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Question Number 53881 by ajfour last updated on 26/Jan/19

Commented by ajfour last updated on 26/Jan/19

$I f θ (0) = θ_{0}, f i n d x (θ) a n d i f p o s s i b l e$ $θ (t) . I s t h e r e a n x_{m a x} ?$
Commented by tanmay.chaudhury50@gmail.com last updated on 27/Jan/19
$when time is t position of rod as shown[as dotted 1) here Torque is variable as m mass move 2)Torque at time t is Mg(L/2)cosθ(t)+mg(L−x(θ)cosθ(t)=(d/dt)(Iw) Right hand side=w(d/dt)[((ML^2 )/3)+m{(L−x(θ)}^2 ]+[((ML^2 )/3)+m{(L−x(θ)}^2 ]α =w{2m(L−x)(dx/dt)}+[M(L^2 /3)+m(L−x)^2 ](dw/dt) continue...$
$w h e n t i m e i s t p o s i t i o n o f r o d a s s h o w n [a s d o t t e d$ $1) h e r e T o r q u e i s v a r i a b l e a s m m a s s m o v e$ $2) T o r q u e a t t i m e t i s$ $M g \frac{L}{2} c o s θ (t) + m g (L - x (θ) c o s θ (t) = \frac{d}{d t} (I w)$ $R i g h t h a n d s i d e = w \frac{d}{d t} [\frac{{M L}^{2}}{3} + m {{(L - x (θ)}}^{2}] + [\frac{{M L}^{2}}{3} + m {{(L - x (θ)}}^{2}] α$ $= w {2 m (L - x) \frac{d x}{d t}} + [M \frac{L^{2}}{3} + m {(L - x)}^{2}] \frac{d w}{d t}$ $c o n t i n u e . . .$
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