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HW16Solutions

# HW16Solutions - rote(cmr2495 homework 23 Turner(58120 This...

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rote (cmr2495) – homework 23 – Turner – (58120) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Two pulley wheels, of respective radii R 1 = 0 . 16 m and R 2 = 0 . 84 m are mounted rigidly on a common axle and clamped together. The combined moment of inertia of the two wheels is I + 4 . 1 kg m 2 . Mass m 1 = 69 kg is attached to a cord wrapped around the first wheel, and another mass m 2 = 20 kg is attached to another cord wrapped around the second wheel: 1 2 m R 1 m R 2 Find the angular acceleration of the system. Take clockwise direction as positive. The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 2 . 82545 rad / s 2 . Explanation: 1 2 m g R 1 m g R 2 T 1 2 T 2 T T 1 Let’s assume m 2 moves downward and take motion downward as positive for m 2 and mo- tion upward as positive for m 1 , take clockwise as positive for the pulley wheels. Since the two masses are connected to the pulleys, their accelerations are a 1 = α R 1 and a 2 = α R 2 . Applying Newton’s second law to m 1 , m 2 , and the pulleys separately, we obtain T 1 - m 1 g = m 1 a 1 = m 1 α R 1 (1) m 2 g - T 2 = m 2 a 2 = m 2 α R 2 (2) τ net = T 2 R 2 - T 1 R 1 = I α . (3) Solving these equations, we find α = m 2 g R 2 - m 1 g R 1 I + m 1 R 2 1 + m 2 R 2 2 = 2 . 82545 rad / s 2 . The fact that α > 0 indicates that our pre- vious assumption that m 2 moves downward is right. If α < 0, our assumption was not correct and m 2 would move upward, but the equations obtained to calculate T 1 and T 2 would be still correct and we need not reas- sume the direction of the motion and recalcu- late. 002 10.0 points A bicycle wheel has a diameter of 54 . 3 cm and a mass of 1 . 41 kg. The bicycle is placed on a stationary stand and a resistive force of 186 N is applied tangent to the rim of the tire. Assume that the wheel is a hoop with all of the mass concentrated on the outside radius. In order to give the wheel an acceleration of 4 . 27 rad / s 2 , what force must be applied by a chain passing over a 1 . 4 cm diameter sprocket? Correct answer: 7277 . 54 N. Explanation: Let : r w = 27 . 15 cm = 0 . 2715 m , M = 1 . 41 kg , F f = 186 N , r s = 0 . 7 m = 0 . 007 m , and α = 4 . 27 rad / s 2 . For a wheel with mass concentrated on the rim, I = M R 2 . Applying rotational equilib- rium summationdisplay τ = I α F r - F f R = I α

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rote (cmr2495) – homework 23 – Turner – (58120) 2 F = F f R + I α r = F f R + M R 2 α r = (186 N)(0 . 2715 m) 0 . 007 m + (1 . 41 kg)(0 . 2715 m) 2 (4 . 27 rad / s 2 ) 0 . 007 m = 7277 . 54 N .
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