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Question Number 174728 by ajfour last updated on 09/Aug/22

Commented by ajfour last updated on 09/Aug/22

$I f r o d i s r e l e a s e d a s s h o w n a n d$ $r o d r o l l s o v e r t h e f i x e d s e m i -$ $c y l i n d e r h o w l o n g d o e s i t t a k e f o r$ $e n d B t o s t r i k e t h e g r o u n d, t h e$ $o t h e r s i d e ?$
Commented by mr W last updated on 14/Aug/22

$i c a n o n l y g e t a p p r o x i m a t e l y$ $T \approx 2.8740 \sqrt{\frac{R}{g}}$
Commented by ajfour last updated on 14/Aug/22

$I m m e n s e w o r k a n d g u i d a n c e,$ $s i r, i s h a l l c h e c k c a r e f u l l y .$
	Answered by mr W last updated on 13/Aug/22

	Commented by mr W last updated on 15/Aug/22

	$A B = L = 4 R$ $A C = a - R θ$ $I_{C} = \frac{L^{2} m}{12} + {(\frac{L}{2} - a + R θ)}^{2} m$ $l e t b = \frac{L}{2} - a$ $I_{C} = \frac{L^{2} m}{12} + {(b + R θ)}^{2} m$ $y_{M} = R \sin θ - (b + R θ) \cos θ$ $\frac{I_{C} ω^{2}}{2} = m g (\frac{4 R}{3} - R \sin θ + (b + R θ) \cos θ)$ $\frac{ω^{2}}{2} [\frac{L^{2} m}{12} + {(b + R θ)}^{2} m] = m g (\frac{4 R}{3} - R \sin θ + (b + R θ) \cos θ)$ $ω^{2} [\frac{4}{3} + {(\frac{b}{R} + θ)}^{2}] = \frac{2 g}{R} [\frac{4}{3} - \sin θ + (\frac{b}{R} + θ) \cos θ]$ $ω = - \frac{d θ}{d t} = \sqrt{\frac{2 g}{R}} \sqrt{\frac{\frac{4}{3} - \sin θ + (\frac{b}{R} + θ) \cos θ}{\frac{4}{3} + {(\frac{b}{R} + θ)}^{2}}}$ $\int_{0}^{T} d t = \sqrt{\frac{R}{2 g}} \int_{θ_{1}}^{θ_{0}} \sqrt{\frac{\frac{4}{3} + {(\frac{b}{R} + θ)}^{2}}{\frac{4}{3} - \sin θ + (\frac{b}{R} + θ) \cos θ}} d θ$ $T = \sqrt{\frac{R}{2 g}} \int_{θ_{1}}^{θ_{0}} \sqrt{\frac{\frac{4}{3} + {(\frac{b}{R} + θ)}^{2}}{\frac{4}{3} - \sin θ + (\frac{b}{R} + θ) \cos θ}} d θ$ $T = \sqrt{\frac{R}{2 g}} \int_{θ_{1}}^{θ_{0}} \sqrt{\frac{\frac{4}{3} + {[θ - (\frac{π + \sqrt{5}}{2} + \sin^{- 1} \frac{2}{3} - 2)]}^{2}}{\frac{4}{3} - \sin θ + [θ - (\frac{π + \sqrt{5}}{2} + \sin^{- 1} \frac{2}{3} - 2)] \cos θ}} d θ$ $T = \sqrt{\frac{R}{2 g}} \int_{1.0097}^{2.3005} \sqrt{\frac{\frac{4}{3} + {(θ - 1.418558)}^{2}}{\frac{4}{3} - \sin θ + (θ - 1.418558) \cos θ}} d θ$ $T \approx 2.8740 \sqrt{\frac{R}{g}}$
	Commented by ajfour last updated on 14/Aug/22
	$let centre of semicircle be O. L=2l I_0 =I_m +m{r^2 +(a−rθ)^2 } I_m =((ml^2 )/3) mg{rcos θ+[l−(a−rθ)]sin θ}dt =d{I_0 ((dθ/dt))} g{rcos θ+(l+rθ−a)sin θ}dθ =((dθ/dt))d{[(l^2 /3)+r^2 +(a−rθ)^2 ]((dθ/dt))} here l=2r , a=((√5)/2)r ((g/r)){cos θ+(θ+2−((√5)/2))sin θ}dθ =ωd{[(7/3)+(((√5)/2)−θ)^2 ]ω} MjS sir may let us know with some great calculator θ(t). Given (g/r)=c, boundary value θ=sin^(−1) (2/3) for ω=0.$
	$l e t c e n t r e o f s e m i c i r c l e b e O .$ $L = 2 l$ $I_{0} = I_{m} + m {r^{2} + {(a - r θ)}^{2}}$ $I_{m} = \frac{{m l}^{2}}{3}$ $m g {r \cos θ + [l - (a - r θ)] \sin θ} d t$ $= d {I_{0} (\frac{d θ}{d t})}$ $g {r \cos θ + (l + r θ - a) \sin θ} d θ$ $= (\frac{d θ}{d t}) d {[\frac{l^{2}}{3} + r^{2} + {(a - r θ)}^{2}] (\frac{d θ}{d t})}$ $h e r e l = 2 r, a = \frac{\sqrt{5}}{2} r$ $(\frac{g}{r}) {\cos θ + (θ + 2 - \frac{\sqrt{5}}{2}) \sin θ} d θ$ $= ω d {[\frac{7}{3} + {(\frac{\sqrt{5}}{2} - θ)}^{2}] ω}$ $M j S s i r m a y l e t u s k n o w$ $w i t h s o m e g r e a t c a l c u l a t o r$ $θ (t) . G i v e n \frac{g}{r} = c, b o u n d a r y v a l u e$ $θ = \sin^{- 1} \frac{2}{3} f o r ω = 0 .$
	Commented by mr W last updated on 14/Aug/22

	Commented by mr W last updated on 14/Aug/22

	$L = 4 R$ $θ_{0} = \frac{π}{2} + \sin^{- 1} \frac{2}{3} \approx 2.3005 (131.8 °)$ $A C = \sqrt{{(\frac{3 R}{2})}^{2} - R^{2}} = \frac{\sqrt{5} R}{2} = a - (\frac{π}{2} + \sin^{- 1} \frac{2}{3}) R$ $\frac{a}{R} = \frac{\sqrt{5}}{2} + (\frac{π}{2} + \sin^{- 1} \frac{2}{3})$ $b = \frac{L}{2} - a$ $\frac{b}{R} = - (\frac{π + \sqrt{5}}{2} + \sin^{- 1} \frac{2}{3} - 2)$ $y_{M} = \frac{R}{\frac{3 R}{2}} \times \frac{L}{2} = \frac{4 R}{3}$
	Commented by mr W last updated on 14/Aug/22

	Commented by mr W last updated on 14/Aug/22

	$y_{B} = R \sin θ - (L - a + R θ) \cos θ$ $y_{B} = R \sin θ - (\frac{L}{2} + b + R θ) \cos θ$ $R \sin θ_{1} - (\frac{L}{2} + b + R θ_{1}) \cos θ_{1} = 0$ $\sin θ_{1} - (2 + \frac{b}{R} + θ_{1}) \cos θ_{1} = 0$ $\sin θ_{1} - (4 - \frac{π + \sqrt{5}}{2} - \sin^{- 1} \frac{2}{3} + θ_{1}) \cos θ_{1} = 0$ $\Rightarrow θ_{1} \approx 1.0097 (57.9 °)$
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