HW23_Solutions

# HW23_Solutions - hiarker(srh959 homework 23 Turner(58220 1...

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Unformatted text preview: hiarker (srh959) homework 23 Turner (58220) 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 = . 15 m and R 2 = 1 . 4 m are mounted rigidly on a common axle and clamped together. The combined moment of inertia of the two wheels is I + 3 kg m 2 . Mass m 1 = 36 kg is attached to a cord wrapped around the first wheel, and another mass m 2 = 5 . 8 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: 1 . 75623 rad / s 2 . Explanation: 1 2 m g R 1 m g R 2 T 1 2 T 2 T T 1 Lets 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 = R 1 and a 2 = R 2 . Applying Newtons second law to m 1 , m 2 , and the pulleys separately, we obtain T 1- m 1 g = m 1 a 1 = m 1 R 1 (1) m 2 g- T 2 = m 2 a 2 = m 2 R 2 (2) net = T 2 R 2- T 1 R 1 = I . (3) Solving these equations, we find = m 2 g R 2- m 1 g R 1 I + m 1 R 2 1 + m 2 R 2 2 = 1 . 75623 rad / s 2 . The fact that &amp;gt; 0 indicates that our pre- vious assumption that m 2 moves downward is right. If &amp;lt; 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 67 . 9 cm and a mass of 1 . 32 kg. The bicycle is placed on a stationary stand and a resistive force of 169 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 3 . 89 rad / s 2 , what force must be applied by a chain passing over a 6 . 8 cm diameter sprocket? Correct answer: 1704 . 92 N. Explanation: Let : r w = 33 . 95 cm = 0 . 3395 m , M = 1 . 32 kg , F f = 169 N , r s = 3 . 4 m = 0 . 034 m , and = 3 . 89 rad / s 2 . For a wheel with mass concentrated on the rim, I = M R 2 . Applying rotational equilib- rium summationdisplay = I F r- F f R = I hiarker (srh959) homework 23 Turner (58220) 2 F = F f R + I r = F f R + M R 2 r = (169 N)(0 . 3395 m) . 034 m + (1 . 32 kg)(0 . 3395 m) 2 (3 . 89 rad / s 2 ) . 034 m = 1704 . 92 N ....
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## This note was uploaded on 04/23/2010 for the course PHY 303K taught by Professor Nui during the Spring '09 term at University of Texas-Tyler.

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HW23_Solutions - hiarker(srh959 homework 23 Turner(58220 1...

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