Chapter 8: Rotational Motion Radians 1) 1 radian = angle subtended by an arc (l) whose length is equal to the radius (r) 2) q = l r r l Convert the following: a) 20 o to radians (0.35 rad) b) 20 o to revolutions (0.056 rev) c) 5000 o to radians (87.3 rad) d) 0.75 rev to radians (4.7 rad) e) 0.40 radians to degrees (23 o ) 3) 360 0 = 2p radians = 1 rev 4) Radians are dimensionless q A bird can only see objects that subtend an angle of 3 X 10-4 rad. How many degrees is that? How small an object can the bird distinguish flying at a height of 100 m? 3 X 10-4 rad 360 o = 0.017 o 2p rad q = l r q r l (approx.) 1
At what height would the bird be able to just distinguish a rabbit that is 30 cm long (and tasty)? q A tiny laser beam is aimed from the earth to the moon (3.8 X 10 8 m). The beam needs to have a diameter of 2.50 m on the moon. What is the angle that the beam can have? r (ANS: 1000 m) l (approx.) (6.6 X 10-9 radians) Convert: a) 0.0200 rev/s to radians/s (0.126 rad/s) b) 30.0 o /s to radians/s (0.524 rad/s) c) 1.40 rad/s to rev/s (0.223 rev/s) d) 3000 rpm to radians/s (314 rad/s) Determine the Arc length of your vision Calculate your arc length of vision at 100 meters. Degrees Revolutions Radians q l (at 100 m) The Mighty Thor swings his hammer at 400 rev/min. Express this in radians/s. v = Dx Dt Angular Velocity w = Dq Dt 400 rev 1 min 2p rad 1 min 60 s 1 rev = 41.9 rad/s Angular Velocity radians an object rotates per second All points on an object rotate at the EXACT same angular velocity All points on an object DO NOT rotate with the same linear speed. (the farther out the faster) 2
Merry-Go-Round Example w a = w b (both rotate through 360 0 in the same time period) v a v b If a person at point b flies off the merrygo-round, will they travel in a curve or straight line? v a < v b (Point b travels a longer distance around the circle) a b b a = Dv Dt Angular Acceleration a = Dw Dt Frequency and Period Frequency = Revolutions per second (rev/s) Period = Time for one complete revolution 1) The angular acceleration of all points on a circle is the same. 2) All points on an object DO NOT experience the same linear acceleration. (the farther out the more acceleration) w = 2p f T = 1 f Converting between Angular and Linear Quantities Linear = Radius X Angular v = rw a tan = ra Note the use of a tan to differentiate from centripetal acceleration, a c or a r : a r a tan One child rides a merry-go-round (2 revolutions per minute) on an inside lion 2.0 m from the center. A second child rides an outside horse, 3.0 m from the center. a) Calculate the frequency and the period. (0.0333 Hz, 30 s) b) Calculate the angular velocity in rad/s. (0.209 rad/s) c) Calculate the linear velocity of each rider. (0.419 m/s, 0.628 m/s) d) Who has a higher angular velocity? 3
A clock has a seconds hand that is 7.00 cm long. a) Calculate the frequency and period. (0.0167 Hz, 60 s) b) Calculate the angular velocity of the second hand in radians/s. (0.105 rad/s) c) Calculate the angular velocity in degrees/s. (6 o /s) d) Calculate the linear velocity at the end of the seconds hand. (7.35 X 10-3 m/s) A baseball bat is found to have a linear acceleration of 13.9 m/s 2 at the sweet spot. The sweet spot is at 59 cm from the handle. a) Calculate the angular acceleration in rad/s 2 (a = 23.6 rad/s 2 ) b) Convert the angular acceleration to rev/s 2 (3.75 rev/s 2 ) A golf club has an angular acceleration of 20.0 rad/s 2 and is 114 cm long. a) Calculate the tangential acceleration of the club. (22.8 m/s 2 ) b) Convert the angular acceleration to rev/s 2. (3.18 rev/s 2 ) c) Calculate the force the club could give to a 45.93 g golf ball. (1.05 N) d) Calculate the velocity of the ball if the club is in contact with the ball for 50 cm. (4.77 m/s) Angular Kinematics A bike wheel starts at 2.0 rad/s. The cyclist accelerates at 3.5 rad/s 2 for the next 2.0 s. a) Calculate the wheel s new angular speed (9.0 rad/s) b) Calculate the number of revolutions. (1.75 revs) 4
A DVD rotates from rest to 31.4 rad/s in 0.892 s. a) Calculate the angular acceleration. (35.2 rad/s 2 ) b) How many revolutions did it make? (2.23 rev) c) If the radius of the disc is 4.45 cm, find the linear speed of a point on the outside edge of the disc. (1.40 m/s) A bicycle slows from v o = 8.4 m/s to rest over a distance of 115 m. The diameter of each wheel is 68.0 cm. a. Calculate the angular velocity of the wheels before braking starts. (24.7 rad/s) b. How many revolutions did each wheel undergo? (HINT: calculate the circumference of the circle first) (53.8 rev) c. Calculate the angular acceleration. (-0.903 rad/s 2 ) d. Calculate the time it took the bike to stop (27.4 s) A spinning bike tire of radius 33.0 cm has an angular speed of 50.0 rad/s. Twenty seconds later, its angular speed is 150.0 rad/s. a) Calculate the angular acceleration. (5.00 rad/s 2 ) b) Calculate the angular displacement over the 20 s. (2000 radians) c) Calculate the revolutions travelled in the 20 s (318 rev) d) Calculate the linear distance the tire travelled in 20 s. (660 m) A pottery wheel turning with an angular speed of 30.0 rev/s is brought to rest in 60.0 revolutions. a) Calculate the radians that the wheel travelled (377 rad) b) Convert the initial speed to rad/s (188 rad/s) c) Calculate the angular acceleration. (-46.9 rad/s 2 ) d) Calculate the time required to stop. (4.00 s) e) If the radius of the wheel is 12.0 cm, calculate the linear distance the outside of the wheel travelled. (45.2 m) A game show wheel with a 90 cm radius is initially turning at 3.0 rev/s. A point on the outside of the wheel travels 147 meters before stopping. a) Calculate how many revolutions it went through. (26 rev) b) Calculate the angular deceleration. (-1.09 rad/s 2 ) c) Calculate how long it took to stop. (17.3 s) d) Calculate the initial linear speed and linear acceleration (tangential) on the outside of the wheel (17.0 m/s, -0.98 m/s 2 ) e) Calculate the initial centripetal acceleration on a point at the edge of the wheel. (319 m/s 2 ) A curve ball is given an initial angular speed of 40.0 rad/s. In the 0.500 s it takes to reach the catcher, it has slowed to 36.0 rad/s. The diameter of the ball is 7.50 cm. a) Calculate the angular acceleration of the ball. (-8 rad/s 2 ) b) Calculate the number of revolutions the ball made. (3.02 rev) c) Calculate the tangential acceleration. (-0.300 m/s 2 ) d) The ball is thrown at an initial velocity of 36.0 m/s. Assume that air drag is 1.00 m/s 2. Calculate the distance to home plate. (17.9 m) 5
A jet turbine (d=65 cm) is spinning with an angular velocity of 10 rad/s. It is accelerated at a rate of 5 rad/ s 2 for 15 s. a. Calculate the final angular velocity of the engine. (85 rad/s) b. Calculate the number of revolutions the turbine made. (113 rev) c. Calculate the final linear speed of a point on the outside of the turbine. (27.6 m/s) d. Calculate the centripetal acceleration at a point on the outside of the turbine at the end. (2344 m/s 2 ) Friction and Rolling Wheels Rolling uses static friction A new part of the wheel/tire is coming in contact with the road every instant A B Torque Braking uses kinetic friction Torque tendency of a force to rotate a body about some axis (the force is always perpendicular to the lever arm) A Point A gets drug across the surface t = Fr pivot r F Torque Sign Conventions Counter-clockwise Torque is positive Torque: Example 1 A wrench is 20.0 cm long and a 200.0 N force is applied perpendicularly to the end. Calculate the torque. 20.0 cm Clockwise Torque is negative 200.0 N t = 40.0 m-n 6
Torque: Example 2 Suppose that same 200.0 N force is now applied at a 60 o angle as shown. Calculate the Torque. Is it greater or less? First we need to resolve the Force vector into x and y components F y 200.0 N 60 o 20.0 cm 200.0 N 60 o Only F y has any effect on the torque (perpendicular) F y = Fsinq = (200.0 N)(sin 60 o ) = 173.2 N t = Fr = (173.2 N)(0.20 m) = 34.6 m-n F x Torque: Example 3 The biceps muscle exerts a 700 N vertical force. Calculate the torque about the elbow. t = Fr = (700 N)(0.050 m) t = 35 m-n A force of 200 N acts tangentially on the rim of a wheel 25 cm in radius. a) Calculate the torque. (50 mn) b) Calculate the torque if the force makes an angle of 40 o to a spoke of the wheel. (32.3 mn) c) If the wheel is mounted vertically, draw a free body diagram of the wheel if the force is the one in (a). Torque: Example 4 Two wheels, of radii r 1 = 30 cm and r 2 =50 cm are connected as shown. Calculate the net torque on this compound wheel when two 50 N force act as shown. r 2 r 1 30 o 50 N First find the horizontal component of the top force: F x = (50 N)(cos 30 o ) = 43 N The top force is pulling clockwise (-) and the bottom force pulls counterclockwise (+) St = F 1 r 1 F 2 r 2 St = (50N)(0.30m) (43 N)(0.50 m) = -6.5 m-n 50 N 7
Two children push on a merry-go-round as shown. Calculate the net torque on the merry-go-round if the radius is 2.0 m. A 60.0 cm diameter wheel is pulled by a 500.0 N force. The force acts at an angle of 65.0 o with respect to the spoke. Assume that a frictional force of 300.0 N opposes this force at a radius of 2.00 cm. Calculate the net torque on the wheel. (+130.0 mn) (+1800 m N) Moment of Inertia (I) Measure of Rotational Inertia An objects resistance to a change in angular velocity Would it be harder to push a child on a playground merry-go-round or a carousel? Deriving I Consider pushing a mass around in a circle (like the child on a merry-go-round) F = ma a = ra F = mra r t = Fr t = mrar t = mr 2 a F m I = moment of inertia I = mr 2 More properly I = Smr 2 = m 1 r 1 2 + m 2 r 2 2 +. St= Ia Moment of Inertia: Example 1 Calculate the moment of inertia (I) for the barbell when rotated about point M. We will assume the barbell is 1.0 m long, and that each weight is a point mass of 45.4 kg. (Ans: 22.7 kg-m 2 ) Would it be harder (require more torque) to twirl a barbell in the middle (pt. M) or the end (Pt. E) E M M 8
Moment of Inertia: Example 2 Now calculate I assuming Mr. Fredericks uses his massive musculature to twirl the barbells from point E. (Ans: 45.4 kg-m 2 ) Moment of Inertia: Example 3 Calculate I for Violet Beauregarde (40.0 kg, radius = 1.20 m). Use the formulas from the book. E I = 23.0 kg-m 2 Calculate the moment of inertia of an 8.00 kg solid wheel with a radius of 25.0 cm.(ans: 0.25 kgm 2 ) A wheel has a moment of inertia of 0.50 kg m 2. a. Calculate the torque (St= Ia) is required to give it an acceleration of 3 rad/s 2. (1.5 mn) b. Calculate its angular speed (from rest) after 5.00 s. (15 rad/s) c. Calculate the number of revolutions it goes through in 5.00 s. (37.5 rad, 5.97 rev) A 15.0 N force is applied to a cord wrapped around a pulley of radius 33.0 cm. The pulley reaches an angular speed (w) of 30.0 rad/s in 3.00 s. a) Calculate the angular acceleration. (10.0 rad/s 2 ) b) Calculate the torque (4.95 m-n) c) Calculate the moment of inertia of the pulley. (0.495 kg-m 2 ) 15.0 N 33.0 cm A 25.0 kg wheel has a radius of 40.0 cm. A 1.20 kg mass is hung on the end of the wheel by a string and falls freely. a) Calculate the moment of inertia of the wheel. (2 kg m 2 ) b) Setup the equation for all of the forces on the mass. c) Setup the equation for the torque on the wheel. d) Using systems of equations, calculate the tension in the cord and the tangential acceleration of the wheel (and bucket) (10.7 N, 0.86 m/s 2 ) 9
A 15.0 N force is applied to a cord wrapped around a pulley of radius 33.0 cm. The pulley reaches an angular speed (w) of 30.0 rad/s in 3.00 s. Since this is a real pulley, there is a frictional torque (t fr = 1.10 m-n) opposing rotation. a) Calculate net torque on the pulley. (3.85 m-n) b) Calculate the moment of inertia of the pulley. (0.385 kg-m 2 ) 33.0 cm Using the same pulley as the previous problem (I=0.385 kg m 2 ), hang a 15.0 N bucket (1.53 kg) from the cord. Calculate the angular and linear acceleration of the bucket. Also calculate T. Remember to include the frictional torque. (t fr = 1.10 m-n) T 33.0 cm 15.0 N We will break this problem into two parts: pulley and bucket. Let s first look at the pulley: St = F T r t fr St = Ia Ia = F T r t fr F T 33.0 cm t fr Now we will deal with the bucket SF = mg F T ma = mg F T 1.53a = 15 F T a= ra1.53ra = 15 F T (1.53)(0.33)a = 15 F T F T mg (0.385)a = F T (0.33) 1.10 0.505a = 15 F T 0.385a = 0.33F T 1.10 a = 7.07 rad/s 2, a = 2.33 m/s 2, F T = 11.5 N Now calculate the rotational speed of the pully (w) and the linear speed of the bucket after 3.00 s. w=w o + at w= 0 + at = (6.98 rad/s 2 )(3.00 s) = 20.9 rad/s v=rw v = (0.330 m)(20.9 rad/s) = 6.90 m/s A 2.00 kg bucket is attached to a 1.00 kg, 4.00 cm radius cylindrical pulley. The bucket is suspended 2.00 m above the floor. a) Calculate the moment of inertia of the cylinder (1/2 MR 2 ) (8X10-4 kgm 2 ) b) Calculate the acceleration of the bucket. (7.84 m/s 2 ) c) Calculate the angular acceleration of the pulley. (196 rad/s 2 ) d) Calculate how long it takes to reach the floor. (0.714 s) T 4.00 cm 10
a) I = 1/2 MR 2 = (1/2)(1.00 kg)(0.04 m) 2 I = 8 X 10-4 kgm 2 b) Bucket Pulley ma = mg F T t = F T R 2a = (2)(9.8) F T Ia = F T R 2a = 19.6 F T a = ar Ia/R = F T R 8 X 10-4 a= F T (0.04) 0.04 c) a = ar a = a/r = 7.84/0.04 = 196 rad/s 2 d) y = y o + vt + ½at 2 y = ½ at 2 t = (2y/a) = 0.714 s a = -7.84 m/s 2 A 1000.0 kg elevator is suspended by a 100.0 kg cylinder pulley of radius 50.0 cm. a) Calculate the moment of inertia of the pulley. (12.5 kgm 2 ) b) Calculate the linear acceleration if the elevator drops. (9.33 m/s 2 ) c) Calculate the angular acceleration of the pulley. (18.7 rad/s 2 ) d) Calculate the tension in the cord. (467 N) e) Are you glad you are not in the elevator? Translational and Rotational Energy Does a rotating helicopter blade have kinetic energy before the helicopter takes off? How about afterwards? Does all of the energy of the fuel go into moving the helicopter? Deriving the Rotational KE Translational Speed (v) speed of the center of a wheel with respect to the ground Can also be called linear speed Use regular K = ½ mv 2 Rotational speed (w) angular speed of the wheel Use K = ½ Iw 2 KE = S½ mv 2 v = rw KE = S ½ m(rw) 2 KE = S ½ mr 2 w 2 I = S mr 2 KE = ½ Iw 2 11
Calculating Work A pottery wheel of I = 0.200 kgm 2 is accelerated from rest to 500 rpm. a) Calculate the final angular velocity in rad/s. (52.4 rad/s) b) Calculate the change in kinetic energy (274 J) c) How much work is done in the process (274 J) d) If the wheel takes 30 seconds to get up to speed, calculate the angular acceleration (1.75 rad/s 2 ) Jupiter has a mass of 1.9 X 10 27 kg and a radius of 7.0 X 10 7 m. Its day is only 9.92 hours long. a) Calculate Jupiter s moment of inertia, treating it as a sphere. (3.72 X 10 42 kgm 2 ) b) Calculate Jupiter s rotational speed in rad/s. (1.76 X 10-4 rad/s) c) Calculate Jupiter s kinetic energy of rotation. (5.76 X 10 34 J) d) Jupiter is 7.78 X 10 11 m from the sun. Calculate Jupiter s moment of inertia, assuming it is a point mass from the sun. (1.15 X 10 51 kgm 2 ) e) Calculate Jupiter s orbital speed around the sun if it takes11.9 year to orbit the sun. (1.67 X 10-8 rad/s) f) Calculate the kinetic energy about the sun. (1.60 X 10 35 J) Law of Conservation of Mechanical Energy (K t + K r + U t ) i = (K t + K r + U t ) f Rotational KE: Example 1 What will be the translational speed of a log (100 kg, radius = 0.25 m, I= ½ mr 2 ) as it rolls down a 4 m ramp from rest? (ANS: v = 7.2 m/s) Rotational work W r = tdq 4 m Now we can calculate the angular speed v = rw so w = v/r w = 7.23 m/s = 29 rad/s 0.25 m 29 rad 1 rev = 4.6 rev/s s 2p rad Rotational KE: Example 2 What will be the translational speed of Bouncing Boy (75 kg, radius = 1.2 m) as he rolls down a 3 m ramp from rest? (Treat him as a sphere) 3 m 12
(KE t + KE r + PE t ) i = (KE t + KE r + PE t ) f (0 + 0 + mgy) i = ( ½ mv 2 + ½ Iw 2 + 0) f mgy = ½ mv 2 + ½ Iw 2 2mgy = mv 2 + Iw 2 (multiplied both sides by 2) v = rw so w = v/r 2mgy = mv 2 + Iv 2 r 2 I = 2/5 mr 2 2mgy = mv 2 + 2mr 2 v 2 5r 2 2mgy = mv 2 + 2mr 2 v 2 2gy = v 2 + 2v 2 5 5r 2 2gy = 5v 2 + 2v 2 5 5 2gy = 7v 2 5 v = 10gy = 10(9.8m/s 2 )(3.00 m) = 6.50 m/s 7 7 Now we can calculate the angular speed v = rw so w = v/r w = 6.50 m/s = 5.4 rad/s 1.2 m 5.4 rad 1 rev = 0.86 rev/s s 2p rad What would his speed be if he just slid down the ramp? (KE t + KE r + PE t ) i = (KE t + KE r + PE t ) f (0 + 0 + mgy) i = ( ½ mv 2 + 0 + 0) f mgy = ½ mv 2 gy = ½ v 2 v 2 = 2gy v = \/2gy = (2 X 9.8 m/s X 3.00 m) 1/2 v = 7.7 m/s Why is this larger than if he rolls? A wooden dowel (mass = 0.50 kg, r = 1.00 cm) rolls down a 1.25 m tall ramp. a) Calculate the moment of inertia (2.5 X 10-5 kgm 2 ) b) Calculate the linear speed at the bottom. (4.04 m/s) c) Calculate the angular speed at the bottom. (404 rad/s, 64.3 rev/s) A bike tire can be modeled as a thin hoop (I=mr 2 ). The wheel has a diameter of 66.0 cm and a mass of 2.00 kg. The wheel is allowed to roll down a ramp and has a linear (translational) speed of 2.00 m/s at the bottom of the ramp. a) Calculate the moment of inertia of the wheel. (0.218 kgm 2 ) b) Calculate the height of the ramp (40.8 cm) 13
Conservation of Angular Momentum The total angular momentum of a rotating body remains constant if the net torque acting on it is zero Angular Momentum (L) is conserved p = mv (linear momentum) L = Iw (angular momentum) I i w i = I f w f The skater in more detail For now, let s just consider the skater s arms: I i w i = I f w f I = mr 2 Her mass will not change when she moves her arms Arms out (initial) Arms in (final) mr i2 w i = mr f2 w f r i2 w i = r f2 w f Since r f will be smaller (arms are in) w f must increase to compensate I wanna go fast! A child sits on a merry-go-round at point A. If the child wants to go faster, should he walk towards the center or the outside? A Cons. of Angular Momentum: Ex. 1 A skater holds her arms at a length of 56 cm. She spins at 9.43 rad/s. What will be her new speed if she pulls her arms tight to her body, 20 cm? I i w i = I f w f I = mr 2 mr i2 w i = mr f2 w f r i2 w i = r f2 w f w f = r i2 w i /r f 2 w f = [(0.56 m) 2 (9.43 rad/s)]/(0.20 m) 2 = 74 rad/s 14
Cons. of Angular Momentum: Example 2 A mass m is attached to the end of a string which passes through a hole in a table. Initially the r 1 is 0.80 m and the block moves at 2.4 m/s. What will be the new speed as the radius of the string is reduced to 0.48 m? I i w i = I f w f I = mr 2 mr i2 w i = mr f2 w f r i2 w i = r f2 w f v = rw so w = v/r r i2 v i = r f2 v f r i r f r i v i = r f v f v f = r i v i = (0.80m)(2.4 m/s) = 4.0 m/s r f (0.48 m) Old record players rotate at 33.3 rev/min. Assume a record has a radius of 15.24 cm and a mass of 100.0 grams. a) Calculate the angular speed in rad/s (3.49 rad/s) b) Calculate the moment of inertia. (1/2 mr 2 ) (1.16 X 10-3 kgm 2 ) c) Calculate angular momentum of the record. (4.05 X 10-3 kg m 2 /s) d) A second record drops on top of the first. Calculate the new moment of inertia and new angular speed. (2.32 X 10-3 kg m 2, 1.74 rad/s) Vectors and Angular Momentum Direction of the vector w is defined by the right-hand rule. L = Iw L is defined as going in the same direction. Merry-Go-Round What will happen if you start to walk on a still merry-go-round? Can you walk around it without it moving? Assume this merry-goround is initially at rest (L total = 0) As the man walks counterclockwise, the platform spins clockwise Total Angular Momentum is still zero. 15
What will happen if the student flips the bicycle wheel while standing on the merrygo-round? A merry-go-round (500 kg) has a radius of 1.5 m. A 40.0 kg student is sitting on the merry-go-round as it spins at 30.0 rev/min. a) Calculate the angular speed (3.14 rad/s) b) Calculate the moment of inertia of the merry-go-round and student, assuming she sits at the edge. (652.5 kg m 2 ) c) Calculate the angular momentum. (2049 kgm 2 /s) d) A second student (80.0 kg) now jumps on the merrygo-round. Calculate the new moment of inertia and angular speed. (832.5 kgm 2, 2.46 rad/s) A 20.0 kg box is placed at the top of a 25.0 o inclined plane. It is connected to a rope wrapped around a 5.00 kg pulley with radius of 20.0 cm. a) Calculate the acceleration of the box and the tension in the cord. (3.68 m/s 2, 9.21 N) b) Assume a frictional torque of 2.00 mn. Recalculate acceleration and tension. (3.24 m/s 2, 18.1 N) 2. R Sun = 6.5 X 10 5 km 4. 2900 m 6. 188 rad/s 8. 1.9 rad/s 2 10. -260 rad/s 2 12. 3300 revolutions 14. a) 1.99 X 10-7 rad/s b) 7.27 X 10-5 rad/s 16. 3.6 X 10 4 rpm 20. 0.56 rad/s 2 22. a) -84 rad/s 2 b) 150 rev 24. 38 m 22. a) -84 rad/s 2 b) 150 rev 24. 38 m 26. a) 0.58 rad/s 2 b) 12 s 30. a) 38 m N b) 33 m N 32. 650 m N 34. 1.89 kg m 2 36. a) 6.1 kg m 2 b) 0.61 kg m 2 38. a) 0.851 kg m 2 b) 0.0720 m N 40.1.99 X 10-3 kg m 2 b) 0.0682 m N 42.590 m N, 240 N 46. 2250 kg m 2, 8800 mn 42. 590 m N, 240 N 44. a) 7.5 m N b) 300 N 46. 2250 kg m 2, 8800 mn 50. 94 J 52. a) 2.6 X 10 29 J b) 2.7 X 10 33 J 54.a) 8.37 m/s, 41.8 rad/s b) 2.50 (mass independent) 16
58. a) angular momentum conserved b) 1.6 60. 0.77 kg m 2 (pull arms close to body) 62. a) 14 kg m2/s b) -2.7 m N 66.a) 1.2 rad/s b) 2000 J, 1200 J (loss of 800J, 40% loss) 32. t = F fr r = m s mgr t = (0.75)(1/4)(1080 kg)(9.8 m/s2)(0.33 m) = 650 mn 34. I = (2/5)MR 2 = (2/5) (12.2)(0.623) 2 = 1.89 kg m 2 35. I = MR 2 = (1.25)(0.333) 2 = 0.139 kg m 2 36. I x = S MR 2 = (1.8)(0.5) 2 + (3.1)(0.5) 2 + (1.8)(1) 2 + (3.1)(1) 2 I x = 6.1 kg m 2 I y = S MR 2 = (1.8)(0.25) 2 + (3.1)(0.25) 2 + (1.8)(0.25) 2 + (3.1)(0.25) 2 I y = 0.61 kg m 2 37. 1.20 X 10-10 m 38. a) I = MR 2 = (1.05)(0.900) 2 = 0.851 kg m 2 b) t net = 0 0 =t applied t friction t applied = F fr R = (0.0800 N)(0.900 m) = 0.0720 mn 53.w = (1rev/8s)(2p/1rev) = 0.785 rad/s I = 1/2mr 2 = (1/2)(1640)(8.20) 2 = 5.51 X 10 4 kgm 2 W = 1/2Iw 2 = (1/2)(5.51 X 10 4 kgm 2 )(0.785 rad/s) 2 W = 1.70 X 10 4 J 54. a) 8.37 m/s, 41.8 rad/s b) 2.50 (mass independ.) 17
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