A) \[\frac{[I+m{{R}^{2}}]g}{(m+M)R}\]
B) \[\frac{(M+m)g{{R}^{2}}}{I-(M+m)I{{R}^{2}}}\]
C) \[\frac{(M+m)gR}{(m+M)I}\]
D) \[\frac{(M-m)g{{R}^{2}}}{I+(M+m){{R}^{2}}}\]
Correct Answer: D
Solution :
Consider the situation Suppose, the acceleration of blocks is a therefore the angular acceleration of the pully will be \[\alpha =\frac{a}{R}\] For motion of \[M,\]\[Mg-{{T}_{1}}=Ma\] ...(i) For motion of \[m,\]\[{{T}_{2}}-mg=ma\] ...(ii) Also net torque on pully, \[\tau ={{T}_{1}}R-{{T}_{2}}R=I\alpha =\frac{Ia}{R}\] ?(iii) Putting \[{{T}_{1}}\] and \[{{T}_{2}}\] from Eqs. (i), (ii) and (iii) we get, \[[M(g-a)-m\,(g+a)]R=I\,a/R\] Which gives, \[a=\frac{(M-m)\,g{{R}^{2}}}{[I+(M+m){{R}^{2}}]}\]You need to login to perform this action.
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