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qz5sol_1243Hs11

# qz5sol_1243Hs11 - Applying one of the kinematic equations...

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TA: Tomoyuki Nakayama Wednesday, March 2, 2011 PHY 2048: Physic 1, Discussion Section 1243H Quiz 5 (Homework Set #7) Name: UFID: Formula sheets are not allowed. Do not store equations in your calculator. You have to solve problems on your own; memorizing final algebraic expressions from homework assignments and just plugging numbers into them will not give you full credit. Leave all your work. ________________________________________________________________________________ In the figure below, block 1 has mass m 1 = 5.00 kg, block 2 has mass m 2 = 8.00 kg, and the pulley is on a frictionless horizontal axle and has radius R = 0.30 m. When released from rest, block 2 falls 2.00 m in 6.00 s (without the cord slipping on the pulley). a) What is the angular acceleration of the pulley?
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Unformatted text preview: Applying one of the kinematic equations to block 2, we get the acceleration of the block. d = (1/2)at 2 ⇒ a = 2d/t 2 = 0.111 m/s 2 Since the cord does not slip on the pulley, the tangential acceleration of the pulley is equal to the acceleration of the blocks. The angular acceleration of the pulley is α R = a t = a α = a/R = 0.370 rad/s 2 b) What is the pulley's rotational inertia? Newton’s 2 nd law applied to the blocks yield tensions T 1 and T 2 . m 1 a = T 1 – m 1 g T 1 = m 1 (a + g) = 54.55 N m 2 a = m 2 g – T 2 T 2 = m 2 (g – a) = 77.51 N Applying Newton’s 2 nd law to the pulley, the rotational inertia of the pulley is I α = T 2 R – T 1 R I = (T 2- T 1 )R/ α = 18.6 kg · m 2...
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