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Question Number 23272 by ajfour last updated on 28/Oct/17

Commented by ajfour last updated on 28/Oct/17

$I f t h e m o u n t e d w h e e l s a r e k e p t$ $s p i n n i n g a t a n a n g u l a r v e l o c i t y$ $ω (a s s h o w n i n d i a g r a m), f i n d$ $t h e t i m e p e r i o d o f s m a l l o s c i l l a t i o n s$ $o f a r o d o f m a s s M k e p t o n t h e m,$ $i f i t b e d i s p l a c e d s l i g h t l y f r o m i t s$ $e q u i l i b r i u m p o s i t i o n . A s s u m e$ $c o e f f i c i e n t o f f r i c t i o n b e t w e e n$ $r o d a n d w h e e l s t o b e μ .$
	Answered by mrW1 last updated on 29/Oct/17

	Commented by mrW1 last updated on 29/Oct/17

	$N_{1} = \frac{\frac{d}{2} + x}{d} \times Mg = (\frac{1}{2} + \frac{x}{d}) Mg$ $F_{1} = μ N_{1} = μ Mg (1 + \frac{x}{d}) (direction \leftarrow)$ $N_{2} = \frac{\frac{d}{2} - x}{d} \times Mg = (\frac{1}{2} - \frac{x}{d}) Mg$ $F_{2} = μ N_{2} = μ Mg (1 - \frac{x}{d}) (direction \to)$ $F = Σ F = F_{2} - F_{1} = - \frac{2 μ Mg}{d} \times x = - kx$ $period T = 2 π \sqrt{\frac{M}{k}} = 2 π \sqrt{\frac{d}{2 μ g}}$
	Commented by ajfour last updated on 29/Oct/17

	$T h a n k s a l o t, S i r .$
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