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Question Number 187216 by ajfour last updated on 14/Feb/23

Commented by mr W last updated on 15/Feb/23

$t i l l t h e r o d s t r i k s t h e g r o u n d, t h e b a l l$ $m o v e s u p w a r d s a n d t h e s t r i n g i s$ $a l w a y s u n d e r t e n s i o n . a t t h e m o m e n t$ $a s t h e r o d s t r i k e s t h e g r o u d t h e b a l l$ $h a s a c e r t a i n v e l o c i t y u p w a r d s a n d$ $t h e t e n s i o n i n s t r i n g g e t s z e r o . t h e n$ $t h e b a l l m o v e s a s f r e e b o d y u p w a r d s$ $w i t h t h i s i n i t i a l v e l o c i t y .$ $a l l i n a l l t h e b a l l m o v e s o n l y i n$ $v e r t i c a l d i r e c t i o n, s o i {d o n}^{'} t$ $u n d e r s t a n d y o u r d i a g r a m w h i c h$ $s h o w s a s i f t h e b a l l a l s o m o v e s$ $s i d e w a r d s .$
	Answered by mr W last updated on 15/Feb/23

	Commented by mr W last updated on 15/Feb/23

	Commented by mr W last updated on 15/Feb/23

	$w h e n t h e r o d f a l l s, t h e t e n s i o n T$ $c h a n g e s a n d t h e i n c l i n a t i o n o f l e f t$ $s t r i n g c h a n g e s, b u t t h e r i g h t s t r i n g$ $r e m a i n s a l w a y s i n v e r t i c a l p o s i t i o n,$ $i t h i n k .$
	Commented by mr W last updated on 15/Feb/23
	$θ=30° tan ϕ=((L(1−((√3)/2)))/((L/2)+a))=((2−(√3))/(1+((2a)/L)))=k, say v cos ϕ=ωL ⇒ω=((v cos ϕ)/L)=(v/(L(√(1+k^2 )))) h+((L/2)+a)(1/(cos ϕ))=2a h=2a−((L/2)+a)(√(1+k^2 )) ((Iω^2 )/2)+((mv^2 )/2)=((MgL)/2) sin θ−mg(a−h) ((L^2 ω^2 )/3)+((mv^2 )/M)=((gL)/2)−((2mg)/M)[((L/2)+a)(√(1+k^2 ))−a] [(1/(3(1+k^2 )))+(m/M)]v^2 =((gL)/2){1−((4m)/M)[((1/2)+(a/L))(√(1+k^2 ))−(a/L)]} ⇒v=(√((((gL)/2){1−((4m)/M)[((1/2)+(a/L))(√(1+k^2 ))−(a/L)]})/((1/(3(1+k^2 )))+(m/M))))$
	$θ = 30 °$ $\tan φ = \frac{L (1 - \frac{\sqrt{3}}{2})}{\frac{L}{2} + a} = \frac{2 - \sqrt{3}}{1 + \frac{2 a}{L}} = k, s a y$ $v \cos φ = ω L$ $\Rightarrow ω = \frac{v \cos φ}{L} = \frac{v}{L \sqrt{1 + k^{2}}}$ $h + (\frac{L}{2} + a) \frac{1}{\cos φ} = 2 a$ $h = 2 a - (\frac{L}{2} + a) \sqrt{1 + k^{2}}$ $\frac{I ω^{2}}{2} + \frac{{m v}^{2}}{2} = \frac{M g L}{2} \sin θ - m g (a - h)$ $\frac{L^{2} ω^{2}}{3} + \frac{{m v}^{2}}{M} = \frac{g L}{2} - \frac{2 m g}{M} [(\frac{L}{2} + a) \sqrt{1 + k^{2}} - a]$ $[\frac{1}{3 (1 + k^{2})} + \frac{m}{M}] v^{2} = \frac{g L}{2} {1 - \frac{4 m}{M} [(\frac{1}{2} + \frac{a}{L}) \sqrt{1 + k^{2}} - \frac{a}{L}]}$ $\Rightarrow v = \sqrt{\frac{\frac{g L}{2} {1 - \frac{4 m}{M} [(\frac{1}{2} + \frac{a}{L}) \sqrt{1 + k^{2}} - \frac{a}{L}]}}{\frac{1}{3 (1 + k^{2})} + \frac{m}{M}}}$
	Commented by ajfour last updated on 15/Feb/23

	$I s h o u l d h a v e m e n t i o n e d t h e$ $r a d i u s o f p u l l e y s a y R .$ $[b u t a s s o o n a s r o d s t r i k e s$ $g r o u n d, s t r i n g g e t s h o r i z o n t a l, s o$ $t h a t s a n a d d i t i o n a l c o n d i t i o n g i v e n$ $t h a t m i g h t h e l p f i n d R = f (L, a)]$ $U n d e r v a r y i n g t e n s i o n i t h i n k$ $a n g l e o f c o n t a c t o f r o p e w i t h p u l l e y$ $c i r c u m f e r e n c e s h o u l d c h a n g e . .$ $t h a n k s f o r s o l u t i o n o f t h i s s i m p l e r$ $c a s e, S i r .$
	Commented by mr W last updated on 15/Feb/23

	$w h y s h o u l d t h e s t r i n g g e t h o r i z o n t a l$ $w h e n t h e r o d s t r i k e s t h e g r o u n d ?$
	Commented by ajfour last updated on 15/Feb/23

	$T h a n k y o u s i r, y o u a r e r i g h t, i$ $g o t m i s g u i d e d b y t h e r e a l m a s s$ $s t r i n g a n d c e n t r i f y g a l f o r c e o n$ $i t a s i t t u r n s r o u n d t h e p u l l e y . .$
	Commented by ajfour last updated on 16/Feb/23
	$τ=TL+T(((L(√3))/2)−2R)−mg((L/2))cos θ T=mg+m(dv/dt) v=−L(dθ/dt) (((ML^2 )/3))(dθ/dt)+mv(((L(√3))/2)−2R) ={mg+m(dv/dθ)((dθ/dt))}(L+((L(√3))/2)−2R) −mg((L/2))cos θ say M=pm, R=qL −(p/3)ω+ω(((√3)/2)−2q)+(g/(2L))cos θ =((g/L)+((ωdω)/dθ))(1+((√3)/2)−2q) say 1+((√3)/2)−2q=k (k−1−(p/3))ω+(g/(2L))(cos θ−2k) =((kωdω)/dθ) ....$
	$τ = T L + T (\frac{L \sqrt{3}}{2} - 2 R) - m g (\frac{L}{2}) \cos θ$ $T = m g + m \frac{d v}{d t}$ $v = - L \frac{d θ}{d t}$ $(\frac{{M L}^{2}}{3}) \frac{d θ}{d t} + m v (\frac{L \sqrt{3}}{2} - 2 R)$ $= {m g + m \frac{d v}{d θ} (\frac{d θ}{d t})} (L + \frac{L \sqrt{3}}{2} - 2 R)$ $- m g (\frac{L}{2}) \cos θ$ $s a y M = p m, R = q L$ $- \frac{p}{3} ω + ω (\frac{\sqrt{3}}{2} - 2 q) + \frac{g}{2 L} \cos θ$ $= (\frac{g}{L} + \frac{ω d ω}{d θ}) (1 + \frac{\sqrt{3}}{2} - 2 q)$ $s a y 1 + \frac{\sqrt{3}}{2} - 2 q = k$ $(k - 1 - \frac{p}{3}) ω + \frac{g}{2 L} (\cos θ - 2 k)$ $= \frac{k ω d ω}{d θ}$ $. . . .$
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