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Question Number 143868 by ajfour last updated on 19/Jun/21

Commented by ajfour last updated on 19/Jun/21

$F i n d m i n i m u m h s u c h t h a t$ $s t i c k j u s t l o s e s c o n t a c t w i t h$ $t a b l e u p o n b e i n g h i t b y a s m a l l$ $b a l l o f m a s s m r e l e a s e d a$ $h e i g h t h a b o v e t h e t a b l e t o p$ $l e v e l . (A s s u m e s u f f i c i e n t$ $f r i c t i o n a t t h e t a b l e c o r n e r)$
	Answered by ajfour last updated on 21/Jun/21

	$A n g u l a r m o m e n t u m$ $c o n s e r v a t i o n d u r i n g e l a s t i c$ $c o l l i s i o n o f b a l l w i t h s t i c k :$ $2 m \sqrt{2 g h} (k L) = I ω_{0}$ $A f t e r c o l l i s i o n s t i c k r o t a t e s;$ $n o w s a y a t s o m e p o i n t o f$ $t i m e t h e r e a f t e r, s a y t = t,$ $a n g l e o f s t i c k f r o m v e r t i c a l$ $i s θ; t h e n$ $r i s e i n p o t e n t i a l e n e r g y$ $= f a l l i n k i n e t i c e n e r g y$ $M g (\frac{L}{2} - k L) \cos θ = \frac{1}{2} I (ω_{0}^{2} - ω^{2})$ $A n d l e t N o r m a l r e a c t i o n$ $t h e n v a n i s h e s \Rightarrow$ $s a y \frac{L}{2} - k L = r$ $M g r \sin θ = - I α$ $\Rightarrow M g r \sin θ = - I (\frac{ω d ω}{d θ}) &$ $ω^{2} r \sin θ = - (\frac{ω d ω}{d θ}) r \cos θ &$ $ω^{2} r \cos θ - (\frac{ω d ω}{d θ}) r \sin θ = g$ $\Rightarrow ω^{2} r = \frac{{M g r}^{2} \cos θ}{I}$ $ω^{2} r \cos θ + \frac{{M g r}^{2} \sin^{2} θ}{I} = g$ $. . . . . . . .$
	Answered by TheHoneyCat last updated on 21/Jun/21

	${Let}^{'} s assume that the bar of mass M will not slide$ $To fall, it therefore needs to end up vertical$ $The ball will fall on the bar and transfer all of its kinetic energy .$ $that energy beeing its potential energy at the beginning because, at that time, it had no speed, and since then, it has been in free fall .$ $Choosing z = 0 for the {bar}^{'} s height .$ $E_{m_{i n i t i a l l y}} = E_{m_{b a l l i n i t i a l l y}} + E_{m_{b a r i n i t i a l l y}}$ $= m g h + (M g) . 0$ $= m g h$ $E_{m_{a f t e r c o n t a c t}} = E_{m_{b a l l l e f t}} + E_{m_{b a r}}$ $=^{1} /_{2} {m v}_{b a l l l e f t}^{2} + E_{p_{v e r t i c a l b a r}} + E_{c_{b a r}}$ $=^{1} /_{2} {m v}_{b a l l l e f t}^{2} + \int_{b a r} \frac{δ {M v}^{2}}{2} + \int_{b a r} δ M g z$ $s u p p o s i n g t h a t t h e m a s s i s u n i f o r m a l y d i s t r i b u t e d$ $=^{1} /_{2} {m v}_{b a l l l e f t}^{2} + \int_{b a r} \frac{δ {m v}^{2}}{2} + \frac{M}{L} g \int_{b a r} z d z$ $=^{1} /_{2} {m v}_{b a l l l e f t}^{2} + \int_{b a r} \frac{δ {m v}^{2}}{2} + \frac{M}{2 L} g ({(1 - k)}^{2} - k^{2})$ $=^{1} /_{2} {m v}_{b a l l l e f t}^{2} + \int_{b a r} \frac{δ {m v}^{2}}{2} + \frac{M}{2 L} g (1 - 2 k)$ $⩾ \frac{M}{2 L} g (1 - 2 k)$ $by the conservation of energy$ $E_{i n i t i a l l y} = E_{a f t e r}$ $so : m g h ⩾ \frac{M}{2 L} g (1 - 2 k)$ $\Leftrightarrow 2 L m h ⩾ M {(1 - 2 k)}_{◼}$
	Commented by TheHoneyCat last updated on 21/Jun/21

	$s o o b v i o u s l y t h e m i n i m u m v a l u e o f h i s :$ $\frac{M (1 - 2 k)}{2 L m}$
	Commented by ajfour last updated on 21/Jun/21

	$T h e s t i c k {d o e s n}^{'} t s l i d e g r a n t e d,$ $b u t s t i c k m a y j u m p d u r i n g$ $c o l l i s i o n a n d l e a v e c o n t a c t$ $a t o n c e; i s e e a f a i r p o s s i b i l i t y$ $o f t h a t .$
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