# hw36 - 1 x C = F m t- gt M C = I C f R = mR 2 2 C mgR = mR...

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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 + a a A 0 C 1 Integrating again, x C = 5 g sin β 14 t 2 + a a A 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
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 + a a A 0 C

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Unformatted text preview: 1 x C = F m t- gt M C = I C f R = mR 2 2 C mgR = mR 2 2 C C = 2 g R Integrating, C = 2 g R t + a a A C 2 Now, velocity of point B, x B = x C-R = F m t- gt-R 2 g R t = F m t- gt-2 gt = F m t-3 gt 3 If cylinder is slipping, then the point B is dragging and hence, its velocity is to the right x B &gt; F m t-3 gt &gt; F mg &gt; 3 4...
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## This note was uploaded on 05/24/2010 for the course TM 212 taught by Professor Richard during the Spring '10 term at University of Illinois, Urbana Champaign.

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hw36 - 1 x C = F m t- gt M C = I C f R = mR 2 2 C mgR = mR...

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