question_answer

Figure shows a pully having moment of inertia \[I\] and radius \[r\]. A string is passing over the fully and is connected to two masses m and M atits ends on either side of the pully. If the string is massless and there is no slippage of rope on the pully, then the accelerations of either of the blocks will be

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, \[a=\frac{(M-m)\,g{{R}^{2}}}{[I+(M+m){{R}^{2}}]}\]