This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: alexander (jra2623) homework 22 Turner (92510) 1 This printout should have 10 questions. Multiplechoice 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 = . 12 m and R 2 = 1 . 1 m are mounted rigidly on a common axle and clamped together. The combined moment of inertia of the two wheels is I + 2 . 9 kg m 2 . Mass m 1 = 92 kg is attached to a cord wrapped around the first wheel, and another mass m 2 = 17 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: 3 . 02757 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 = 3 . 02757 rad / s 2 . The fact that &gt; 0 indicates that our pre vious assumption that m 2 moves downward is right. If &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 58 . 3 cm and a mass of 2 . 09 kg. The bicycle is placed on a stationary stand and a resistive force of 104 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 . 97 rad / s 2 , what force must be applied by a chain passing over a 6 . 8 cm diameter sprocket? Correct answer: 912 . 384 N. Explanation: Let : r w = 29 . 15 cm = 0 . 2915 m , M = 2 . 09 kg , F f = 104 N , r s = 3 . 4 m = 0 . 034 m , and = 3 . 97 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 alexander (jra2623) homework 22 Turner (92510) 2 F = F f R + I r = F f R + M R 2 r = (104 N)(0 . 2915 m) . 034 m + (2...
View
Full
Document
This note was uploaded on 10/20/2011 for the course PHY 302K taught by Professor Kaplunovsky during the Summer '08 term at University of Texas.
 Summer '08
 Kaplunovsky
 Work

Click to edit the document details