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Final Exam Solutions

Immediately after the collision 240 the wreckage

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Unformatted text preview: 2.0 − 0) + m2 g (0 − 2.0) + (m1 + m2 )v 2 + I ω 2 …[1] 2 2 where I = 1 MR 2 and ω = v / R 2 R M g m1 m2 Thus, from eq.[1] 0 = 1.5*10.0*2.0  ­ 2.5*10.0*2.0 + 0.5(1.5+2.5)v2+0.5v2 0 =  ­ 20 + 2.5 v2 or v = [8.0]1/2 = 2.83 m/s [b]( 4 pts) What is the magnitude of the radial acceleration of a point such as P on the rim of the pulley? ac = v 2/R = 8.0/0.2 = 40.0 m/s2. [c]( 4 pts) What is the magnitude of the tangential acceleration of point P on the rim of the pulley? The magnitude of the tangential acceleration of point P is the same as the acceleration of m1 or m2. v2 = 2atΔy or at = v2/2Δy = 8.0/2*2.0 = 2.0 m/s2 [d]( 4 pts)What is the magnitude of the net torque on the pulley? τ = Iα = Iat/R = 0.5(2.0)(0.2)2(2.0/0.2) = 0.4 N.m Prob. 5 [a] (10pts)A car with mass M travelling...
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