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hw36 - 1 ˙ x C = F m t-μ gt ∑ M C = I C α f R = mR 2 2...

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TAM212 Introductory Dynamics Spring 2005 S. Balachandar Homework 36 (4.62, 4.78) 4.62 N f mg y x Equations of motion F x = m ¨ x C mg sin β - f = m ¨ x C M c = I C α fr = 2 5 mr 2 ¨ θ f = 2 5 mr ¨ θ ¨ x C = r ¨ θ Solving the three equations, we get ¨ θ = 5 g sin β 7 r ¨ x C = 5 g sin β 7 1
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Integrating ¨ x C , we get ˙ x C = 5 g sin β 7 t + a0 a0 a18 0 C 1 Integrating again, x C = 5 g sin β 14 t 2 + a0 a0 a18 L C 2 x C ( t ) = 5 g sin β 14 t 2 + L 4.78 (a) F 0 C mg f N Assume rolling; the equations of motion are then F x = m ¨ x C F 0 - f = m ¨ x C M C = I C α f R = mR 2 2 ¨ θ C f = mR 2 ¨ θ C ¨ x C = R ¨ θ Solving, the three equations, we get F 0 = 3 mR ¨ θ C 2 F 0 = 3 f 2
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For rolling, 0 f μ N f μ mg F 0 3 μ mg F 0 mg 3 μ So, if F 0 mg > 3 μ , it will slip (b) Assume slipping, i.e. f = μ N . So the equations for motion become, F x = m ¨ x C F 0 - μ N = m ¨ x C F 0 - μ mg = m ¨ x C Integrating, m ˙ x C = F 0 t - μ mgt
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Unformatted text preview: 1 ˙ x C = F m t-μ gt ∑ M C = I C α f R = mR 2 2 ¨ θ C μ mgR = mR 2 2 ¨ θ C ¨ θ C = 2 μ g R Integrating, ˙ θ C = 2 μ g R t + a a A C 2 Now, velocity of point B, ˙ x B = ˙ x C-R ˙ θ = F m t-μ gt-R 2 μ g R t = F m t-μ gt-2 μ gt = F m t-3 μ gt 3 If cylinder is slipping, then the point B is dragging and hence, its velocity is to the right ˙ x B > F m t-3 μ gt > F mg > 3 μ 4...
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