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Unformatted text preview: Answer key for third exam version B 1 1. Free credit. I decided that this problem was too tricky, since people might assume that the wheel would slow down and stop. You can work out a solution where the wheel goes through 0 angular velocity and starts rotation in the op posite direction. It will eventually (after 79 seconds) achieve an angular velocity which is √ 2 ω i , meaning that it has double its initial energy. 2. The spinning iceskater effect was demonstrated and explained in class with the professorholdingdumbbellswhileseatedinspinningchair demonstration. Answer = D, conservation of angular momentum 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. This problem says that a steel wire of length 0.50 m and radius 0.00075 m stretches by 0.0011 m when a weight w is suspended with this wire. According to the Young’s Modulus definition Y ≡ ( F/A ) Δ L/L = ⇒ F = Y A Δ L L = 2 . × 10 11 π (7 . 5 × 10 4 ) 2 1 . 1 × 10 3 . 50 = 778 N Answer = D 5. The net torque is related to the angular acceleration by τ = Iα with I = 2 M R 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 + 2 α ( θ θ ), where ω = 2 . 40 * 2 π rad/sec, and θ θ = 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 Nm Answer = A Answer key for third exam version B 2 6. 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....
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 Spring '08
 Maguire
 Physics, Energy, Kinetic Energy, Work, exam version, Wblock, c.m. xcm, +120o

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