The acceleration of the cylinder in the downward direction is determined as follows:From the figure, \( mg-2T=ma \)______ (1) \( 2Tr=\frac{1}{2}m{r}^{2}\left(\frac{a}{r}\right)\) \( 2T=\frac{1}{2}ma\) _______ (2)Substituting eqn. (2) in (1), we get \( mg=ma+\frac{1}{2}ma\) \( mg=\frac{3}{2}ma\) \( a=\left(\frac{2}{3}\right)g\) Since the centre of the cylinder moving away from rest position, the velocity \( v\) after it falls from distance \( h\) is, \( {v}^{2}=2\left(\frac{2}{3}g\right)h\) \( v=\sqrt{\frac{4gh}{3}}\) |
A cylinder of mass M is suspended through two strings wrapped around it as shown in figure then
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