 447. m/s. The velocit vector: v = 894.4e θ + 447.e z = ( ) dθ r + 1 4 = 1.5, V z from which dθ = 1 1.5 = 894.4 rad/s, from which the transverse velocit is v θ = r ( ) dθ = 894.4 m/s. 88

Problem 13.16 At t =, two projectiles A and B are simultaneousl launched from O with the initial velocities and elevation angles shown. Determine the velocit of projectile A relative to projectile B at t =.5 s and at t = 1s. A 1 m/s B 1 m/s 6 O 3 v A = (9.81 m/s j)t + (1 m/s)(cos 6 i + sin 6 j) v B = (9.81 m/s j)t + (1 m/s)(cos 3 i + sin 3 j) v A/B = v A v B = (1 m/s)(.366i +.366j) v A/B = ( 3.66i + 3.66j) m/s Since v A/B doesn t depend on time, the answer is the same for both times v A/B = ( 3.66i + 3.66j) m/s Problem 13.163 Relative to the earth-fied coordinate sstem, the disk rotates about the fied point O at 1 rad/s. What is the velocit of point A relative to point B at the instant shown? A 1 rad/s m O B v A = (1 rad/s)( m)i = ( m/s)i v B = (1 rad/s)( m)j = ( m/s)j v A/B = v A v B = ( i j) m/s Problem 13.164 Relative to the earth-fied coordinate sstem, the disk rotates about the fied point O with a constant angular velocit of 1 rad/s. What is the acceleration of point A relative to point B at the instant shown? a A = (1 rad/s) (m)j = ( m/s )j a B = (1 rad/s) (m)i = ( m/s )i a A/B = a A a B = (i j) m/s 89

/ Problem 13.165 The train on the circular track is traveling at 15 m/s. The train on the straight track is traveling at 6 m/s. In terms of the earth-fied coordinate sstem shown, what is the velocit of passenger A relative to passenger B? 15 m O B A 15 m/ s 6 m s v A = ( 6j) m/s, v B = ( 15j) m/s v A/B = v A v B = ( 1j) m/s Problem 13.166 The train on the circular track is traveling at a constant speed of 15 m/s. The train on the straight track is traveling at 6 m/s and is increasing its speed at.6 m/s. In terms of the earth-fied coordinate sstem shown, what is the acceleration of passenger A relative to passenger B? a A = ( j), a B = ( 15 m/s ).6 m/s i = ( 1.48 ) m/s 15 m a A/B = a A a B = ( 1.48i.6j) m/s 9

Problem 13.167 In Active Eample 13.16, suppose that the velocit of the current increases to 3 m/s flowing east. If the helmsman wants to travel northwest relative to the earth, what direction must he point the ship? What is the resulting magnitude of the ships velocit relative to the earth? m/ /s B A W N S E The ship is moving at 5 m/s relative to the water. Use relative velocit concepts v A = v B + v A/B O v A ( cos 45 i + sin 45 j) = (3 m/s)i + (5 m/s)( cos θi + sin θj) Breaking into components we have v A cos 45 = (3 m/s) (5 m/s) cos θ, v A sin 45 = (5 m/s) sin θ. Solving these equations we find θ = 19.9, v A =.41 m/s. 7.1 west of north,.41 m/s. Problem 13.168 A private pilot wishes to fl from a cit P to a cit Q that is km directl north of cit P. The airplane will fl with an airspeed of 9 km/h. At the altitude at which the airplane will be fling, there is an east wind (that is, the wind s direction is west) with a speed of 5 km/h. What direction should the pilot point the airplane to fl directl from cit P to cit Q? How long will the trip take? W N S E Q km 5 km/h Assume an angle θ, measured ccw from the east. V Plane/Ground = V Plane/Air + V Air/Ground V Plane/Air = (9 km/h)(cos θi + sin θj) P V Air/Ground = (5 km/h)i V Plane/Ground = [(9 cos θ 5)i + (9 sin θ)j] km/h We want the airplane to travel due north therefore ( ) 5 9 cos θ 5 = θ = cos 1 = 8.7 9 Thus the heading is The ground speed is now The time is 9 8.7 = 9.93 east of north v = (9 km/h) sin(8.1 ) = 85.6 km/h t = d v = km =.7 h = 4. min 85.6 km/h 91

Problem 13.169 The river flows north at 3 m/s. (Assume that the current is uniform.) If ou want to travel in a straight line from point C to point D in a boat that moves at a constant speed of 1 m/s relative to the water, in what direction should ou point the boat? How long does it take to make the crossing? Assume an angle θ, measured ccw from the east. V Boat/Ground = V Boat/Water + V Water/Ground V Boat/Water = (1 m/s)(cos θi + sin θj) V Water/Ground = (3 m/s)j 3 m/s D V Boat/Ground = [(1 cos θ)i + (3 + 1 sin θ)j] m/s We want the boat to travel at an angle tan φ = 4 5 4 m N Therefore 3 + 1 sin θ 1 cos θ Thus the heading is = 4 5 θ = 5.11 W S E The ground speed is now 5.11 north of east C 5 m The time is v = (1 cos θ) + (3 + 1 sin θ) = 11.6 m/s t = d v = 5 + 4 m 11.6 m/s = 55. s Problem 13.17 The river flows north at 3 m/s (Assume that the current is uniform.) What minimum speed must a boat have relative to the water in order to travel in a straight line form point C to point D? How long does it take to make the crossing? 3 m/s D Strateg: Draw a vector diagram showing the relationships of the velocit of the river relative to the earth, the velocit of the boat relative to the river, and the velocit of the boat relative to the earth. See which direction of the velocit of the boat relative to the river causes it magnitude to be a minimum. 4 m W N S E The minimum velocit occurs when the velocit of the boat relative to the water is 9 from the velocit of the boat relative to the earth. v B = v W + v B/W. The angle of travel is ( ) 4 m θ = tan 1 = 38.7. 5 m Using the triangle that is drawn we have v B/W = (3 m/s) cos 38.7 =.34 m/s, C 5 m v B = (3 m/s) sin 38.7 = 1.87 m/s. The time is given b (4 m) t = + (5 m) = 34 s. v B v B/W =.34 m/s. t = 34 s. 9

Problem 13.171 Relative to the earth, the sailboat sails north with speed v = 6 m/s and then sails east at the same speed. The tell-tale indicates the direction of the wind relative to the boat. Determine the direction and magnitude of the wind s velocit (in m/s) relative to the earth. 6 Tell-tale v wind/ground = v wind/boat + v boat/ground N In position one we have v wind/ground = v wind/boat1 i + (6 m/s)j In position two we have W S E v wind/ground = v wind/boat ( cos 6 i + sin 6 j) + (6 m/s)i Since the wind has not changed these two epressions must be the same. Therefore v wind/boat1 = v wind/boat cos 6 } + 6 m/s 6 m/s = v wind/boat sin 6 { vwind/boat1 =.536 m/s v wind/boat = 6.98 m/s Using either position one or position two we have v wind/ground = (.536i + 6j) m/s v wind/ground = (.536) + (6) m/s = 6.51 m/s ( ).536 direction = tan 1 =.91 east of north 6 Problem 13.17 Suppose ou throw a ball straight up at 1 m/s and release it at m above the ground. What maimum height above the ground does the ball reach? How long after release it does the ball hit the ground? (c) What is the magnitude of its velocit just before it hits the ground? The equations of motion for the ball are a = g = 9.81 m/s, v = v gt = 1 9.81t (m/s), and The ball hits the ground when = m.tofind out when this occurs, we set = m into the equation and solve for the time(s) when this occurs. There will be two times, one positive and one negative. Onl the positive time has meaning for us. Let this time be t = t. The equation for t is = = + 1t 9.81t / (m). Solving, we get t =. s. = + v t gt / = + 1t 9.81t / (m). The maimum height occurs when the velocit is zero. Call this time t = t 1. It can be obtained b setting velocit to zero, i.e., v = = 1 9.81t 1 (m/s). Solving, we get t 1 = 1. s. Substituting this time into the equation, we get a maimum height of MAX = 7.1 m. (c) The velocit at impact is determined b substituting t =. s into the equation for v. Doing this, we find that at impact, v = 11.8 m/s 93

Problem 13.173 Suppose that ou must determine the duration of the ellow light at a highwa intersection. Assume that cars will be approaching the intersection traveling as fast as 14.6. km/h, that the drivers reaction times are as long as.5 s, and that cars can safel achieve a deceleration of at least.4g. How long must the light remain ellow to allow drivers to come to a stop safel before the light turns red? What is the minimum distance cars must be from the intersection when the light turns ellow to come to a stop safel at the intersection? The speed-time equation from initial speed to stop is given b integrating the equation d s =.4g. From which ds =.4gt + V, and s(t) =.gt + V t, where V is the initial speed and the distance is referenced from the point where the brakes are applied. The initial speed is: V = 14.6+1/36 = 9. 5 m/s. The time required to come to a full stop The distance traveled after brake application is traveled from brake application to full stop is given b s(t) =.gt + V t, from which s(t ) = 17.6 m. The distance traveled during the reaction time is d = V (.5) = 9. 5(.5) = 14. 53 m, from which the total distance is d t = 9. 5 + 14. 53= 43. 58m ds(t ) = ist = V.4g = 9. 5 (.4)( 9. 81) = 7.4 s. The driver s reaction time increases this b.5 second, hence the total time to stop after observing the ellow light is T = t +.5 = 7.9 s Problem 13.174 The acceleration of a point moving along a straight line is a = 4t + m/s. When t = s, its position is s = 36 m, and when t = 4 seconds, its position is s = 9 meters. What is its velocit when t = 4s? The position-time equation is given b integrating d s ds = 4t +, from which = t + t + V, and s(t) = ( ) t 3 + t + V t + d, 3 where V,d are the initial velocit and position. From the problem conditions: s() = from which ( ) 3 + ( ) + V () + d = 36, 3 (1) V + d = ( ) 8.s(4) = 3 ( ) 4 3 + (4 ) + V (4) + d = 9, 3 from which () 4V + d = ( ) 94. 3 Subtract (1) from () to obtain ( ) 94 8 V = =.33 m/s. 6 The velocit at t = 4 seconds is [ ] ds(t) = [t + t + V ] t=4 = 3 + 8 +.33 = 4.33 m/s t=4 94

Problem 13.175 A model rocket takes off straight up. Its acceleration during the s its motor burns is 5 m/s. Neglect aerodnamic drag, and determine the maimum velocit of the rocket during the flight and the maimum altitude the rocket reaches. The strateg is to solve the equations of motion for the two phases of the flight: during burn t s seconds, and after burnout: t>s. Phase 1: The acceleration is: d s ds = 5, from which = 5t, and s(t) = 1.5t, since the initial velocit and position are zero. The velocit at burnout is V burnout = (5)() = 5 m/s. The altitude at burnout is h burnout = (1.5)(4) = 5 m. Phase. The acceleration is: d s ds = g, from which = g(t ) + V burnout (t ), and s(t) = g(t ) / + V burnout (t ) + h burnout,(t ). The velocit during phase 1 is constantl increasing because of the rocket s positive acceleration. Maimum occurs at burnout because after burnout, the rocket has negative acceleration and velocit constantl decreases until it reaches zero at maimum altitude. The velocit from maimum altitude to impact must be constantl increasing since the rocket is falling straight down under the action of gravit. Thus the maimum velocit during phase occurs when the rocket impacts the ground. The issue of maimum velocit becomes this: is the velocit at burnout greater or less than the velocit at ground impact? The time of flight is given b = g(t flight ) / + V burnout (t flight ) + h burnout, from which, in canonical form: (t flight ) + b(t flight ) + c =, where b = (V burnout /g) and c = (h burnout /g). The solution (t flight ) = b ± b c = 11.11, =.9 s. Since the negative time is not allowed, the time of flight is t flight = 13.11 s. The velocit at impact is V impact = g(t flight ) + V burnout = 59 m/s which is higher in magnitude than the velocit at burnout. The time of maimum altitude is given b ds = = g(t ma alt ) + V burnout, from which t ma alt = V burnout g t ma alt = 7.1 s. The maimum altitude is = 5.1 s, from which h ma = g (t ma alt ) + V burnout (t ma alt ) + h burnout = 177.4 m 95

Problem 13.176 In Problem 13.175, if the rocket s parachute fails to open, what is the total time of flight from takeoff until the rocket hits the ground? The solution to Problem 13.175 was (serendipitousl) posed in a manner to ield the time of flight as a peripheral answer. The time of flight is given there as t flight = 13.11 s Problem 13.177 The acceleration of a point moving along a straight line is a = cv 3, where c is a constant. If the velocit of the point is v, what distance does the point move before its velocit decreases to v? The acceleration is dv = cv 3. Using the chain rule, dv = dv ds = v dv ds ds = cv3. Separating variables and integrating: dv v = cds, from which 1 v = cs + C. Ats =, v = v, from which 1 v = cs 1 v, and v = v 1 + v cs. Invert: v cs = v v 1. When v = v ( ) 1, s = cv Problem 13.178 Water leaves the nozzle at above the horizontal and strikes the wall at the point indicated. What was the velocit of the water as it left the nozzle? Strateg: Determine the motion of the water b treating each particle of water as a projectile. 3.66 m 6.1 m 1.67 m equations: dv d = g and dv Denote θ =. The path is obtained b integrating the = gt + V n sin θ, d =, from which = V n cos θ. Substitute: = g ( ) 1.67 + 1.67 tan θ. 44, V n cos θ ( ) 1.67 g from which V n = cos θ ( 1.67 tan θ.44) =.9 m/s = g t + (V n sin θ)t +. = (V n cos θ)t +. Choose the origin at the nozzle so that =, and =. When the stream is (t impact ) = 6.1 3.66 =.44 m, the time is = g (t impact) + (V n sin θ)t impact.44. At this same time the horizontal distance is 1.67 (t impact ) = 1.67 = (V n cos θ)t impact, from which t impact = Vn cos θ. 96

Problem 13.179 In practice, a quarterback throws the football with a velocit v at 45 above the horizontal. At the same instant, a receiver standing 6.1 m in front of him starts running straight down field at 3.5 m/s and catches the ball. Assume that the ball is thrown and caught at the same height above the ground. What is the velocit v? 45 3.5 m/s 6.1 m Denote θ = 45. The path is determined b integrating the equations; d d = g, d =, from which = gt + v sin θ, d = v cos θ. v 45 1 ft/s = g t + (v sin θ)t, = (v cos θ)t, ft where the origin is taken at the point where the ball leaves the quarterback s hand. When the ball reaches the receiver s hands, v sin θ =, from which t flight =. g At this time the distance down field is the distance to the receiver: = 3.5 t flight + 6.1. But also = (v cos θ)t flight, from which 6.1 t flight = (v cos θ 3.5). Substitute: 6.1 (v cos θ ) = v sin θ, from which 3.5 g 37.1 g = v sin θ(v cos θ 3.5). The function f(v ) = v sin θ(v cos θ 3.5) 37.1 g was graphed to find the zero crossing, and the result refined b iteration: v = 11. 1 m/s. Check: The time of flight is t = 1.7 s and the distance down field that the quarterback throws the ball is d = 3.87 + 6.1 = 1 m, which seem reasonable for a short, lob pass. check. 97

Problem 13.18 The constant velocit v = m/s. What are the magnitudes of the velocit and acceleration of point P when =.5 m? =. sin π P 1 m Let = t m/s. Then =.5 m at t =.15 s. We know that v = m/s and a =. From =. sin(πt), we obtain d d =.4π cos(πt) and =.8π sin(πt). At t =.15 s, d d =.141 m and = v =.889 m/s and = a = 5.58 m/s. Therefore v = v + v =.19 m/s, a = a + a = 5.58 m/s. Problem 13.181 The constant velocit v = m/s. What is the acceleration of point P in terms of normal and tangential components when =.5 m? See the solution of Problem 13.18. The angle θ between the ais and the path is ( ) ( ) v.889 θ = arctan = arctan = 4.. Then v a t = a cos θ + a sin θ = + ( 5.58) sin 4. =.7 m/s, a θ a t a a N = a sin θ a cos θ = ( 5.58) cos 4. = 5.1 m/s. a N The instantaneous radius is ρ = v + v = () + (.889) =.939 m. a N 5.1 98

Problem 13.18 The constant velocit v = m/s. What is the acceleration of point P in terms of polar coordinates when =.5 m? See the solution of Problem 13.19. The polar angle θ is ( ) ( ).141 θ = arctan = arctan = 9.5. Then.5 a θ a a r a r = a cos θ + a sin θ = + ( 5.58) sin 9.5 =.75 m/s, a θ = a sin θ + a cos θ = + ( 5.58) cos 9.5 = 4.86 m/s. θ a Problem 13.183 A point P moves along the spiral path r = (.1)θ m, where θ is in radians. The angular position θ = t rad, where t is in seconds, and r = at t =. Determine the magnitudes of the velocit and acceleration of P at t = 1s. r P θ The path r =.t m, θ = t rad. The velocit components are v r = dr =. m/s,v θ = r dθ = (.t) =.4t. At t = 1 seconds the magnitude of the velocit is v = vr + v θ =. +.4 =.447 m/s The acceleration components are: a r = d r r a θ = r ( dθ ( d ) θ + ) = (.t)( ) m/s, ( dr )( ) dθ = (.)() =.8 m/s. The magnitude of the acceleration is a = ar + a θ = 1.13 m/s 99

Problem 13.184 In the cam-follower mechanism, the slotted bar rotates with a constant angular velocit ω = 1 rad/s, and the radial position of the follower A is determined b the profile of the stationar cam. The slotted bar is pinned a distance h =. m to the left of the center of the circular cam. The follower moves in a circular path of.4 m radius. Determine the velocit of the follower when θ = 4 in terms of polar coordinates, and in terms of cartesian coordinates. h θ r A The first step is to get an equation for the path of the follower in terms of the angle θ. This can be most easil done b referring to the diagram at the right. Using the law of cosines, we can write R = h + r hr cos θ. This can be rewritten as r hr cos θ + (h R ) =. We need to find the components of the velocit. These are v r =ṙ and v θ = r θ. We can differentiate the relation derived from the law of cosines to get ṙ. Carring out this differentiation, we get rṙ hṙ cos θ + hr θ sin θ =. Solving for ṙ, we get O h r θ c R P ṙ = hr θ sin θ (h cos θ r). Recalling that ω = θ and substituting in the numerical values, i.e., R =.4 m, h =. m,ω = 1 rad/s, and θ = 4, we get r =.553 m, v r =.13 m/s, and v θ = 6.64 m/s The transformation to cartesian coordinates can be derived from e r = cos θi + sin θj, and e θ = sin θi + cos θj. Substituting these into v = v r e r + v θ e θ, we get v = (v r cos θ v θ sin θ)i + (v r sin θ + v θ cos θ)j. Substituting in the numbers, v = 5.9i + 3.71j (m/s) Problem 13.185* In Problem 13.184, determine the acceleration of the follower when θ = 4 in terms of polar coordinates and in terms of cartesian coordinates. Information from the solution to Problem 13.184 will be used in this solution. In order to determine the components of the acceleration in polar coordinates, we need to be able to determine all of the variables in the right hand sides of a r = r r θ and that a θ = r θ + ṙ θ. We alread know everthing ecept r and θ. Since ω is constant, θ = ω =. We need onl to find the value for r and the value for r at θ = 4. Substituting into the original equation for r, wefind that r =.553 m at this position on the cam. To find r, we start with ṙ = v r. Taking a derivative, we start with rṙ hṙ cos θ + hr θ sin θ = from Problem 13.184 (we divided through b ). Taking a derivative with respect to time, we get Evaluating, we get r = 46.17 m/s. Substituting this into the equation for a r and evaluating a n,wegeta r = 15.81 m/s and a θ = 51. m/s The transformation of cartesian coordinates can be derived from e r = cos θi + sin θj, and e θ = sin θi + cos θj. Substituting these into a = a r e r + a e e θ, we get a = a r e r + a θ e θ, we get a = (a r cos θ a θ sin θ)i + (a r sin θ + a θ cos θ)j. Substituting in the numbers, we get a = 63.46 i 1.1 j(m/s ). r = ṙ + hr θ sin θ + hr θ cos θ + hr θ sin θ. (h cos θ r) 1