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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 = 0.2 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 R = a t = a = a/R = 1.00 rad/s 2 b) What is the pulley's rotational inertia? Newtons 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) = 20.0 N m 2 a = m 2 g T 2 T 2 = m 2 (g a) = 48.0 N Applying Newtons 2 nd law to the pulley, the rotational inertia of the pulley is I = T 2 R T 1 R I = (T 2 T 1 )R/ = 5.60 kg m 2...
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This note was uploaded on 12/10/2011 for the course PHY 2048 taught by Professor Field during the Spring '08 term at University of Florida.
 Spring '08
 Field

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