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Question Number 96817 by peter frank last updated on 05/Jun/20

Commented by mr W last updated on 06/Jun/20

$b o t h r e s u l t s i n q u e s t i o n a r e w r o n g!$ $p l e a s e c h e c k i f t h e y a r e y o u r t y p o s o r$ $q u e s t i o n i n b o o k i s r e a l l y w r o n g .$
Commented by peter frank last updated on 07/Jun/20

$yes your right i see$
	Answered by mr W last updated on 06/Jun/20

	Commented by mr W last updated on 06/Jun/20

	$\frac{1}{2} {m v}^{2} = \frac{1}{2} {m u}_{0}^{2} + m g r (1 - \cos α)$ $\Rightarrow v^{2} = u_{0}^{2} + 2 g r (1 - \cos α)$ $m g \cos α - N = \frac{{m v}^{2}}{r}$ $\Rightarrow N = m (g \cos α - \frac{v^{2}}{r}) = m [g \cos α - \frac{u_{0}^{2}}{r} - 2 g (1 - \cos α)]$ $\Rightarrow N = m [(3 \cos α - 2) g - \frac{u_{0}^{2}}{r}]$ $w h e n N = 0, i . e . c o n t a c t l o o s e s :$ $(3 \cos α - 2) g - \frac{u_{0}^{2}}{r} = 0$ $\Rightarrow \cos α = \frac{1}{3} (2 + \frac{u_{0}^{2}}{r g})$ $v^{2} = u_{0}^{2} + 2 g r (1 - \frac{2}{3} - \frac{u_{0}^{2}}{3 r g})$ $v^{2} = u_{0}^{2} + \frac{2}{3} (g r - u_{0}^{2})$ $v^{2} = \frac{u_{0}^{2} + 2 g r}{3}$ $\Rightarrow v = \sqrt{\frac{u_{0}^{2} + 2 g r}{3}}$
	Commented by peter frank last updated on 07/Jun/20

	$thank you$
	Commented by peter frank last updated on 07/Jun/20

	$my be when N = 0 the$ $initil velocity u_{o} also 0$ $\Rightarrow v = \sqrt{\frac{u_{0}^{2} + 2 g r}{3}}$ $u = 0$ $\Rightarrow v = \sqrt{\frac{2 g r}{3}}$
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