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Question Number 175000 by mr W last updated on 15/Aug/22

Commented by mr W last updated on 16/Aug/22

$a n u n i f o r m r o d w i t h l e n g t h L r o l l s$ $o v e r t h e f i x e d s e m i c y l i n d e r w i t h$ $r a d i u s R .$ $f i n d t h e p e r i o d o f t h e o s c i l l a t i o n .$
	Answered by mr W last updated on 16/Aug/22

	Commented by ajfour last updated on 16/Aug/22
	$mg{−r+rcos θ+(rθ)sin θ} +(m/2)((l^2 /(12))+r^2 θ^2 )((dθ/dt))^2 =K g{−r+rcos θ+(rθ)sin θ} +(l^2 /2)((1/(12))+(r^2 /l^2 )θ^2 )((dθ/dt))^2 =k If θ is small gr{−2sin^2 (θ/2)+θ^2 }+(l^2 /(24))((dθ/dt))^2 =k ⇒ ω^2 =((24)/l^2 ){k−((gr)/2)θ^2 } It seems (to me, as well, Sir) T=((πl)/( (√(3gr)))) .$
	$m g {- r + r \cos θ + (r θ) \sin θ}$ $+ \frac{m}{2} (\frac{l^{2}}{12} + r^{2} θ^{2}) {(\frac{d θ}{d t})}^{2}$ $= K$ $g {- r + r \cos θ + (r θ) \sin θ}$ $+ \frac{l^{2}}{2} (\frac{1}{12} + \frac{r^{2}}{l^{2}} θ^{2}) {(\frac{d θ}{d t})}^{2} = k$ $I f θ i s s m a l l$ $g r {- 2 \sin^{2} \frac{θ}{2} + θ^{2}} + \frac{l^{2}}{24} {(\frac{d θ}{d t})}^{2} = k$ $\Rightarrow ω^{2} = \frac{24}{l^{2}} {k - \frac{g r}{2} θ^{2}}$ $I t s e e m s (t o m e, a s w e l l, S i r)$ $T = \frac{π l}{\sqrt{3 g r}} .$
	Commented by mr W last updated on 16/Aug/22

	$M C = R θ$ $I_{C} = \frac{L^{2} m}{12} + m {(R θ)}^{2}$ $\frac{d}{d t} (I_{C} \frac{d θ}{d t}) = - m g (R θ) \cos θ$ $[\frac{L^{2} m}{12} + m {(R θ)}^{2}] \frac{d^{2} θ}{{d t}^{2}} + 2 {m R}^{2} θ {(\frac{d θ}{d t})}^{2} = - m g (R θ) \cos θ$ $[\frac{L^{2}}{12} + {(R θ)}^{2}] \frac{d^{2} θ}{{d t}^{2}} = - R θ [g \cos θ + 2 R {(\frac{d θ}{d t})}^{2}]$ $f o r v e r y s m a l l θ, w e c a n s a y$ $\approx \frac{L^{2}}{12} \times \frac{d^{2} θ}{{d t}^{2}} = - g R θ$ $\Rightarrow \frac{d^{2} θ}{{d t}^{2}} + \frac{12 g R}{L^{2}} θ = 0$ $ω = \sqrt{\frac{12 g R}{L^{2}}}$ $\Rightarrow T = \frac{2 π}{ω} = \frac{π L}{\sqrt{3 g R}}$
	Commented by mr W last updated on 17/Aug/22

	$y e s s i r, t h a n k s!$
	Commented by Tawa11 last updated on 17/Aug/22

	$Great sir$
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