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Question Number 214047 by ajfour last updated on 25/Nov/24

Commented by ajfour last updated on 25/Nov/24

$T h e L s h a p e d r i g i d s c a l e c a n t u r n$ $a b o u t t h e p i v o t o n a f r i c t i o n l e s s t a b l e .$ $T h e s m a l l m a s s m h i t s a n d a f t e r$ $r e b o u n d i n g h a p p e n s t o f i n d t h e$ $o t h e r e n d . F i n d θ, i f α i s m e a s u r e d .$ $(T a k e c o e f f i c i e n t o f r e s t i t u t i o n e)$
	Answered by mr W last updated on 25/Nov/24

	Commented by mr W last updated on 27/Nov/24

	$u = v e l o c i t y o f b a l l b e f o r e c o l l i s i o n$ $v = v e l o c i t y o f b a l l a f t e r c o l l i s i o n$ $ω = a n g u l a r v e l o c i t y o f L - f r a m e$ $a f t e r c o l l i s i o n$ $u \sin θ = v \sin ϕ . . . (i)$ $e = \frac{v \cos ϕ + \frac{ω l}{2}}{u \cos θ}$ $\Rightarrow e u \cos θ = v \cos ϕ + \frac{l ω}{2} . . . (i i)$ $m u \cos θ \times \frac{l}{2} = \frac{{M l}^{2} ω}{3} - m v \cos ϕ \times \frac{l}{2}$ $\Rightarrow u \cos θ + v \cos ϕ = \frac{2 M l ω}{3 m} . . . (i i i)$ $t = \frac{α}{ω}$ $\frac{l}{2} - l \sin α = v \sin ϕ \times t$ $\frac{l ω}{α} (\frac{1}{2} - \sin α) = v \sin ϕ$ $\Rightarrow (\frac{1}{2 α} - \frac{\sin α}{α}) l ω = u \sin θ . . . (i v)$ $l \cos α = v \cos ϕ \times t$ $\frac{l ω \cos α}{α} = v \cos ϕ . . . (v)$ $w e h a v e 5 e q u a t i o n s f o r 6 v a r i a b l e s :$ $u, v, ω, θ, ϕ, α$ $(i i i) :$ $u \cos θ + \frac{l ω \cos α}{α} = \frac{2 M l ω}{3 m}$ $(\frac{2 M}{3 m} - \frac{\cos α}{α}) l ω = u \cos θ . . . (v i)$ $(v) i n t o (i i) :$ $e u \cos θ = \frac{l ω \cos α}{α} + \frac{l ω}{2}$ $\Rightarrow (\frac{\cos α}{α} + \frac{1}{2}) \frac{l ω}{e} = u \cos θ$ $f r o m (v i) :$ $(\frac{\cos α}{α} + \frac{1}{2}) \frac{l ω}{e} = (\frac{2 M}{3 m} - \frac{\cos α}{α}) l ω$ $(1 + \frac{1}{e}) \frac{\cos α}{α} = \frac{2 M}{3 m} - \frac{1}{2 e}$ $\Rightarrow \frac{\cos α}{α} = \frac{\frac{2 M}{3 m} - \frac{1}{2 e}}{1 + \frac{1}{e}} . . . (v i i)$ $s i n c e α < \frac{π}{6}, \frac{\frac{2 M}{3 m} - \frac{1}{2 e}}{1 + \frac{1}{e}} > \frac{\sqrt{3} \times 6}{2 \times π}$ $\Rightarrow \frac{M}{m} > \frac{9 \sqrt{3} (1 + \frac{1}{e})}{2 π} + \frac{3}{4 e}$ $(i v) / (v i) :$ $\frac{\frac{1}{2 α} - \frac{\sin α}{α}}{\frac{2 M}{3 m} - \frac{\cos α}{α}} = \tan θ$ $\Rightarrow θ = \tan^{- 1} \frac{\frac{1}{2} - \sin α}{\frac{2 α M}{3 m} - \cos α} . . . (v i i i)$ $f r o m (i v) :$ $\frac{l ω}{u} = \frac{α \sin θ}{\frac{1}{2} - \sin α} . . . (i x)$ ${(i)}^{2} + {(v)}^{2} :$ ${(u \sin θ)}^{2} + {(\frac{l ω \cos α}{α})}^{2} = v^{2}$ $\frac{v^{2}}{u^{2}} = \sin^{2} θ + {(\frac{l ω}{u})}^{2} {(\frac{\cos α}{α})}^{2}$ $\Rightarrow \frac{v}{u} = \sin θ \sqrt{1 + {(\frac{α}{\frac{1}{2} - \sin α})}^{2} {(\frac{\frac{2 M}{3 m} - \frac{1}{2 e}}{1 + \frac{1}{e}})}^{2}} . . . (x)$ $(i) / (v) :$ $\frac{u \sin θ}{\frac{l ω \cos α}{α}} = \tan ϕ$ $\tan ϕ = \frac{\sin θ}{(\frac{l ω}{u}) (\frac{\cos α}{α})} = (\frac{\frac{1}{2} - \sin α}{α}) (\frac{1 + \frac{1}{e}}{\frac{2 M}{3 m} - \frac{1}{2 e}})$ $\Rightarrow ϕ = \tan^{- 1} (\frac{\frac{1}{2} - \sin α}{α}) (\frac{1 + \frac{1}{e}}{\frac{2 M}{3 m} - \frac{1}{2 e}}) . . . (x i)$
	Commented by mr W last updated on 27/Nov/24

	$e x a m p l e :$ $\frac{M}{m} = 8, e = 0.75$ $\Rightarrow α \approx 0.3370 \approx 19.31 °$ $\Rightarrow θ \approx 0.1958 \approx 11.22 °$ $\Rightarrow ϕ \approx 0.2462 \approx 14.11 °$ $\Rightarrow \frac{v}{u} \approx 0.7986$ $\Rightarrow \frac{l ω}{u} \approx 0.3873$
	Commented by ajfour last updated on 26/Nov/24

	$s^{2} = l^{2} + \frac{l^{2}}{4} - \frac{l^{2}}{2} \sin α$ $s^{2} = \frac{l^{2}}{4} (3 - 2 \sin α) = v^{2} (\frac{α^{2}}{ω^{2}})$ $\Rightarrow \frac{s}{α l} = \frac{v}{ω l} = \frac{\sqrt{3 - 2 \sin α}}{2 α}$ $m u (\frac{l}{2} \cos θ) = \frac{{M l}^{2} ω}{3} - m v (\frac{l}{2}) \cos ϕ$ $u \sin θ = v \sin ϕ$ $&$ $m (\frac{u}{v}) (\frac{l}{2} \cos θ) = \frac{{M l}^{2}}{3} (\frac{ω}{v}) - m \cos ϕ (\frac{l}{2})$ $\frac{\sin ϕ \cos θ}{\sin θ} + \cos ϕ = \frac{2 M}{3 m} (\frac{α l}{s})$ $\cot θ (\frac{l}{2} - l \sin α) + l \cos α = (\frac{2 M}{3 m}) α l$ $\cot θ = \frac{(\frac{2 M}{3 m}) α - \cos α}{\frac{1}{2} - \sin α}$ $\tan θ = \frac{\frac{1}{2} - \sin α}{(\frac{2 M}{3 m}) α - \cos α}$
	Commented by ajfour last updated on 26/Nov/24
	yeah! And I got the same, another way, Sir.
	Commented by mr W last updated on 26/Nov/24

	$c a n y o u p l e a s e c h e c k f o l l o w i n g :$ $α i s n o t i n d e p e n d e n t . i t i s g i v e n$ $t h r o u g h$ $\frac{\cos α}{α} = \frac{\frac{2 M}{3 m} - \frac{1}{2 e}}{1 + \frac{1}{e}}$
	Commented by ajfour last updated on 26/Nov/24

	$h i n g e r e a c t i o n i s a l s o t h e r e, i t h i n k$ $f r a m e i s n o t f r e e, s o t h i s s h o u l d n o t$ $a p p l y .$
	Commented by mr W last updated on 26/Nov/24

	$s e e a b o v e$
	Commented by ajfour last updated on 27/Nov/24

	$N i c e s i r, e d e c i d e s a l l a n g l e s$ $f o r t h i s c a s e!$
	Commented by mr W last updated on 27/Nov/24

	$y e s!$ $b e s i d e {i t}^{'} s p o s s i b l e o n l y i f$ $\frac{M}{m} > \frac{9 \sqrt{3} (1 + 1 / e)}{3 π} + \frac{3}{4 e}$
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