TIME OF COMPLETION NAME SOLUTION DEPARTMENT OF NATURAL SCIENCES. PHYS 1111, Exam 3 Section 1 Version 1 December 6, 2005 Total Weight: 100 points


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1 TIME OF COMPLETION NAME SOLUTION DEPARTMENT OF NATURAL SCIENCES PHYS 1111, Exam 3 Section 1 Version 1 December 6, 2005 Total Weight: 100 points 1. Check your examination for completeness prior to starting. There are a total of nine (9) problems on seven (7) pages. 2. Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor). 3. You will have 50 minutes to complete the examination. 4. The total weight of the examination is 100 points. 5. There are five (5) multiple choice and four calculation problems. Work all multiple choice problems and 4 calculation problems. Show all work; partial credit will be given for correct work shown. 6. If you have any questions during the examination, see your instructor who will be located in the classroom. 7. Start: 10:30 a.m. Stop: 11:45 a.m PROBLEM POINTS CREDIT TOTAL 100 PERCENTAGE
2 CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT. 1. Rolling without slipping depends on a. Kinetic friction between the rolling object and the ground. b. Static friction between the rolling object and the ground. c. Tension between the rolling object and the ground. d. The force of gravity between the rolling object and the Earth. 2. A car is negotiating a flat circular curve of radius 50 m with a speed of 20 m/s. The maximum centripetal force (provided by static friction) is N. What is the centripetal acceleration of the car? a m/s 2 b m/s 2 c. 4.0 m/s 2 d. 8.0 m/s 2 a cp = V t 2 /R = (20.0 m/s) 2 /(50.0 m) 3. The gravitational force between two objects is proportional to a. The distance between the two objects. b. The square of the distance between the two objects. c. The product of the masses of the two objects. d. The square of the product of the masses of the two objects. 4. Two equal forces are applied to a door at the doorknob. The first force is applied
3 perpendicular to the door; the second force is applied at 30 to the plane of the door. Which force exerts the greater torque? a. The first applied perpendicular to the door. b. The second applied at an angle. c. Both exert equal nonzero torques. d. Both exert zero torques. 5. Two uniform solid spheres have the same mass, but one has twice the radius of the other. The ratio of the larger sphere's moment of inertia to that of the smaller sphere is a. 4/5. b. 8/5. c. 2. d. 4. (2/5 M (2R) 2 )/(2/5 M R 2 ) = 4 6. Three children are trying to balance on a seesaw, which consists of a fulcrum rock, acting as a pivot at the center, and a very light board 3.6 m long. Two playmates are already on either end. Boy A has a mass of 50 kg, and girl B a mass of 35 kg. a. Where should girl C, whose mass is 25 kg, place herself so as to balance the seesaw? = (50.0kg)(9.81 m/s 2 )(1.80 m) = 883 Nm =  (35.0kg)(9.81 m/s 2 )(1.80 m) = Nm C =  (25.0kg)(9.81 m/s 2 )(x) =  (245N)(x) = Nm Nm (245 N)(x) = 0
4 x = 1.08 m b. What is the magnitude of the normal force that the rock exerts on the board? N x = 0 N y = N w Ax = 0 w Ay =  m A g w Bx = 0 w By =  m B g w Cx = 0 w Cy =  m C g F x = 0 F y = 0 N  m A g m B g m C g = 0 N = m A g + m B g + m C g = (m A + m B + m C )g = 1079 N 7. Suppose our Sun eventually collapses into a white dwarf, losing about half its mass in the process, and winding up with a radius 1.0% of its existing radius. a. Assuming the lost mass carries away no angular momentum, what would the Sun s new rotation rate be? (Take the Sun s current period to be about 30 days.) I I I = I F F I I = 2/5 M 0 R 0 2 I I = 2/5 (M 0 /2)(R 0 /100) 2 (2/5 M 0 R 0 2 ) I = (2/5 (M 0 /2)(R 0 /100) 2 ) F
5 I = (1/2)(1/100) 2 ) F I = (1/20000) F F = (20000) I I = (2)/T = (2)/(30 days x (24 hours/1 day)(3600 s/1hour)) = 2.42 x 106 rad/s F = 4.84 x 102 rad/s (To practice, convert the new angular velocity into the new period) b. What would be its final KE in terms of its initial KE of today? K = ½ I 2 K I = ½ (2/5 M 0 R I K F = ½ (2/5 (M 0 /2)(R 0 /100) 2 2 F K F /K I = [½ (2/5 (M 0 /2)(R 0 /100) 2 ) 2 F ] / [½ (2/5 (M 0 )(R 0 ) 2 ) 2 I ] K F /K I = [(1/2)(1/100) 2 ) 2 F ] / [ 2 I ] = (1/20000)( 2 F / 2 I ) = The rings of Saturn are composed of chunks of ice that orbit the planet. The inner radius of the rings is 73,000 km, while the outer radius is 170,000 km. Find the period of an orbiting chunk of ice at the inner radius and the period of a chunk at the outer radius. Compare your numbers with Saturn s mean rotation period of 10 hours and 39 minutes. The mass of Saturn is kg. T 2 O = ((4 2 3 )/(GM S ))R O T 2 O = ((4 2 )/(6.673 x Nm 2 /kg 2 )(5.70 x kg)))(170,000,000 m) 3 = 5.06 x 10 9 s 2 T O = 7.11 x 10 4 s = 19.8 hours T 2 I = ((4 2 3 )/(GM S ))R I T 2 I = ((4 2 )/(( x Nm 2 /kg 2 )(5.70 x kg)))(73,000,000 m) 3 = 4.00 x 10 8 s 2
6 T I = 2.00 x 10 4 s = 5.56 hours TIME OF COMPLETION NAME SOLUTION DEPARTMENT OF NATURAL SCIENCES PHYS 1111, Exam 3 Section 1 Version 1 December 4, 2006 Total Weight: 100 points 1. Check your examination for completeness prior to starting. There are a total of nine (10) problems on seven (7) pages. 2. Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor). 3. You will have 75 minutes to complete the examination. 4. The total weight of the examination is 100 points. 5. There are six (6) multiple choice and four calculation problems. Work five (5) multiple choice problems and four calculation problems. Show all work; partial credit will be given for correct work shown. 6. If you have any questions during the examination, see your instructor who will be located in the classroom. 7. Start: 6:00 p.m. Stop: 7:15 p.m PROBLEM POINTS CREDIT
7 TOTAL 100 PERCENTAGE CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT. 1. Tripling the mass of the bob on a simple pendulum will cause a change in the frequency of the pendulum swing by what factor? a b c d What condition or conditions are necessary for static equilibrium? a. ΣF x = 0 b. ΣF x = 0, ΣF y = 0, Σ = 0 c. Σ = 0 d. ΣF x = 0, ΣF y = 0 3. Two equal forces are applied to a door at the doorknob. The first force is applied perpendicular to the door; the second force is applied at 30 to the plane of the door. Which force exerts the greater torque? a. The first applied perpendicular to the door. b. The second applied at an angle. c. Both exert equal nonzero torques. d. Both exert zero torques.
8 4. A solid cylinder of mass 10 kg is pivoted about a frictionless axis thought the center O. A rope wrapped around the outer radius R 1 = 1.0 m, exerts a force F 1 = 5.0 N to the right. A second rope wrapped around another section of radius R 2 = 0.50 m exerts a force F 2 = 6.0 N downward. What is the angular acceleration of the cylinder? a. 1.0 rad/s 2 b rad/s 2 c rad/s 2 d rad/s 2 5. A mass on a spring undergoes SHM. When the mass is at its maximum displacement from equilibrium, its instantaneous velocity a. Is maximum. b. Is less than maximum, but not zero. c. Is zero. d. Cannot be determined from the information given. 2. The moment of inertia of a rigid body a. Depends on the axis of rotation. b. Cannot be negative. c. Depends on mass distribution. d. All of the above.
9 7. A centrifuge rotor rotating at 10,300 rpm is shut off and is eventually brought uniformly to rest by a frictional torque of 1.20 m N. If the mass of the rotor is 4.80 kg and it can be approximated as a solid cylinder of radius m, a. What is the rotor s moment of inertia? I = ½ MR 2 = ½ (4.80 kg)( m) 2 = kgm 2 b. What is the angular acceleration of the rotor as it is brought to rest? /I = (1.20 Nm)/( kgm 2 ) = 99.2 rad/s 2 c. Through how many revolutions will the rotor turn before coming to rest? 2 = ) = 10,300 rpm = 1079 rad/s = 0 )=  2 / (2 ) = (1079 rad/s) 2 /(2 x 99.2 rad/s 2 ) = 5868 rad = 934 rev 8. A uniform disk turns at 2.4 rev s around a frictionless spindle. A nonrotating rod, of the same mass as the disk and length equal to the disk s diameter, is dropped onto the freely spinning disk. They then both turn around the spindle with their centers superposed. What is the angular frequency in rev s of the combination? I i = ½ MR 2
10 I f = ½ MR 2 + (1/12) ML 2 = ½ MR 2 + (1/12) M(2R) 2 = ½ MR 2 + (1/3) MR 2 = (5/6) MR 2 I i i = I f f f = I i i / I f f = (2.40 rev/s) (½ MR 2 ) /( 5/6 MR 2 ) = 1.44 rev/s 9. A kg toy is undergoing simple harmonic motion on the end of a horizontal spring with spring constant k = 300 N/m. The amplitude of the motion is m. a. Find the total mechanical energy of the toy at any point of its motion. E = ½ ka 2 = ½ (300 N/m) (0.200 m) 2 = 6.00 J c. Find the maximum speed attained by the toy. ½ m V 2 max = ½ ka 2 m V 2 max = ka 2 V 2 max = (k/m)a 2 V max = ((300 N/m)/(0.150 kg)) ½ (0.200 m) = 8.94 m/s d. How many oscillations per second does the toy undergo? T = 2 (m/k) ½ T = 2 ((0.150 kg)/(300 N/m)) ½ T = s F = 1/T = 7.11 Hz e. What is the speed of the toy at 10.0 cm to the right from the equilibrium position? V = + (k/m(a 2 x 2 )) ½ V = + ((300 N/m)/(0.150 kg)((0.200 m) 2 (0.100 m) 2 )) ½ = m/s
11 V = 7.75 m/s 7. Pictured below is a very light wooden plank with two masses, 10.0 kg each, on top of it. Find the reaction forces at points A and B. N Ax = N A cos (90 o ) = 0 N Ay = N A sin (90 o ) = N A N Bx = N B cos (90 o ) = 0 N By =N B sin (90 o ) = N B w 1x = w 1 cos (90 o ) = 0 w 1y = w 1 sin (90 o ) =  m 1 g =  (10.0 kg)(9.81 m/s 2 ) = N w 2x = w 2 cos (90 o ) = 0 w 2y = w 2 sin (90 o ) =  m 2 g =  (10.0 kg)(9.81 m/s 2 ) = N F x = 0 => 0 = 0 F y = 0 => N A + N B 98.1 N 98.1 N = 0 N A + N B = 196 N Assuming that the axis of rotation is at point A:
12 NA = 0 NB = N B (3.00 m) w1 =  (98.1 N) (0.750 m) = Nm sp =  (98.1 N) (2.50 m) = Nm = 0 N B (3.00 m) 73.6 Nm 245 Nm = 0 N B = 106 N N A = 196 N 106 N = 90.0 N TIME OF COMPLETION NAME SOLUTION DEPARTMENT OF NATURAL SCIENCES PHYS 1111, Exam 3 Section 1 Version 1 November 28, 2007 Total Weight: 100 points 1. Check your examination for completeness prior to starting. There are a total of ten (10) problems on seven (7) pages. 2. Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor). 3. You will have 75 minutes to complete the examination. 4. The total weight of the examination is 100 points. 5. There are six (6) multiple choice and four calculation problems. Work all calculation problems and 5 (five) multiple choice. Show all work; partial credit will be given for correct work shown. 6. If you have any questions during the examination, see your instructor who will be located in the classroom.
13 7. Start: 6:00 p.m. Stop: 7:15 p.m PROBLEM POINTS CREDIT TOTAL 100 PERCENTAGE CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT. 1. A solid disk rolls down the slope as shown below. What is the direction of the disc s angular momentum? (A) Down the slope. (B) Up the slope. (D) Into the page. (C) Out of the page. (E) Clockwise. 2. A very light rod with two identical masses attached to it is rotating in the counterclockwise direction around the axis which goes through one of the rod s ends as shown below. Which one of the masses has the most kinetic energy?
14 (A) A. (B) B. (C) Both masses have the same amount of kinetic energy. (D) Neither one of the masses has any kinetic energy. 3. Given the two masses from Problem 2, as the mass A undergoes the angular displacement of 1.40 rad, the mass B undergoes the angular displacement of (A) 0 rad. (B) rad. (C) 1.40 rad. (D) 2.80 rad. (E) 5.60 rad. 4. A merrygoaround spins feely when Janice moved quickly to the center along the radial direction. Is it true to say that same. (A) The moment of inertia of the system decreases and the angular speed increases. (B) The moment of inertia of the system decreases and the angular speed decreases. (C) The moment of inertia of the system decreases and the angular speed remains the (D) The moment of inertia of the system increases and the angular speed increases. (E) The moment of inertia of the system increases and the angular speed decreases. 5. What are the dimensions of the moment of inertia? (A) M L. (B) M L 2.
15 (C) M L 3. (D) M/L 2. (E) L/T What is the correct expression for torque, in terms of the magnitude of the force, F, the radial distance from the axis of rotation, r, and the angle between the force and the radial direction,? (A) = F r sin. (B) = F r cos. (C) = F r tan. (D) = F r cot. (E) = F r. 7. Two disks of identical masses but different radii (r and 2r) are spinning on frictionless bearings at the same angular speed of 1.00 rad/s but in opposite direction. The two disks are brought slowly together. The frictional force between the surfaces eventually brings them to a common angular velocity. What is the magnitude of the final angular velocity? L I = L F L I = (1/2 m r 2 ) + (1/2 m (2r) 2 ) ( ) L F = (1/2 m r 2 ) F + (1/2 m (2r) 2 ) F (1/2 m r 2 ) + (1/2 m (2r) 2 ) ( ) = (1/2 m r 2 ) F + (1/2 m (2r) 2 ) F r 2 + (2r) 2 ( ) = r 2 F + (2r) 2 F r 2 + 4r 2 ( ) = r 2 F + 4r 2 F + 4( ) = F + 4 F F =  3/5 I
16 F = rad/s 8. A uniform beam of mass m and length l is hung from two cables, one at the end of the beam and the other 5l/8 of the way to the other end as shown below. If m = 10.0 kg and l = 2.00 m, determine the magnitudes of the forces the cable exerts on the beam. Note: we will use standard xy coordinate system with the positive xaxis pointing to the right and positive yaxis pointing upward. T Lx = 0 T Ly = T L T Rx = 0 T Ry = T R w x = 0 w y =  m g F x = 0
17 F y = 0 T L + T R m g = 0 T L + T R = (10.0 kg)(9.81 m/s 2 ) = 98.1 N TL = 0 TR = T R (1.25 m) w =  (10.0 kg)(9.81 m/s 2 )(1.00 m) = Nm = 0 T R (1.25 m) 98.1 Nm = 0 T R = 78.5 N T L = 19.6 N 9. An automotive engine shaft revs up from 3.00 x 10 3 rev/min to 4.50 x 10 3 rev/min during a 1.20 s interval. (a) Calculate the magnitude of the angular acceleration of the shaft assuming it is constant during the interval. = (3.00 x 10 3 rev/min)(2 rad/1 rev)(1 min/60 s) = x 10 3 rad/s = (4.50 x 10 3 rev/min)(2 rad/1 rev)(1 min/60 s) = x 10 3 rad/s = + t = (  ) t = 131 rad/s 2 (b) Determine the number of revolutions of the shaft during the interval. = + 2 ( ) ( ) = (  )/ (2 = (469 rad)(1 rev/2 rad) = 74.6 rev (c) Assuming that the shaft can be approximated as a solid cylinder with a radius of 5.00 cm and mass of 10.0 kg, what is the change in the shaft s kinetic energy during this motion?
18 I = ½ m R 2 I = ½ (10.0 kg) ( m) 2 = kgm 2 K i = ½ I = ½ ( kgm 2 )(0.314 rad/s) 2 = 616 J K f = ½ I = ½ ( kgm 2 )(0.471 rad/s) 2 = 1387 J K = 771 J 10. Three forces are applied to a wheel of radius m, as shown below. One force is perpendicular to the rim, one is tangent to it, and the third makes a 40.0 o angle with the radius. (a) What is the net torque on the wheel due to these three forces for the axis perpendicular to the wheel and passing through its center? = (11.9 N) (0.350 m) sin(0 o ) = 0 = (14.6 N) (0.350 m) sin(40.0 o ) = Nm = (8.50 N) (0.350 m) sin(90 o ) = 2.98 Nm = Nm I = ½ m R 2 (b) Given that the mass of the wheel is 5.00 kg, what is its moment of inertia? I = ½ (5.00 kg) (0.350 m) 2 = kgm 2 = I (c) Find the angular acceleration that the wheel has as a result of these three forces. I = rad/s 2
19 (d) Does the wheel rotate clockwise or counterclockwise? Clockwise. TIME OF COMPLETION NAME SOLUTION DEPARTMENT OF NATURAL SCIENCES PHYS 1111, Exam 3 Section 1 Version 1 April 27, 2005 Total Weight: 100 points 1. Check your examination for completeness prior to starting. There are a total of ten (10) problems on seven (7) pages. 2. Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor). 3. You will have 75 minutes to complete the examination. 4. The total weight of the examination is 100 points. 5. There are six (6) multiple choice and four calculation problems. Work five (5) multiple choice and four calculation problems. Show all work; partial credit will be given for correct work shown. 6. If you have any questions during the examination, see your instructor who will be located in the classroom. 7. Start: 6:00 p.m. Stop: 7:15 p.m PROBLEM POINTS CREDIT
20 TOTAL 100 PERCENTAGE
21 CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT. 1. The moment of inertia of a rigid body e. Depends on the axis of rotation. f. Cannot be negative. g. Depends on mass distribution. h. All of the above. 2. If the period of a system undergoing a SHM is doubled, the frequency of the system is a. Doubled. b. Halved. c. For times as large. d. Onequarter as large. 3. Which one of the following quantities is zero when an object in simple harmonic motion is at its maximum displacement? a. Potential energy. b. Acceleration. c. Kinetic energy. d. Frequency.
22 4. What condition or conditions are necessary for rotational equilibrium? a. ΣF x = 0. b. ΣF x = 0, ΣΤ = 0. c. ΣΤ = 0. d. ΣF x = 0, ΣF y = What is the quantity used to measure an object's resistance to changes in rotation? a. Mass. b. Moment of inertia. c. Torque. d. Angular velocity. 6. If a net torque is applied to an object, that object will experience: a. A constant angular velocity. b. An angular acceleration. c. A constant moment of inertia. d. An increasing moment of inertia. 7. A person of mass 75.0 kg stands at the center of a rotating merrygoround of radius 3.00 m and moment of inertia of 920 kgm 2. The merrygoround rotates without friction with angular velocity of 2.00 rad/s. The person walks radially to the edge of the merrygoround. I i = 920 kgm 2 a. Calculate the angular velocity of the merrygoround when the person reaches the edge. I f = 920 kgm 2 + (75.0 kg)(3.00 m) 2 = 1595 kgm 2
23 i rad/s L i = L f I i i I f f f = I i i I f f = (920 kgm 2 ) 2.00 rad/s 1595 kgm 2 ) = 1.15 rad/s b. Calculate the rotational kinetic energy of the system before and after the person s walk. 2 KE i = ½ I i i ½ (920 kgm 2 ) (2.00 rad/s) 2 = 1840 J 2 KE f = ½ I f f ½ (1595 kgm 2 ) (1.15 rad/s) 2 = 1055 J d. If the merrygoround rotates in a clockwise direction as viewed from above, what is the direction of its angular momentum? Inward. (Right Hand Rule!) 8. A shop sign weighing 245 N is supported by a uniform beam as shown below. Find the tension in the wire and the horizontal and vertical forces exerted by the hinge on the beam. Remember to draw a freebody diagram first! (w) x = w cos( 90 o ) = 0 (w) y = w sin( 90 o ) = N (F V ) x = F V cos(90 o ) = 0 (F V ) y = F V sin(90 o ) = F V (F H ) x = F H cos(0 o ) = F H
24 (F H ) y = F H sin(0 o ) = 0 (T) x = T cos(145 o ) (T) y = T sin(145 o ) F x = 0 F H + T cos(145 o ) = 0 F y = N + F V + T sin(145 o ) = 0 w =  (245 N)(1.70 m) = Nm T = + (T)(1.350 m) sin(35 o ) FH = 0 FV = 0 = Nm + (T)(1.350 m) sin(35 o ) = 0 T = 538 N F H = 441 N F V = N 9. Neptune is an average distance of 4.50 x 10 9 km from the Sun. Estimate the length of the Neptunian year given that the Earth is 1.50 x 10 8 km from the Sun on the average. T N 2 = (4 2 /GM S )R N 3 T E 2 = (4 2 /GM S )R E 3 T N 2 /T E 2 = R N 3 /R E 3 T N /T E = (R N 3 /R E 3 ) 1/2 T N /T E = ((4.50 x 10 9 ) 3 /(1.50 x 10 8 ) 3 ) 1/2 = 164 T N = 164 years What is the Neptune s orbital speed?
25 V = 2R/T V = 2(4.50 x m)/(164x365x24x3600 s) = 5467 m/s 10. A massspring system undergoes SHM in horizontal direction. If the mass is kg, the spring constant is 12.0 N/m, and the amplitude is 15.0 cm, a. What would be the maximum speed of the mass? V max = {(k/m) A 2 } ½ V max = {[(12.0 N/m)/(0.250 kg)](0.150 m) 2 } ½ = 1.03 m/s b. Where would it x= 0 c. What would be the speed at a halfamplitude position? V = {(k/m)(a 2 x 2 )} ½ V = {[(12.0 N/m)/(0.250 kg)][(0.150 m) 2  ( m) 2 ]} ½ = m/s d. How many oscillations per second does the system go through? f = (1/2 (k/m) ½ = (1/2 {(12.0 N/m)/(0.250 kg)} ½ = 1.10 Hz e. What would be the magnitude of the mass acceleration at the halfamplitude position? a = F/m F = k x a = k x/m a = (12.0 N/m)( m)/(0.250 kg) = 3.60 m/s 2 TIME OF COMPLETION NAME SOLUTION
26 DEPARTMENT OF NATURAL SCIENCES PHYS 1111, Exam 3 Section 1 Version 1 April 26, 2006 Total Weight: 100 points 1. Check your examination for completeness prior to starting. There are a total of nine (9) problems on seven (7) pages. 2. Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor). 3. You will have 75 minutes to complete the examination. 4. The total weight of the examination is 100 points. 5. There are five (5) multiple choice and four calculation problems. Work all multiple choice problems and 4 calculation problems. Show all work; partial credit will be given for correct work shown. 6. If you have any questions during the examination, see your instructor who will be located in the classroom. 7. Start: 10:30 a.m. Stop: 11:45 a.m PROBLEM POINTS CREDIT TOTAL 100 PERCENTAGE CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR
27 PARTIAL CREDIT. 1. An ice skater performs a pirouette (a fast spin) by pulling in his outstretched arms close to his body. What happens to his moment of inertia about the axis of rotation? a. It does not change. b. It increases. c. It decreases. d. It changes, but it is impossible to tell which way. 2. A wheel with moment of inertia 3.00 kg m 2 has a net torque of 3.50 N m applied to it. What angular acceleration does it experience? a rad/s 2 b rad/s 2 c rad/s 2 d rad/s 2 3. A fan has blades m long rotating at 20.0 rpm. What is the centripetal acceleration of a point on the outer tip of a blade? a m/s 2. b m/s 2. c m/s 2. d m/s According to Kepler s first law, Halley s comet circles the sun in an elliptical path with the Sun at one focus of the ellipse. What is at the other focus of the ellipse? a. Nothing. b. The Earth.
28 c. The comet itself passes through the other focus. d. The tail of the comet stays at the other ellipse. 5. The quantity moment of inertia is equivalent, in terms of fundamental quantities of mass, length, and time, to: a. ML 2 T 2. b. ML. c. ML 2. d. ML 1 T A 60.0kg painter is on a uniform 25kg scaffold supported from above by ropes. There is a 4.0kg pail of paint to one side, as shown. What are the magnitudes of the tension forces exerted by the ropes on the scaffold? (20) T Lx = T L cos (90 o ) = 0 T Ly = T L sin (90 o ) = T L T Rx = T R cos (90 o ) = 0 T Ry = T R sin (90 o ) = T R w px = w p cos (90 o ) = 0
29 w py = w p sin (90 o ) =  m p g =  (60.0 kg)(9.81 m/s 2 ) = N w sx = w s cos (90 o ) = 0 w sy = w s sin (90 o ) =  m s g =  (25.0 kg)(9.81 m/s 2 ) = N w pail x = w pail cos (90 o ) = 0 w pail y = w pail sin (90 o ) =  m pail g =  (4.00 kg)(9.81 m/s 2 ) = N F x = 0 => 0 = 0 F y = 0 => T R + T L 589 N 245 N 39.2 N = 0 T R + T L = 873 N Assuming that the axis of rotation is at the point where the left cable is attached to the scaffold: TL = 0 TR = T R (4.00 m) wp =  (589 N) (2.00 m) = Nm sp =  (245 N) (2.00 m) = Nm wp =  (39.2 N) (1.00 m) = Nm = 0 T R (4.00 m) Nm 490 Nm 39.2 Nm = 0 T R = 427 N T L = 873 N 417 N = 446 N 7. A potter s wheel is rotating around a vertical axis through its center at an angular velocity of 1.50 rad/s. The wheel can be considered a uniform disk of mass 5.00 kg and diameter m. The potter then throws a 1.00kg chunk of clay onto the rim of the rotating wheel. What is the angular velocity of the wheel after the clay sticks to it?
30 I i = ½ m w R 2 = ½ (5.00 kg)(0.400 m) 2 = kg m 2 I f = ½ m w R 2 + m c R 2 = ½ (5.00 kg)(0.400 m) 2 + (1.00 kg)(0.400 m) 2 = kg m 2 I i i = I f f f = (I i /I f ) i = 1.07 rev/s A satellite of mass 5500 kg orbits the Earth mass kg and has a period of 6200 s. a. Find the magnitude of the Earth s gravitational force on the satellite. Hint: find answer to b first and use it to calculate the gravitational force. F = G m 1 m 2 /r 2 F = G M s M E /R 2 = (6.673 x Nm 2 /kg 2 ) (5500 kg) (6.00 x kg )/(7.26 x 10 6 m) 2 = 41,779 N b. What is the orbital radius of the satellite? T 2 = (4 2 /GM E )R 3 T 2 GM E = 4 2 R 3 R 3 = T 2 GM E /(4 2 ) R = (T 2 GM E /(4 2 )) 1/3 R = ((6200 s) 2 (6.673 x Nm 2 /kg 2 )(6.00 x kg) /(4 2 )) 1/3 = 7.26 x 10 6 m 9. A teenager pushes tangentially on a small handdriven merrygoround and is able to accelerate it from rest to an angular velocity of 1.75 rad/s in 10.0 s. Assume the merrygoround is a uniform disk of radius 2.50 m and has a mass of 760 kg, and two children (each with a mass of 25.0 kg) sit opposite each other on the edge. a. What angular acceleration is required to reach 1.75 rad/s velocity in 10.0 s? = 0 + t
31 = (  0 ) /t = (1.75 rad/s) / (10.0 s) = rad/s 2 c. What is the moment of inertia of the merrygoround with two children on it? I = ½ M R 2 + m R 2 + m R 2 = ½ (760 kg)(2.50 m) (25.0 kg)(2.50 m) 2 = 2688 kg m 2 b. Calculate the torque required to produce the acceleration, neglecting frictional torque. = I = (2688 kgm 2 )(0.175 rad/s 2 ) = 470 Nm = F R Bonus (5 points): What is the magnitude of the tangential force required to produce this torque? F = /R = (470 Nm)/(2.50 m) = 188 N TIME OF COMPLETION NAME SOLUTION DEPARTMENT OF NATURAL SCIENCES PHYS 1111, Exam 3 Section 1 Version 1 April 25, 2007 Total Weight: 100 points 1. Check your examination for completeness prior to starting. There are a total of ten (10) problems on seven (7) pages. 2. Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor). 3. You will have 75 minutes to complete the examination. 4. The total weight of the examination is 100 points. 5. There are six (6) multiple choice and four calculation problems. Work all calculation
32 problems and 5 (five) multiple choice. Show all work; partial credit will be given for correct work shown. 6. If you have any questions during the examination, see your instructor who will be located in the classroom. 7. Start: 6:00 p.m. Stop: 7:15 p.m PROBLEM POINTS CREDIT TOTAL 100 PERCENTAGE CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT. 1. Two children are riding on a merrygoround. Child A is at greater distance from the axis of rotation than child B. Which child has the larger angular displacement? a. Child A. b. Child B. c. Both have the same angular displacement. d. Cannot be determined based on the given information. 2. What is the quantity used to measure an object s resistance to changes in rotational motion?
33 a. Mass. b. Moment of inertia. c. Angular velocity. d. Angular acceleration. 3. Two forces are applied to a doorknob, perpendicular to the door. The first force is twice as large as the second force. The ratio of the torque of the first force to the torque of the second is: a. 1/4. b. 1/2. c. 2. d When you ride a bicycle, what is the direction of the angular velocity of the wheels? a. To your left. b. To your right. c. Forward. d. Backward. 5. You sit on a freely rotating stool (no friction). When you extend your arms, a. Your moment of inertia decreases and your angular velocity increases. b. Your moment of inertia increases and your angular velocity increases.
34 c. Your moment of inertia decreases and your angular velocity decreases. d. Your moment of inertia increases and your angular velocity decreases. 6. Two identical spheres, each of mass M and radius R just touch each other. What is the magnitude of the gravitational force that they exert on each other? a. (GM 2 )/(R 2 ). b. (GM 2 )/(2R 2 ). c. (GM 2 )/(4R 2 ). d. (2GM 2 )/(R 2 ). 7. Shekema and Rodney are carrying a 40.0 kg block on a 2.00 m board as shown below. The mass of the board is 5.00 kg. Shekema insists that she is very strong and could have carried the block all by herself, so they place the block m away from her and 1.50 m away from Rodney. Find the forces (in Newtons) exerted by each to carry the block. N Rx = N R cos (90 o ) = 0
35 N Ry = N R sin (90 o ) = N R N Sx = N S cos (90 o ) = 0 N Sy = N S sin (90 o ) = N S W board x = w board cos (90 o ) = 0 W board y = w board sin (90 o ) =  m board g =  (5.00 kg)(9.81 m/s 2 ) = N W block x = w block cos (90 o ) = 0 W block y = w block sin (90 o ) =  m block g =  (40.0 kg)(9.81 m/s 2 ) = N F x = 0 => 0 = 0 F y = 0 => N R + N S 589 N 49.1 N 392 N = 0 N R + N S = 442 N Assuming that the axis of rotation is at the point where Rodney is holding the board: NR = 0 NS = N S (2.00 m) w board =  (49.1 N) (1.00 m) = Nm w block =  (392 N) (1.50 m) = Nm = 0 N S (2.00 m) 49.1 Nm 588 Nm = 0 N S = 319 N N R = 442 N 319 N = 123 N 8. A compact disc rotates from rest to 500 rev/min in 5.50 s. a. What is its angular acceleration, assuming that it is constant? 0 = = (500 rev/min)(1 min/60.0 s)(2 rad/1 rev) = 52.4 rad/s
36 = 0 + t = (  0 ) /t = (52.4 rad/s) / (5.50 s) = 9.52 rad/s 2 b. How many revolutions does it make in 5.50 s? = t +t 2 = ½ (52.4 rad/s 2 )(5.50 s) 2 = 793 rad = 126 rev c. At the moment the disk reaches the angular velocity of 500 rev/min, what is the tangential velocity of a point on the rim of the disc 6.00 cm from the center of rotation? V t = R = (52.3 rad/s)( m) = 3.14 m/s 9. Penny goes bowling with her friends. She notices that a bowling ball of radius 11.0 cm and mass m = 7.20 kg is rolling without slipping on a horizontal ball return at 2.00 m/s. It then rolls without slipping up a hill to a height h before momentarily coming to rest. Find h. E i = E f m g y i + ½ m V 2 i + ½ I i = m g y f + ½ m V f + ½ I f y i = 0 V i = 2.00 m/s i = V i /R y f = h V f = 0 f = 0 ½ m V i 2 + ½ I i 2 = m g y f m V 2 i + I 2 i = 2 m g y f
37 m V 2 i + (2/5 m R 2 )(V i /R) 2 = 2 m g y f V i 2 + (2/5) V i 2 = 2 g y f (7/10) V i 2 = g y f y f = m 10. Amanda, David, and Rotasha play with two children who want a ride on a merrygoround. The merrygoround is a solid disc that rotates freely (without friction) around its center and has a moment of inertia of 500 kgm 2 relative to this axis. The children have the same masses of 25.0 kg and sit at distances of 1.00 m and 2.00 m away from the center of merrygoround. The radius of marrygoround is 3.00 m. a. What is the moment of inertia of the merrygoround with the children on it relative to the center of rotation? I = 500 kgm 2 +(25.0 kg)(1.00 m) 2 + (25.0 kg)(2.00 m) 2 = 625 kgm 2 b. Amanda, David, and Rotasha pull the merrygoround with tangential forces as shown in the figure. (The magnitudes are: A = 300 N, D = 150 N, R = 200 N) What is the net torque that they exert on the merrygoround? = (300 N)(3.00 m) + (200 N)(3.00 m) (150 N)(3.00 m) = 1050 Nm c. What is the angular acceleration of the merrygoround? /I = (1050 Nm)/(625 kgm 2 ) = 1.68 rad/s 2
38 TIME OF COMPLETION NAME SOLUTION DEPARTMENT OF NATURAL SCIENCES PHYS 1111, Exam 3 Section 1 Version 1 April 28, 2008 Total Weight: 100 points 1. Check your examination for completeness prior to starting. There are a total of ten (10) problems on seven (7) pages. 2. Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor). 3. You will have 75 minutes to complete the examination. 4. The total weight of the examination is 100 points. 5. There are six (6) multiple choice and four calculation problems. Work all calculation problems and 5 (five) multiple choice. Show all work; partial credit will be given for correct work shown. 6. If you have any questions during the examination, see your instructor who will be located in the classroom. 7. Start: 6:00 p.m. Stop: 7:15 p.m PROBLEM POINTS CREDIT TOTAL 100
39 PERCENTAGE CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT. 1. A particle is traveling in a horizontal circle at a constant speed. One can conclude that is constant. (A) Velocity. (B) Acceleration. (C) Net force. (D) All of the above. (E) None of the above. 2. The dimension of torque is the same as (A) Impulse. (B) Momentum. (C) Force. (D) Energy. (E) None of the above. 3. Two forces are applied to a doorknob, perpendicular to the door. The first force is twice as large as the second force. The ratio of the torque of the first force to the torque of the second is: (A) 1/4. (B) 1/2. (C) 2. (D) 4. (E) 8.
40 4. Two children are riding on a merrygoround. Child A is at greater distance from the axis of rotation than child B. Which child has the larger linear displacement? (A) Child A. (B) Child B. (C) Both have the same zero linear displacement. (D) Both have the same nonzero linear displacement. (E) Cannot be determined based on the given information. 5. Two children are riding on a merrygoround. Child A is at greater distance from the axis of rotation than child B. It is true statement that (A) Child A has greater angular velocity than child B. (B) Child B has greater angular velocity than child B.. (C) Both have the same zero angular velocity. (D) Both have the same nonzero angular velocity. (E) Cannot be determined based on the given information. 6. You sit on a rotating stool which rotates clockwise (seen from above). What is the direction of your angular acceleration if the stool is slowing down? (A) Vertically upward. (B) Vertically downward. (C) Horizontally and to your left. (D) Horizontally and to your right. (E) Your angular acceleration has no direction.
41 7. A wheel starts to rotate around its center with a constant angular acceleration of 2.60 rad/s 2. The wheel starts from rest and rotates for 6.00 s. At the end of that time a. What is its angular velocity? t 2.60 rad/s 2 s) = 15.6 rad/s b. Through what angle has the wheel turned? ) ) = ) ) = (15.6 rad/s) 2 /(2 x 2.60 rad/s 2 ) = 46.8 rad c. How many revolutions has it made? 46.8 rad/(2 rad/rev) = 7.45 rev d. What is the magnitude of tangential velocity and tangential acceleration of a point m from the axis of rotation? V t = r = (15.6 rad/s)(0.300 m) = 4.68 m/s a t = r = (2.60 rad/s 2 )(0.300 m) = m/s 2 8. To determine the location of his center of mass, a physics student lies on a lightweight plank supported by two scales 2.50 m apart. The left scale reads 290 N, and the right scale reads 122 N. (Hint: the readings correspond to the normal forces exerted on the plank.)
42 a. Draw a freebody diagram. b. Find the student s mass. N Lx = N L cos (90.0 o ) = 0 N Ly = N L sin (90.0 o ) = N L = 290 N N Rx = N R cos (90.0 o ) = 0 N Ry = N R sin (90.0 o ) = N R = 122 N w x = m g cos (90.0 o ) = 0
43 w y = m g sin (90.0 o ) =  m g F x = 0 F y = N N  mg = 0 mg = 412 N m = 42.0 kg c. Find the distance from the student s head to his center of mass. NL = 0 NR = (122 N) (2.50 m) = 305 Nm w = w (x) = (412 N) (x) = Nm (412 N) (x) = 0 x = m 9. A merrygoround of radius 2.00 m is rotating about its center making one revolution every 5.00 seconds. A child of mass 25.0 kg stands at the rim. a. What is the magnitude of the angular velocity of merrygoround in rad/s? = (2 rad)/(5.00 s) = 1.26 rad/s b. What is the magnitude of centripetal acceleration of the child? a cp = r = (1.26 rad/s) 2 (2.00 m) = 3.18 m/s 2
44 c. The merrygoround is very slippery (frictionless) and the child stays on it by holding onto a rope attached to the pivot at the center of merrygoround. What has to be the magnitude of tension in the rope for the child to stay on merrygoround? T = m a cp = (25.0 kg)(3.19 m/s 2 ) = 79.4 N d. The rope suddenly breaks. What is the motion of the child immediately after it? The child continues to move in the direction which was tangential to the circle the moment the rope broke, until some other force changes his velocity. 10. A square metal plate m on each side is pivoted about an axis through point O at its center and perpendicular to the plate. Calculate the net torque about this axis due to the three forces shown in the figure if the magnitudes of the forces are F 1 = 18.0 N, F 2 = 26.0 N, and F 3 = 14.0 N. The plate and all forces are in the plane of the page. r 1 = r 2 = r 3 = (( m) 2 + ( m) 2 ) 1/2 /2 = m 1 = F 1 r 1 sin( 1 ) = (18.0 N)(0.127 m) sin (45.0 o ) = 1.62 Nm 2 =  F 2 r 2 sin( 2 ) =  (26.0 N)(0.127 m) sin (45.0 o ) = Nm 3 = F 3 r 3 sin( 3 ) = (14.0 N)(0.127 m) sin (90.0 o ) = 1.78 Nm
45 = 1.62 Nm Nm Nm = 1.07 Nm TIME OF COMPLETION NAME SOLUTION DEPARTMENT OF NATURAL SCIENCES PHYS 1111, Exam 3 Section 1 Version 1 August 1, 2005 Total Weight: 100 points 1. Check your examination for completeness prior to starting. There are a total of ten (10) problems on eight (8) pages. 2. Authorized references include your calculator with calculator handbook, and the Reference Data Pamphlet (provided by your instructor). 3. You will have 80 minutes to complete the examination. 4. The total weight of the examination is 100 points. 5. There are six (6) multiple choice and four calculation problems. Work all problems. Show all work; partial credit will be given for correct work shown. 6. If you have any questions during the examination, see your instructor who will be located in the classroom. 7. Start: 11:30 a.m. Stop: 12:50 p.m PROBLEM POINTS CREDIT TOTAL 100
46 PERCENTAGE CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR PARTIAL CREDIT. 2. Which one of the following quantities is zero when an object in simple harmonic motion is at its maximum displacement? (5) e. Potential energy. f. Acceleration. g. Kinetic energy. h. Frequency. 2. Two equal forces are applied to a door. The first force is applied at the midpoint of the door; the second force is applied at the doorknob. Both forces are applied perpendicular to the door. Which force exerts the greater torque? (5) a. The first at the midpoint. b. The second at the doorknob. c. Both exert equal nonzero torques. d. Both exert zero torques. 3. The quantity angular momentum (in terms of the fundamental quantities of mass, length, and time) is equivalent to: a. MLT 2. (5) b. ML 2 T 1. c. ML 2 T 3. d. ML 3 T. 4. Two moons orbit a planet in nearly circular orbits. Moon A has orbital radius r, and moon B
47 has orbital radius 4r. Moon A takes 20 days to complete one orbit. How long does it take moon B to complete an orbit? a. 20 days. (5) b. 80 days. c. 160 days. d. 320 days. 5. The moment of inertia of a spinning body about its spin axis depends on its: a. Angular speed, shape, and mass. (5) b. Angular acceleration, mass, and axis position. c. Mass, shape, axis position, and size. d. Mass, size, shape, and speed. 6. A boy and a girl are riding on a merrygoround which is turning at a constant rate. The boy is near the outer edge, and the girl is closer to the center. Who has the greater linear speed? a. The boy. (5) b. The girl. c. Both have the same nonzero translational velocity. d. Both have zero translational velocity. 7. A 64.0kg person stands on a weightless diving board supported by two pillars, one at the enof the board, the other 1.10m away. The pillar at the end of the board exerts a downward force of 828 N. The board has a length of PHYS 1111 Exam 3, Version 1 Fall
48 a. How far from that pillar is the person standing? N 1x = N 1 cos (90 o ) = 0 N 1y = N 1 sin (90 o ) =  N 1 = 828 N N 2x = N 2 cos (90 o ) = 0 N 2y = N 2 sin (90 o ) = N 2 w x = w cos (90 o ) = 0 w y = w sin (90 o ) =  w =  m g =  (64.0 kg)(9.80 m/s 2 ) = N F x = 0 => 0 = 0 F y = 0 => (828 N) + N 2 (627 N) = 0 N 2 = 1455 N N2 = N 2 (1.10 m) = (1455 N)(1.10 m) = 1601 Nm N1 = 0 w =  (627 N) (x) = 0  (627 N) x N m = 0 x = 2.55 m PHYS 1111 Exam 3, Version 1 Fall
49 b. Find the force exerted by the second pillar. N 2 = 1455 N 8. A potter s wheel is rotating around a vertical axis through its center at a frequency of 1.5 rev s. The wheel can be considered a uniform disk of mass 5.00 kg and diameter m. The potter then throws a 3.10kg chunk of clay, approximately shaped as a flat disk of radius 8.00 cm, onto the center of the rotating wheel. a. What is the frequency of the wheel after the clay sticks to it? I i = ½ m w R 2 = ½ (5.00 kg)(0.200 m) 2 = kg m 2 I f = ½ m w R 2 + ½ m c R 2 = ½ (5.00 kg)(0.200 m) 2 + ½ (3.10 kg)( m) 2 = kg m 2 I i i = I f f f = (I i /I f ) i = 1.36 rev/s b. (Bonus) If the wheel rotates in clockwise direction as viewed from above, what is the direction of the wheel s angular momentum? 9. A kg mass resting on a horizontal frictionless surface is connected to a fixed spring with PHYS 1111 Exam 3, Version 1 Fall
50 spring constant of 80.0 N/m. The mass is displaced 16.0 cm from its equilibrium position and released. a. What is the amplitude of oscillations? A = m b. What is the maximum speed of the mass? V max = (k/m) 1/2 A = ((80.0 N/m)/(0.500 kg)) 1/2 (0.160 m) = 2.02 m/s c. What is the total mechanical energy of the system? TE = ½ m V max 2 = ½ (0.500 kg) (2.02 m/s) 2 = 1.02 J d. What is the frequency of oscillations? f = (1/(2)(k/m) 1/2 = 2.01 Hz e. What is the period of oscillations? T = 1/f = s 10. A uniform disk of mass 1.00 kg and radius 10.0 m has four symmetrically placed masses, each of mass kg fastened at positions 5.00 m from the center. The disk is free to rotate around its center. d. What is the moment of inertia of the disk relative to its center? I = ½ m disk R m r 2 = ½ (1.00 kg)(10.0 m) (0.250 kg)(5.00 m) 2 = 75.0 kg m 2 PHYS 1111 Exam 3, Version 1 Fall
51 e. Two tangential forces are applied to the disk as shown in the figure. What is the net torque acting on the disk? 1 = F 1 d 1 = (40.0N)(10.0m) = 400 Nm 2 =  F 2 d 2 =  (20.0N)(5.00m) = 100 Nm = 400 Nm 100 Nm = 300 Nm f. What is the angular acceleration of the disk? = I = / I = (300 Nm)/ (75.0 kgm 2 ) = 4.00 rad/s 2 g. Assuming that initially the system was at rest, what is the kinetic energy of the disk 10.0 s after the start of the motion? KE = ½ I 2 = ½ I ( t) 2 = ½ (75.0 kgm 2 )((4.00 rad/s 2 )(10.0 s)) 2 = 60,000 J PHYS 1111 Exam 3, Version 1 Fall
52 PHYS 1111 Exam 3, Version 1 Fall
53 PHYS 1111 Exam 3, Version 1 Fall
54 9. After you pick up a spare, your bowling ball rolls without slipping back toward the ball rack with a linear speed of 2.85 m/s. To reach the rack, the ball rolls up the ramp that gives the ball a m vertical rise. What is the speed of the ball when it reaches the top of the ramp? E i = m g y i + ½ mv i + ½ I i = 0 + ½ mv i + ½ I i = ½ mv i + ½ (2/5mR 2 2 ) i 2 2 mv i + ½ (2/5)mV i = ½ E f = m g y f + ½ mv f 2 + ½ I f 2 = m g y f + ½ mv f 2 + ½ (2/5mR 2 ) f 2 = m g y f + ½ mv f 2 + ½ (2/5)mV f 2 E i = E i V i + (2/5)V i = 2 g y f + V f + (2/5)V f 2 2 (7/5)V i = 2 g y f + (7/5)V f 2 2 V i = (5/7)(2 g y f )+ V f 2 2 V f =  (5/7)(2 g y f )+ V i V f = ( (5/7)(2 g y f )+ V 2 i ) 1/2 V f = ( (5/7)(2 (9.81 m/s 2 ) (0.530 m))+ (2.85 m/s) 2 ) 1/2 V f = m/s PHYS 1111 Exam 3, Version 1 Fall
55 Bonus (5 points): What is the direction of the ball s angular momentum? PHYS 1111 Exam 3, Version 1 Fall
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