HW 10 solution

# HW 10 solution - gutierrez(ig3472 – HW10 – ditmire...

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Unformatted text preview: gutierrez (ig3472) – HW10 – ditmire – (58216) 1 This print-out should have 25 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points An Atwood machine is constructed using a hoop with spokes of negligible mass. The 2 . 4 kg mass of the pulley is concentrated on its rim, which is a distance 21 . 3 cm from the axle. The mass on the right is 0 . 92 kg and on the left is 1 . 57 kg. 3 . 2 m 2 . 4 kg 21 . 3 cm ω 1 . 57 kg . 92 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: 1 . 30266 m / s 2 . Explanation: Let : M = 2 . 4 kg , R = 21 . 3 cm , m 1 = 0 . 92 kg , m 2 = 1 . 57 kg , h = 3 . 2 m , and v = ω R. Consider the free body diagrams 1 . 57 kg . 92 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 = ( mr 2 ) parenleftBig a r parenrightBig = mr a ma = T 1 − T 2 ma = ( m 1 − m 2 ) g − ( m 1 + m 2 ) a ma + ( m 1 + m 2 ) a = ( m 1 − m 2 ) g a = ( m 1 − m 2 ) g m + m 1 + m 2 = (0 . 92 kg − 1 . 57 kg) (9 . 8 m / s 2 ) 2 . 4 kg + 0 . 92 kg + 1 . 57 kg = 1 . 30266 m / s 2 . 002 10.0 points A simple pendulum consists of a 2.6 kg point mass hanging at the end of a 4.3 m long light string that is connected to a pivot point. Calculate the magnitude of the torque (due to the force of gravity) around this pivot gutierrez (ig3472) – HW10 – ditmire – (58216) 2 point when the string makes a 4 . 4 ◦ angle with the vertical. The acceleration of gravity is 9 . 81 m / s 2 . Correct answer: 8 . 41422 N · m. Explanation: Let : m = 2 . 6 kg , d = 4 . 3 m , θ = 4 . 4 ◦ , and g = 9 . 81 m / s 2 . Solution: τ = F d sin θ = mg d sin θ = (2 . 6 kg) ( 9 . 81 m / s 2 ) (4 . 3 m) sin4 . 4 ◦ = 8 . 41422 N · m . 003 (part 1 of 5) 10.0 points A disk has mass 8 kg and outer radius 20 cm with a radial mass distribution (which may not be uniform) so that its moment of inertia is 3 4 mR 2 . The disk is given a hard kick (impulse) along a horizontal surface at time t . The kicking force acts along a horizontal line through the disk’s center, so the disk acquires a linear velocity 1 . 2 m / s but no initial angular velocity. The coefficient of friction between the disk and the surface is 0 . 06 . The kinetic friction force between the sur- face and the disk slows down its linear motion while at the same time making the disk spin on its axis at an accelerating rate. Eventually, the disk’s rotation catches up with its linear motion, and the disk begins to roll at time t rolling without slipping on the surface....
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HW 10 solution - gutierrez(ig3472 – HW10 – ditmire...

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