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solution3ExamASP08

# solution3ExamASP08 - Answer key for third exam version A 1...

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Answer key for third exam version A 1 1. Say that the disk has mass M 1 and radius R 1 , while the sphere has mass M 2 and R 2 . At the top of the plane, the disk has potential energy M 1 gh , while the sphere has potential energy M 2 gh , where h is the height of inclined plane. At the bottom of the inclined plane the potential energies are both defined to be zero, and the kinetic energies equal to the initial potential energies K 1 = M 1 gh = 1 2 M 1 v 2 1 + 1 2 I 1 ω 2 = 1 2 M 1 v 2 1 + 1 4 M 1 R 2 1 v 2 1 R 2 1 = 3 4 M 1 v 2 1 = v 1 = v u u t 4 gh 3 speed of c.m. of disk K 2 = M 2 gh = 1 2 M 2 v 2 2 + 1 2 I 2 ω 2 = 1 2 M 2 v 2 2 + 1 5 M 2 R 2 2 v 2 2 R 2 2 = 7 10 M 2 v 2 2 = v 2 = v u u t 10 gh 7 speed of c.m. of sphere The sphere is going faster at the bottom of the inclined plane for any height h , and this does not depend on the mass nor the radius of the sphere. We did a class demonstration using a solid disk and a hoop to show the same qualitative result: the object with the smaller moment of inertia will win the race to the bottom because proportionately less kinetic energy is in rotational motion and proportionately more kinetic energy is in translational motion. Answer C 2. The mass M X of planet X is not given, but it can be obtained from knowing the radius R X and the surface acceleration g X (see Lecture 16, pages 1–2): g X = G M X R 2 X = GM X = g X R 2 X For the weight of a mass m at a distance h above the surface of planet X to be equal to the weight on the Earth’s surface we have mg E = GM X m ( R X + h ) 2 = ( R X + h ) 2 = GM x g E = g X R 2 X g E h = R X v u u t g X g E - 1 = 5 . 16 × 10 7 v u u t 13 . 2 9 . 8 - 1 = 8 . 30 × 10 6 meters Answer E

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Answer key for third exam version A 2 3. We can solve this problem by using either conservation total energy, or by conservation of angular momentum, between the minimum distance r 1 and the maximum distance r 2 . The angular momentum method is easier L 1 = L 2 = mv 1 r 1 = L 2 = mv 2 r 2 = v 1 = r 2 r 1 v 2 The speed v 2 at the maximum distance is given as 908 m/s. This gives the maximum speed at the closest distance as v 1 = 5 . 46 × 10 4 m/s. Answer C 4. The net torque is related to the angular acceleration by τ = with I = 2 MR 2 / 5 in this case ( M = 1 . 85 kg, R = 0 . 45 / 2 m). We can compute the angular acceleration from ω 2 = ω 2 0 + 2 α ( θ - θ 0 ), where ω 0 = 2 . 40 * 2 π rad/sec, and θ - θ 0 = 18 . 2 * 2 π radians. Finally, we will have τ = 2 5 (1 . 85)(0 . 225) 2 (2 . 40 × 2 π ) 2 2 × 18 . 2 × 2 π = 0 . 0372 N-m Answer = B 5. This answer was actually given in the MasteringPhysics Assignment 12, item 13 for the Satellite in Orbit . The extra hint for this item 1 emphasized that MasteringPhysics comment, and specifically mentioned the Space Shuttle astronauts appearing “weightless” was continuous falling but never reaching the Earth.
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solution3ExamASP08 - Answer key for third exam version A 1...

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