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larsen (sl44887) – rotational dynamics – fox – (c12181) 1 This print-out should have 49 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0points A rod can pivot at one end and is free to rotate without friction about a vertical axis, as shown. A force vector F is applied at the other end, at an angle θ to the rod. L m F θ If vector F were to be applied perpendicular to the rod, at what distance d from the axis of rotation should it be applied in order to produce the same torque vector τ ? 1. d = L cos θ 2. d = L tan θ 3. d = L 4. d = L sin θ correct 5. d = √ 2 L Explanation: The force generates a torque of τ = F L sin θ , so the distance is L sin θ . 002 10.0points A system of two wheels fixed to each other is free to rotate about a frictionless axis through the common center of the wheels and per- pendicular to the page. Four forces are ex- erted tangentially to the rims of the wheels, as shown below. F 2 F F F 2 R 3 R What is the magnitude of the net torque on the system about the axis? 1. τ = 14 F R 2. τ = 2 F R correct 3. τ = F R 4. τ = 5 F R 5. τ = 0 Explanation: The three forces F apply counter-clockwise torques while the other force 2 F applies a clockwise torque, so τ = summationdisplay F i R i = ( − 2 F ) (3 R ) + F (3 R ) + F (3 R ) + F (2 R ) = 2 F R . 003 10.0points A non-uniform beam with a mass 3 M and length L is in stable equilibrium when placed at the edge of a table with half its length sticking off the table edge. But when a point- like mass M is placed at the far end of the part of the beam off the table, the beam is brought to unstable equilibrium; i.e. , it is right on the verge of tipping off the table. L 2 M 3 M At what distance to the left of the table edge is the center of mass of the beam located?

larsen (sl44887) – rotational dynamics – fox – (c12181) 2 1. 2 L 5 2. 3 L 4 3. L 7 4. 4 L 5 5. 5 L 6 6. L 2 7. L 6 correct 8. L 4 9. L 5 10. 2 L 3 Explanation: Let the fulcrum be at the edge of the table. If the center of mass of the beam is at distance x from the edge, then summationdisplay τ ccw = summationdisplay τ cw (3 M ) x = M · L 2 x = L 6 . 004 10.0points A friend incorrectly says that a body cannot be rotating when the net torque acting on it is zero. What is the correct statement? 1. The original statement made by the friend is actually correct. 2. A body can have an angular velocity only when a non-zero net torque is acting on it. 3. Once a body starts rotating the net torque is zero. 4. A body’s angular velocity cannot change if the net torque acting on it is zero and the moment of inertia does not change. correct Explanation: The rate and direction of rotation of a body cannot change when a zero net torque acts on it. Once started rotating, a body will continue rotating even when no torque acts on it. Again, emphasize change . 005 10.0points When you pedal a bicycle with a constant downward force, is maximum torque pro- duced when the pedal sprocket arms are in the horizontal position, in the vertical posi- tion, or in the diagonal position? 1. All torques are the same. 2. vertical position 3. diagonal position 4. horizontal position correct Explanation: The maximum lever arm is when the pedal sprocket arm is horizontal (lever = radius), with the minimum at the vertical orientation (lever arm = 0). 006 10.0points A spool (similar to a yo-yo) is pulled in three ways as shown. There is sufficient friction for rotation. Note that in each case, the cord is wrapped tightly around the inner part of the spool, and does not slip. The tension force is of course parallel to the cord, and along the dashed line sketched. a b c In what direction will each spool move (in the order spool a, spool b, spool c)? 1. right; right left

larsen (sl44887) – rotational dynamics – fox – (c12181) 3 2. right; left; left 3. left; right; right 4. right; right; right correct 5. right; left; right 6. None of these Explanation: In (a) there is a clockwise torque about the point of contact with the table, so the spool rolls to the right. In (b) the line of action extends through the point of table contact, yielding no lever arm and therefore no torque; with a force component to the right, the spool slides to the right without rolling. In (c) the torque produces clockwise rota- tion so the spool rolls to the right. 007 10.0points A bucket filled with water has a mass of 31 kg and is hanging from a rope that is wound around a stationary cylinder of radius 0 . 042 m. If the cylinder does not rotate and the bucket hangs straight down, what is the mag- nitude of the torque produced by the bucket around the center of the cylinder? The accel- eration of gravity is 9 . 81 m / s 2 . Correct answer: 12 . 7726 N · m. Explanation: Let : m = 31 kg , r = 0 . 042 m , and g = 9 . 81 m / s 2 . Since θ = 90 ◦ , sin θ = 1 and τ = F d = ( m g ) r = (31 kg)(9 . 81 m / s 2 )(0 . 042 m) = 12 . 7726 N · m . 008(part1of2)10.0points The arm of a crane at a construction site is 16.0 m long, and it makes an angle of 14 . 5 ◦ with the horizontal. Assume that the max- imum load the crane can handle is limited by the amount of torque the load produces around the base of the arm. What maximum torque can the crane with- stand if the maximum load the crane can handle is 620 N? Correct answer: 9604 . 02 N · m. Explanation: Let : d = 16 . 0 m and W max = 620 N . θ = 90 . 0 ◦ − 14 . 5 ◦ = 75 . 5 ◦ , so τ max = F d sin θ = W max d sin θ = (620 N)(16 m)(sin 75 . 5 ◦ ) = 9604 . 02 N · m . 009(part2of2)10.0points What is the maximum load for this crane at an angle of 30 . 1 ◦ with the horizontal? Correct answer: 693 . 811 N. Explanation: Let : θ = 90 . 0 ◦ − 30 . 1 ◦ = 59 . 9 ◦ We have the same maximum torque, so W = τ max d sin θ = 9604 . 02 N · m (16 m) sin 59 . 9 ◦ = 693 . 811 N . 010 10.0points A wooden bucket filled with water has a mass

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larsen (sl44887) – rotational dynamics – fox – (c12181) 4 of 70 kg and is attached to a rope that is wound around a cylinder with a radius of 0.058 m. A crank with a turning radius of 0.41 m is attached to the end of the cylinder. What minimum force directed perpendicu- larly to the crank handle is required to raise the bucket? The acceleration of gravity is 9 . 81 m / s 2 . Correct answer: 97 . 1429 N. Explanation: Let : m = 70 kg , r = 0 . 058 m , d = 0 . 41 m and g = 9 . 81 m / s 2 . Since θ = 90 ◦ , sin θ = 1 and τ = F d sin θ = F d τ min = ( m g ) r F min d = m g r F min = m g r d = (70 kg)(9 . 81 m / s 2 )(0 . 058 m) 0 . 41 m = 97 . 1429 N . 011(part1of2)10.0points The hour and minute hands of the clock in the famous Parliament Clock Tower in Lon- don are 1 . 7 m and 2 . 8 m long and have masses of 85 kg and 63 kg , respectively. 3 2 1 12 11 10 9 8 7 6 5 4 Calculate the magnitude of the torque around the center of the clock due to the weight of these hands indicating 2 h and 40 min; i.e. , 2:40 o’clock. Assume the clock hands can be modeled as uniform thin rods and at 3:00 o’clock, the hour hand is precisely 90 ◦ from the vertical. The acceleration of gravity is 9 . 81 m / s 2 . Correct answer: 51 . 3169 N · m. Explanation: Let : ℓ h = 0 . 85 m , ℓ m = 1 . 4 m , m h = 85 kg , and m min = 63 kg . τ = W ℓ sin θ = m g l sin θ . Consider the total mass of each hand to be concentrated at the midpoint of that hand with positive torques directed counter- clockwise. θ m = (40 min) 360 ◦ (60 min) − 180 ◦ = 60 ◦ , so τ m = m m g ℓ m sin θ m = (63 kg) (9 . 81 m / s 2 ) (1 . 4 m) sin 60 ◦ = 749 . 322 N · m . θ h = bracketleftbigg 2 h + (40 min) 1 h 60 min bracketrightbigg 360 ◦ 12 h − 180 ◦ = − 100 ◦ , so τ h = m h g ℓ h sin θ h = (85 kg) (9 . 81 m / s 2 ) (0 . 85 m) × sin( − 100 ◦ ) = − 698 . 005 N · m , and τ = τ hr + τ min = − 698 . 005 N · m + (749 . 322 N · m) = 51 . 3169 N · m bardbl τ bardbl = 51 . 3169 N · m .

larsen (sl44887) – rotational dynamics – fox – (c12181) 5 012(part2of2)10.0points The torque is 1. Cannot be determined from given infor- mation. 2. clockwise. 3. counter-clockwise. correct Explanation: Look at the clock and the sign of the torque in Part 1. keywords: 013 10.0points A circular-shaped object of mass 14 kg has an inner radius of 8 cm and an outer radius of 25 cm. Three forces (acting perpendicular to the axis of rotation) of magnitudes 13 N, 22 N, and 13 N act on the object, as shown. The force of magnitude 22 N acts 28 ◦ below the horizontal. 13 N 13 N 22 N 28 ◦ ω Find the magnitude of the net torque on the wheel about the axle through the center of the object. Correct answer: 4 . 74 N · m. Explanation: Let : a = 8 cm = 0 . 08 m , b = 25 cm = 0 . 25 m , F 1 = 13 N , F 2 = 22 N , F 3 = 13 N , and θ = 28 ◦ . F 1 F 3 F 2 θ ω The total torque is τ = a F 2 − b F 1 − b F 3 = (0 . 08 m) (22 N) − (0 . 25 m) (13 N + 13 N) = − 4 . 74 N · m , with a magnitude of 4 . 74 N · m . 014 10.0points 015(part1of2)10.0points To tighten a bolt, you push with a force of 60 N at the end of a wrench handle that is 0 . 23 m from the axis of the bolt. What torque are you exerting? Correct answer: 13 . 8 N · m. Explanation: Let : F = 60 N and d = 0 . 23 m .

larsen (sl44887) – rotational dynamics – fox – (c12181) 6 Torque is force times lever arm, so τ = (60 N) (0 . 23 m) = 13 . 8 N · m . 016(part2of2)10.0points If you move your hand inward to be only 0 . 17 m from the bolt, what force do you have to exert to achieve the same torque? Correct answer: 81 . 1765 N. Explanation: Let : d = 0 . 17 m . F = τ d = 13 . 8 N · m 0 . 17 m = 81 . 1765 N . These answers assume that you are pushing perpendicular to the wrench handle. Other- wise, you would need to exert more force to get the same torque. 017 10.0points A simple pendulum consists of a small object of mass 3 . 5 kg hanging at the end of a 2 . 9 m long light string that is connected to a pivot point. Find the magnitude of the torque (due to the force of gravity) about this pivot point when the string makes a 3 . 50364 ◦ angle with the vertical. The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 6 . 07881 N · m. Explanation: Let : m = 3 . 5 kg , L = 2 . 9 m , and θ = 3 . 50364 ◦ . 2 . 9 m 3 . 5 kg mg 2 . 9 m The torque is τ = F d = m g L sin θ = (3 . 5 kg)(9 . 8 m / s 2 )(2 . 9 m) sin 3 . 50364 ◦ = 6 . 07881 N · m . 018 10.0points A uniform, rigid rod of mass M and length L is pivoted frictionlessly at its upper end. If the rod is dropped from an initially horizontal position, it swings freely through a vertical position, as shown. A B C D In which of the four positions illustrated is the net torque about the pivot on the rod greatest, and in which of the four positions is it the least? 1. Not enough information is provided. 2. The same in all four positions, since the force M g acts with lever arm L 2 in all four positions. 3. Greatest at A ; least (zero) at D correct 4. Greatest at D ; least (zero) at A Explanation: The only forces acting on the rod are the force of the pivot keeping the rod in place, and the gravitational force. The pivot force

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larsen (sl44887) – rotational dynamics – fox – (c12181) 7 has a zero lever arm and as such exerts no net torque. At A, the lever arm is L 2 and is perpendicular to the gravitational force at A , yielding the maximum torque M g L 2 . As the rod falls, the lever arm shortens until at position D it is 0 since the line of force passes through the pivot point. 019 10.0points A horizontal bar with a mass m suspended from one end is held by a cord with tension T , fastened at a distance D from the end of the bar that is attached to a wall with a frictionless pin. m D θ Tension T Pin What is the torque about the pin exerted on the bar by the cord? 1. D T (1 − tan θ ) 2. D T (1 − sin θ ) 3. D T sin θ 4. D T (1 − cos θ ) 5. D T cos θ correct 6. D T tan θ Explanation: The distance from the pin to the cord is D cos θ , so the torque is T × D cos θ = D T cos θ . 020 10.0points Four forces of the same magnitude but differ- ing directions act at and tangent to the rim of a uniform wheel free to spin about its center of mass (CM). Which statement about the wheel is cor- rect? 1. The net force on the wheel is zero but the net torque about the CM is not zero. 2. The net force on the wheel is not zero but the net torque about the CM is zero. correct 3. The net force on the wheel and the net torque on the wheel about the CM are zero. 4. Neither the net force on the wheel, nor the net torque on the wheel about the CM is zero. 5. None of these Explanation: Two forces point to the right and two forces point toward the top of the page, so the net force summationdisplay F is certainly not zero. However the torque exerted by the force at the top of the wheel exactly cancels the torque exerted by the force at the bottom of the wheel, and the torque exerted by the force at the left of the wheel exactly cancels the torque exerted by the force at the right of the wheel, so the net torque summationdisplay vector τ = 0 . 021(part1of3)10.0points An amusement park builds a ride in which the victim is made to spin about a pole with a rocket strapped on his seat. The rider, box and rocket have an initial total mass of 220 kg. Neglect the mass of the rod of length 4 m .

larsen (sl44887) – rotational dynamics – fox – (c12181) 8 290 m / s 57 N 4 m 220 kg What is the moment of inertia of rider, box and rocket about the pole? The acceleration of gravity is 9 . 8 m / s 2 . Treat the rider, box and rocket as a point mass. Correct answer: 3520 kg · m 2 . Explanation: Let : F = 57 N , v t = 290 m / s , M = 220 kg , and ℓ = 4 m . For a point mass, the inertia is I = M ℓ 2 = (220 kg) (4 m) 2 = 3520 kg · m 2 . 022(part2of3)10.0points The rocket develops a thrust of 57 N perpen- dicular to the path of the rider. What is the initial angular acceleration of the rider? Correct answer: 0 . 0647727 rad / s 2 . Explanation: The torque of this force is τ = ℓ F = Iα α = ℓ F I = (4 m) (57 N) 3520 kg · m 2 = 0 . 0647727 rad / s 2 . 023(part3of3)10.0points After what time t is the rider’s velocity equal to 8 m / s? Neglect the change in mass of the rocket. Correct answer: 30 . 8772 s. Explanation: Let : v = 8 m / s . ω = α t for constant angular acceleration, so v = ℓ ω = ℓ α t t = v ℓ α = 8 m / s (4 m) (0 . 0647727 rad / s 2 ) = 30 . 8772 s . 024 10.0points An Atwood machine is constructed using a hoop with spokes of negligible mass. The 2 . 2 kg mass of the pulley is concentrated on its rim, which is a distance 20 . 8 cm from the axle. The mass on the right is 1 . 58 kg and on the left is 1 . 92 kg. 3 . 7 m 2 . 2 kg 20 . 8 cm ω 1 . 92 kg 1 . 58 kg What is the magnitude of the linear acceler- ation of the hanging masses? The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 0 . 584561 m / s 2 . Explanation: Let : M = 2 . 2 kg , R = 20 . 8 cm , m 1 = 1 . 58 kg ,

larsen (sl44887) – rotational dynamics – fox – (c12181) 9 m 2 = 1 . 92 kg , h = 3 . 7 m , and v = ω R . Consider the free body diagrams 1 . 92 kg 1 . 58 kg T 2 T 1 m 2 g m 1 g a a The net acceleration a = r α is in the direc- tion of the heavier mass m 2 . For the mass m 1 , F net = m 1 a = m 1 g − T 1 T 1 = m 1 g − m 1 a and for the mass m 2 , F net = m 2 a = T 2 − m 2 g T 2 = m 2 a + m 2 g . The pulley’s mass is concentrated on the rim, so I = M r 2 , and τ net = summationdisplay τ ccw − summationdisplay τ cw = I α T 1 r − T 2 r = ( m r 2 ) parenleftBig a r parenrightBig = m r a m a = T 1 − T 2 m a = ( m 1 − m 2 ) g − ( m 1 + m 2 ) a m a + ( m 1 + m 2 ) a = ( m 1 − m 2 ) g a = ( m 1 − m 2 ) g m + m 1 + m 2 = (1 . 58 kg − 1 . 92 kg) (9 . 8 m / s 2 ) 2 . 2 kg + 1 . 58 kg + 1 . 92 kg = 0 . 584561 m / s 2 . 025 10.0points An Atwood machine is constructed using a disk of mass 2 . 3 kg and radius 21 . 4 cm. 1 . 4 m 21 . 4 cm 2 . 3 kg ω 1 . 56 kg 1 . 03 kg What is the acceleration of the system? The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 1 . 38877 m / s 2 . Explanation: Let : m = 2 . 3 kg , R = 21 . 4 cm = 0 . 214 m , m 1 = 1 . 03 kg , m 2 = 1 . 56 kg , and h = 1 . 4 m . Consider the free body diagrams 1 . 56 kg 1 . 03 kg T 2 T 1 m 2 g m 1 g a a The net acceleration a = r α is in the di- rection of the heavier mass m 1 . For the mass m 1 , F net = m 1 a = m 1 g − T 1 T 1 = m 1 g − m 1 a and for the mass m 2 , F net = m 2 a = T 2 − m 2 g T 2 = m 2 a + m 2 g .

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larsen (sl44887) – rotational dynamics – fox – (c12181) 10 The pulley is a solid disk, so I = 1 2 m r 2 and τ net = summationdisplay τ ccw − summationdisplay τ cw = I α T 1 r − T 2 r = parenleftbigg 1 2 m r 2 parenrightbigg parenleftBig a r parenrightBig = 1 2 m r a Multiplying by 2 r gives m a = 2 T 1 − 2 T 2 = 2 ( m 1 − m 2 ) g − 2 ( m 1 + m 2 ) a a = 2 ( m 1 − m 2 ) g m + 2 m 1 + 2 m 2 = 2 (1 . 03 kg − 1 . 56 kg) (9 . 8 m / s 2 ) 2 . 3 kg + 2 (1 . 03 kg) + 2 (1 . 56 kg) = 1 . 38877 m / s 2 . 026(part1of6)10.0points A pulley (in the form of a uniform disk) with mass M p and radius R p is attached to the ceiling in a uniform gravitational field g and rotates with no friction about its pivot. Mass M 2 is larger than mass m 1 , and they are connected by a massless inextensible cord. T 1 , T 2 , and T 3 are magnitudes of the tensions. h R M p ω M 2 m 1 T 2 T 1 T 3 What is the relationship between the ten- sion T 1 and m 1 g ? 1. T 1 = m 1 g 2. Not enough information is available. 3. T 1 > m 1 g correct 4. T 1 < m 1 g Explanation: Consider the free body diagrams M 2 m 1 T 2 T 1 M 2 g m 1 g a a The equation of motion for the mass m 1 is T 1 − m 1 g = m 1 a T 1 = m 1 ( g + a ) > m 1 g . (1) 027(part2of6)10.0points What is the relationship between the magni- tudes of the accelerations? 1. a M 2 < a m 1 . 2. a M 2 = a m 1 . correct 3. a M 2 > a m 1 . 4. Not enough information is available. Explanation: The rope does not stretch, so the accelera- tion of both masses must be equal. 028(part3of6)10.0points The center of mass of m 1 + M 2 1. Not enough information is available. 2. does not accelerate. 3. accelerates up. 4. accelerates down. correct Explanation: Let up be positive; then a M 1 = − a and a m 2 = a .

larsen (sl44887) – rotational dynamics – fox – (c12181) 11 ( M 2 + m 1 ) a cm = M 2 a M 2 + m 1 a m 1 a cm = M 2 a M 2 + m 1 a m 1 M 2 + m 1 = − M 2 a + m 1 a M 2 + m 1 < 0 since M 2 > m 1 . 029(part4of6)10.0points What is the relationship between the magni- tudes of T 1 and T 2 ? 1. Not enough information is available. 2. T 2 = T 1 3. T 2 < T 1 4. T 2 > T 1 correct Explanation: α = a r , so the equation of motion for the torque is ( T 2 − T 1 ) r = I α = parenleftbigg 1 2 M p r 2 parenrightbigg a r T 1 − T 2 = 1 2 M p a T 2 = T 1 + 1 2 M p a > T 1 . 030(part5of6)10.0points What is the relationship between the magni- tudes of T 3 , T 2 , and T 1 ? 1. Not enough information is available. 2. T 1 + T 2 = T 3 3. T 1 + T 2 > T 3 4. T 1 + T 2 < T 3 correct Explanation: Using a free body diagram for the pulley alone T 3 = T 1 + T 2 + M p g > T 1 + T 2 . (2) 031(part6of6)10.0points What is the relation between the magnitudes of T 3 , m 1 , M 2 , and M p ? 1. T 3 < ( m 1 + M 2 + M p ) g correct 2. T 3 = ( m 1 + M 2 + M p ) g 3. Not enough information is available. 4. T 3 > ( m 1 + M 2 + M p ) g Explanation: The equation of motion for M 2 is M 2 g − T 2 = M 2 a T 2 = M 2 ( g − a ) (3) From the three equations of motion, T 3 = T 1 + T 2 + M p g = m 1 ( g + a ) + M 2 ( g − a ) + M p g = ( m 1 + M 2 + M p ) g − ( M 2 − m 1 ) a < ( m 1 + M 2 + M p ) g since M 2 > m 1 . 032(part1of3)10.0points A common physics demonstration consists of a ball resting at the end of a board of length ℓ that is elevated at an angle θ with the hor- izontal. A light cup is attached to the board at r c so that it will catch the ball when the support stick is suddenly removed. θ r l c Cup Support stick Hinged end What is the maximum angle at which the ball, placed at the very end, will immediately lag behind the board in falling (after the stick is suddenly removed)? 1. θ max = 27 . 4 ◦

larsen (sl44887) – rotational dynamics – fox – (c12181) 12 2. None of these 3. θ max = 45 . 0 ◦ 4. θ max = 35 . 3 ◦ correct 5. θ max = 33 . 5 ◦ 6. θ max = 37 . 2 ◦ 7. θ max = 53 . 6 ◦ 8. θ max = 43 . 1 ◦ 9. θ max = 30 . 0 ◦ 10. θ max = 41 . 8 ◦ Explanation: I end = 1 12 m ℓ 2 + m parenleftbigg ℓ 2 parenrightbigg 2 = 1 3 m ℓ 2 . The board just starts to move when summationdisplay τ = I α m g ℓ 2 cos θ = parenleftbigg 1 3 m ℓ 2 parenrightbigg α α = 3 g 2 ℓ cos θ , and the tangential acceleration of the end of the board is a t = α ℓ = 3 2 g cos θ with a vertical component of a y = a t cos θ = 3 2 g cos 2 θ . If this is greater than g , the board will pull ahead of the ball in falling, so the condition for the maximal θ is 3 2 g cos 2 θ ≥ g cos 2 θ ≥ 2 3 cos θ ≥ radicalbigg 2 3 θ ≤ 35 . 3 ◦ . 033(part2of3)10.0points How far from the hinged end should the cup be placed if the system is started with this limiting angle and the ball is at the end of the stick of length ℓ ? 1. None of these 2. r c = 2 ℓ 3 cos θ correct 3. r c = 2 ℓ cos θ 3 4. r c = 2 3 ℓ 5. r c = ℓ sin θ 3 6. r c = 2 ℓ 3 sin θ Explanation: For the cup to be underneath the release point of the ball, r c = ℓ cos θ = ℓ cos 2 θ cos θ = 2 ℓ 3 cos θ . 034(part3of3)10.0points If a ball is at the very end of the stick of length 0 . 7 m at this critical angle, what is r c ? Correct answer: 0 . 571548 m. Explanation: r c = 2 ℓ 3 cos θ = 2 ℓ 3 radicalbigg 3 2 = radicalbigg 2 3 ℓ = radicalbigg 2 3 (0 . 7 m) = 0 . 571548 m . 035(part1of2)10.0points A block of mass 1 kg and one of mass 9 kg are connected by a massless string over a pulley that is in the shape of a disk having a radius of 0 . 17 m, and a mass of 9 kg. In addition, the

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larsen (sl44887) – rotational dynamics – fox – (c12181) 13 blocks are allowed to move on a fixed block- wedge of angle 46 ◦ , as shown. The coefficient of kinetic friction is 0 . 15 for both blocks. 1 kg 9 kg 46 ◦ 0 . 17 m 9 kg What is the acceleration of the two blocks? The acceleration of gravity is 9 . 8 m / s 2 . As- sume the positive direction is to the right. Correct answer: 3 . 64038 m / s 2 . Explanation: Let : m 1 = 1 kg , m 2 = 9 kg , M = 9 kg , and R = 0 . 17 m . m 1 m 2 θ M T 1 T 2 Applying Newton’s law to m 1 , N 1 − m 1 g = m 1 a y = 0 N 1 = m 1 g , where the force of friction on m 1 is f 1 = μ N 1 = μ m 1 g = (0 . 15) (1 kg) (9 . 8 m / s 2 ) = 1 . 47 N and T 1 − f 1 = m 1 a T 1 = m 1 a + f 1 . (1) For the mass m 2 , applying Newton’s law perpendicular to the slanted surface N 2 − m 2 g cos θ = m 2 a perp = 0 N 2 = m 2 g cos θ , so the force of friction is f 2 = μ N 2 = μ m 2 g cos θ = (0 . 15) (9 kg) (9 . 8 m / s 2 ) cos 46 ◦ = 9 . 19033 N . Applying Newton’s law parallel to the surface, m 2 g sin θ − f 2 − T 2 = m 2 a T 2 = − m 2 a + m 2 g sin θ − f 2 (2) Subtraction eq. (1) from, eq. (2), T 2 − T 1 = m 2 g sin θ − ( m 1 + m 2 ) a − f 1 − f 2 . Applying Newton’s law to the pulley, summationdisplay τ = I α + T 1 R − T 2 R = − M R 2 2 a R − T 1 + T 2 = 1 2 M a m 2 g sin θ − ( m 1 + m 2 ) a − f 1 − f 2 = 1 2 M a a = 2 m 2 g sin θ − 2 f 1 − 2 f 2 2 m 1 + 2 m 2 + M . Since 2 m 2 g sin θ − 2 ( f 1 + f 2 ) = 2(9 kg)(9 . 8 m / s 2 )(sin 46 ◦ ) − 2(1 . 47 N + 9 . 19033 N) = 105 . 571 kg · m / s 2 and 2 m 1 + 2 m 2 + M = 2(1 kg) + 2(9 kg) + 9 kg = 29 kg , then a = 2 m 2 g sin θ − 2 f 1 − 2 f 2 2 m 1 + 2 m 2 + M = 105 . 571 kg · m / s 2 29 kg = 3 . 64038 m / s 2 .

larsen (sl44887) – rotational dynamics – fox – (c12181) 14 036(part2of2)10.0points Find the tension in the horizontal part of the string. Correct answer: 5 . 11038 N. Explanation: From eq. (1), T 1 = f 1 + m 1 a = 1 . 47 N + (1 kg) (3 . 64038 m / s 2 ) = 5 . 11038 N . 037 10.0points An Atwood machine is constructed using two wheels (with the masses concentrated at the rims). The left wheel has a mass of 2 . 1 kg and radius 23 . 59 cm. The right wheel has a mass of 2 . 5 kg and radius 29 . 6 cm. The hanging mass on the left is 1 . 6 kg and on the right 1 . 32 kg. 3 m 1 m 4 m 2 m What is the acceleration of the system? The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 0 . 364894 m / s 2 . Explanation: Let : m 1 = 2 . 1 kg , r 1 = 23 . 59 cm , m 2 = 2 . 5 kg , r 2 = 29 . 6 cm , m 3 = 1 . 6 kg , and m 4 = 1 . 32 kg . 3 m 1 1 m 4 m 2 2 2 m T 1 T 3 T a a r r The net acceleration a = r α is in the di- rection of the heavier mass m 3 . Each pul- ley’s mass is concentrated on the rim, so I = m pulley r 2 and τ net = Iα = m r 2 parenleftBig a r parenrightBig = m r a so that for the leftmost pulley T 1 r 1 − T 2 r 1 = m 1 r 1 a m 1 a = T 1 − T 2 (1) and for the rightmost pulley T 2 r 2 − T 3 r 2 = m 2 r 2 a m 2 a = T 2 − T 3 . (2) Applying Newton’s Law to the hanging masses, 3 m 4 4 m 1 T 3 3 T a a m g m g the net forces are m 3 a = m 3 g − T 1 and (3) m 4 a = T 3 − m 4 g . (4) Adding these four equations gives m 1 a + m 2 a + m 3 a + m 4 a = m 3 g − m 4 g a = ( m 3 − m 4 ) g m 1 + m 2 + m 3 + m 4 = (1 . 6 kg − 1 . 32 kg) (9 . 8 m / s 2 ) 2 . 1 kg + 2 . 5 kg + 1 . 6 kg + 1 . 32 kg = 0 . 364894 m / s 2 . 038(part1of2)10.0points A cylindrical fishing reel has a mass of

larsen (sl44887) – rotational dynamics – fox – (c12181) 15 0 . 565 kg and a radius of 4 . 33 cm. A fric- tion clutch in the reel exerts a restraining torque of 1 . 33 N · m if a fish pulls on the line. The fisherman gets a bite, and the reel be- gins to spin with an angular acceleration of 83 . 2 rad / s 2 . What force does the fish exert on the line? Correct answer: 31 . 7337 N. Explanation: Let : m = 0 . 565 kg , r = 4 . 33 cm = 0 . 0433 m , τ r = 1 . 33 N · m , and α = 83 . 2 rad / s 2 . τ net = I α F r − τ r = parenleftbigg 1 2 m r 2 parenrightbigg α F r = τ r + 1 2 m r 2 α F = τ r r + 1 2 m r α = 1 . 33 N · m 0 . 0433 m + (0 . 565 kg) (0 . 0433 m) (83 . 2 rad / s 2 ) 2 = 31 . 7337 N . 039(part2of2)10.0points How much line unwinds in 0 . 816 s? Correct answer: 119 . 939 cm. Explanation: Let : ω 0 = 0 and t = 0 . 816 s . θ = ω 0 + 1 2 α t 2 = 1 2 α t 2 , so s = r θ = 1 2 r α t 2 = 1 2 (4 . 33 cm) (83 . 2 rad / s 2 ) (0 . 816 s) 2 = 119 . 939 cm . 040 10.0points A uniform 6 kg rod with length 18 m has a frictionless pivot at one end. The rod is released from rest at an angle of 22 ◦ beneath the horizontal. 9 m 18 m 6 kg 22 ◦ What is the angular acceleration of the rod immediately after it is released? The moment of inertia of a rod about the center of mass is 1 12 m L 2 , where m is the mass of the rod and L is the length of the rod. The moment of inertia of a rod about either end is 1 3 m L 2 , and the acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 0 . 7572 rad / s 2 . Explanation: Let : m = 6 kg , L = 18 m , and θ = 22 ◦ . The rod’s moment of inertia about its end- point is I = 1 3 m L 2 , so the angular accelera- tion of the rod is α = τ I = 1 2 m g L cos θ 1 3 m L 2 = 3 2 g L cos θ

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larsen (sl44887) – rotational dynamics – fox – (c12181) 16 = 3 2 9 . 8 m / s 2 18 m cos 22 ◦ = 0 . 7572 rad / s 2 . 041 10.0points Consider a thin 18 m rod pivoted at one end. A uniform density spherical object (whose mass is 9 kg and radius is 5 m) is attached to the free end of the rod and the moment of inertia of the rod about an end is I rod = 1 3 m L 2 and the moment of iner- tia of the sphere about its center of mass is I sphere = 2 5 m r 2 . 18 m C mass 9 kg 9 kg radius 5 m 58 ◦ What is the angular acceleration of the rod immediately after it is released from its ini- tial position of 58 ◦ from the vertical? The acceleration of gravity g = 9 . 8 m / s 2 . Correct answer: 0 . 411048 rad / s 2 . Explanation: Let : m = 9 kg , L = 18 m , r = 5 m , and θ = 58 ◦ . The moment of inertia of the rod with re- spect to the pivot point is I rod = 1 3 m r 2 , and the moment of inertia of the spherical mass with respect to the pivot point is I sphere = 2 5 m r 2 + m ( L + r ) 2 . Then, the moment of inertia of the system I is I = I rod + I m = 1 3 m L 2 + 2 5 m r 2 + m ( L + r ) 2 = 1 3 m (18 m) 2 + 2 5 m (5 m) 2 + m [(18 m) + (5 m)] 2 = 5823 kg m 2 . The center of mass of the rod is 1 2 L and the center of mass of the sphere L + r from the pivot point, so the center of mass of the rod plus mass system is CM = m parenleftbigg L 2 parenrightbigg + m ( L + r ) m + m = L 4 + L + r 2 = 3 L 4 + r 2 = 3 (18 m) 4 + 5 m 2 = 16 m . and the angular acceleration of the rod is α = τ I = CM ( m + m ) g sin θ I = (16 m) (18 kg) (9 . 8 m / s 2 ) sin 58 ◦ 5823 kg m 2 = 0 . 411048 rad / s 2 . 042 10.0points A massless rope is wrapped around a uniform cylinder of radius R and mass M .

larsen (sl44887) – rotational dynamics – fox – (c12181) 17 ω R M What is the linear acceleration of the cylin- der? Assume the unwrapped portion of the rope is vertical and the axis of the cylinder remains horizontal. 1. a = g 4 2. a = 2 g 5 3. a = 3 g 4 4. a = 3 g 5 5. a = 2 g 3 correct 6. a = g 2 7. a = 2 g 8. a = g 3 9. a = g 5 10. a = g Explanation: Let down be positive. M g T α R a a = R α and I = 1 2 M R 2 . From linear kinematics M g − T = M a and from rotational kinematics T R = I α ( M g − M a ) R = parenleftbigg 1 2 M R 2 parenrightbigg parenleftBig a R parenrightBig M g − M a = 1 2 M a g = 3 2 a a = 2 3 g . 043(part1of3)10.0points A block of mass 2 . 33 kg is suspended above the ground at a height 6 . 68 m by a spool with two arms. The spool-with-arms arrange- ment is a combination of a solid uniform cylin- der of mass 2 . 02 kg and radius 0 . 292 m, and two rods, each of length 0 . 619 m and mass 0 . 146 kg. h m l R What is the moment of inertia of the spool- with-arms arrangement around the rotation axis which passes through its center of mass? The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 0 . 0954402 kg · m 2 . Explanation: The moment of inertia of a rod around its center is 1 12 M L 2 , and the moment of inertia of a solid uniform cylinder is 1 2 M R 2 . Let : m a = 0 . 146 kg , m = 2 . 33 kg , ℓ = 0 . 619 m , M = 2 . 02 kg , and R = 0 . 292 m .

larsen (sl44887) – rotational dynamics – fox – (c12181) 18 Therefore, the total moment of inertia is I = 2 parenleftbigg 1 12 m a ℓ 2 parenrightbigg + 1 2 M R 2 = 1 6 (0 . 146 kg) (0 . 619 m) 2 + 1 2 (2 . 02 kg)(0 . 292 m) 2 = 0 . 0954402 kg · m 2 . 044(part2of3)10.0points If I is the moment of inertia of the spool with arms, α the clockwise angular acceleration of the spool, T the tension in the string and a the downward acceleration of the block, which statement is incorrect ? 1. m R a = Iα + TR correct 2. m g − T = m a 3. α R = a 4. a = m g R 2 m R 2 + I 5. I α = TR Explanation: Refer to the figure below: m R mg T T α The linear velocity at the edge of the spool is a = R α , Newton’s 2nd law for the block gives m g − T = m a T = mg − ma Applying torque on the spool, from the ten- sion in the string acting at right angles, τ = I α = TR I parenleftBig a R parenrightBig = ( m g − m a ) R a = m g R 2 m R 2 + I The remaining choice is incorrect. 045(part3of3)10.0points If the system is released from rest, how long will it take for the block to hit the ground? Correct answer: 1 . 42063 s. Explanation: The distance fallen by the mass is s = s 0 + v 0 t − 1 2 a t 2 . When it reaches the ground, 0 = h − 1 2 a t 2 fall t fall = radicalbigg 2 h a = radicalBigg 2 h ( m R 2 + I ) m g R 2 = radicalBigg 2 (6 . 68 m) (2 . 33 kg) (9 . 8 m / s 2 ) (0 . 292 m) 2 × radicalBig (2 . 33 kg) (0 . 292 m) 2 + 0 . 0954402 kg · m 2 = 1 . 42063 s . keywords: 046 10.0points A cylindrical flywheel is initially at rest and is free to pivot with negligible friction about the z -axis of the cylinder. Its moment of inertia with respect to the z -axis is 1 2 m r 2 , where its mass is 0 . 23 kg and its radius is 0 . 068 m. The

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larsen (sl44887) – rotational dynamics – fox – (c12181) 19 figure below shows the value of an applied torque as a function of time. − 2 − 1 0 1 2 3 4 5 0 1 2 3 4 5 6 7 8 9 10 time ( s ) ∆ t Torque ( N · m ) Calculate the kinetic energy K of the cylin- der when it reaches 5 s. Correct answer: 76162 . 2 J. Explanation: The momentum of intertia is I = 1 2 m r 2 = 1 2 (0 . 23 kg) (0 . 068 m) 2 = 0 . 00053176 kg · m 2 the angular velocity is ω = ω 0 + α t and τ = α I . Since the torque is constant over the time intervals (∆ t = 1 second) and the initial an- gular velocity ω 0 = 0 radians per second, we have K = 1 2 I ω 2 f = 1 2 I parenleftBigg ω 0 + 5 summationdisplay i =1 ∆ ω i parenrightBigg 2 = 1 2 I parenleftBigg 5 summationdisplay i =1 α i ∆ t parenrightBigg 2 = 1 2 I parenleftBigg 5 summationdisplay i =1 τ i I parenrightBigg 2 (∆ t ) 2 = (∆ t ) 2 2 I parenleftBigg 5 summationdisplay i =1 τ i parenrightBigg 2 = (∆ t ) 2 2 I bracketleftBig τ 01 + τ 12 + τ 23 + τ 34 + τ 45 bracketrightBig 2 = (1 s) 2 2 (0 . 00053176 kg) bracketleftBig (4 N m) + (1 N m) + (5 N m) + ( − 2 N m) + (1 N m) bracketrightBig 2 K = (1 s) 2 0 . 00106352 kg m 2 parenleftBig 9 N m parenrightBig 2 = 76162 . 2 J . 047(part1of3)10.0points The blocks shown in figure are connected by an inextensible string of negligible mass. The string passes over a frictionless pulley without slipping on the rim of the pulley. The block on the frictionless incline is moving with a constant acceleration up the incline. 24 kg T 1 R = 0 . 34 m M 6 kg μ = 0 T 2 6 . 5 m / s 2 24 ◦ Determine the tension in the vertical part of the string. The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 79 . 2 N. Explanation: Let : a = 6 . 5 m / s 2 , m 1 = 24 kg and g = 9 . 8 m / s 2 . Considering a free-body diagram for the hanging mass m 1 , m 1 g − T 1 = m 1 a T 1 = m 1 ( g − a ) = (24 kg) (9 . 8 m / s 2 − 6 . 5 m / s 2 ) = 79 . 2 N .

larsen (sl44887) – rotational dynamics – fox – (c12181) 20 048(part2of3)10.0points Determine the tension in the string parallel to the inclined plane. Correct answer: 62 . 9161 N. Explanation: Let : R = 0 . 34 m , θ = 24 ◦ , m 2 = 6 kg , T 2 W bardbl N W ⊥ W θ T 2 = m 2 ( a + g sin θ ) = (6 kg) bracketleftBig 6 . 5 m / s 2 + (9 . 8 m / s 2 ) sin 24 ◦ bracketrightBig = 62 . 9161 N . 049(part3of3)10.0points Find the mass of the pulley. It has uniform density and is shaped like a narrow cylindrical disk. Correct answer: 5 . 01043 kg. Explanation: Applying torque to the pulley, ( T 1 − T 2 ) R = Iα = I parenleftBig a R parenrightBig I = ( T 1 − T 2 ) R 2 a = (79 . 2 N − 62 . 9161 N) (0 . 34 m) 2 6 . 5 m / s 2 = 0 . 289603 kg m 2 . For the disk I disk = 1 2 M R 2 M = 2 I R 2 = 2 (0 . 289603 kg m 2 ) (0 . 34 m) 2 = 5 . 01043 kg .

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