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Unformatted text preview: Version 060 Midterm 03 Yao (58180) 1 This printout should have 18 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points A solid ball of mass m and radius r has I CM = 2 5 mr 2 , while a hollow ball of mass M and radius R has I CM = 2 3 M R 2 . If a hollow ball P is placed side by side at the top of a ramp with a solid ball Q , and they do not have the same mass nor the same radius , which ball will reach the bottom of the ramp first, if they roll without slipping? 1. Both balls will arrive at the bottom at the same time. 2. It is impossible to say because it depends on how the masses and the radii compare. 3. Ball Q will always beat ball P . correct 4. Ball P will always beat ball Q . Explanation: The CM speed v 1 I , so the object with largest rotational inertia (the hollow ball) will always lose the race. 002 10.0 points A long rod is pivoted (without friction) at one end. It is released from rest at A in a horizontal position and swings down, passing positions B and C . A B C b Compare the angular accelerations at B and C . 1. There is not enough information. 2. B &lt; C 3. B = C 4. B &gt; C correct Explanation: Using the rotational analogue to Newtons second law = mvectorg vectorr = I = m I vectorg vectorr = m I g r sin . r mg b The angular acceleration decreases as the rod falls and the angle decreases, so B &gt; C . 003 10.0 points A cylinder with moment of inertia I 1 rotates with angular speed about a frictionless ver tical axle. A second cylinder, with moment of inertia I 2 = 3 4 I 1 , initially not rotating, drops onto the first cylinder. Since the surfaces are rough, the two eventually reach the same an gular speed . 3 4 I 1 I 2 I 1 Before After Calculate the final kinetic energy K with respect to the initial kinetic energy K . 1. K = 1 3 K Version 060 Midterm 03 Yao (58180) 2 2. K = 3 4 K 3. K = 2 3 K 4. K = 4 7 K correct 5. None of these 6. K = 4 5 K 7. K = 3 7 K 8. K = 3 5 K 9. K = 2 5 K Explanation: summationdisplay vector L = const From conservation of angular momentum I 1 = ( I 1 + I 2 ) , so = I 1 I 1 + I 2 (1) = I 1 I 1 + 3 4 I 1 = 4 7 . Substituting from Eq. 1, the final rota tional kinetic energy is K = 1 2 ( I 1 + I 2 ) 2 = 1 2 ( I 1 + I 2 ) I 2 1 ( I 1 + I 2 ) 2 2 = 1 2 I 2 1 I 1 + I 2 2 , (2) and the initial rotational kinetic energy is K = 1 2 I 1 2 2 = 1 1 2 I 1 K (3) Substituting 2 from Eq. 3 into Eq. 2, we have K = 1 2 I 2 1 I 1 + I 2 1 2 I 1 K = I 1 I 1 + I 2 K = I 1 I 1 + 3 4 I 1 K = 4 7 K , since K &lt; K , kinetic energy is lost....
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 Spring '11
 Dutt

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