Question

The pulley in figure are identical, each having a radius R and moment of inertia I. Find the acceleration of the block M.

Answer 1

As per the given criteria,

Assume, Acceleration = a

angular acceleration, $\alpha =\frac{a}{R}$

for M , tension = T

and For m, tension = T'

Now, the tension between the pulley = T''

Therefore ,

$Mg-T=Ma⟹T=M(g-a)$

Now ,

$T^{\prime }-mg=ma⟹T^{\prime }=m(g+a)$

Now For the pulleys,

$\alpha =\frac{a}{R}⟹\alpha =(T-T^{\prime \prime })\frac{R}{I}⟹\frac{a}{R}=(T-T^{\prime \prime })\frac{R}{I}⟹a=(T-T^{\prime \prime })\frac{R^2}{I}$

Now solving these two values of a , we get

$2a=(T-T^{\prime })\frac{R^2}{I}$

Now, put the value of T & T', we get

$2a=\left(M\right(g-a)-m(g+a\left)\right)\frac{R^2}{l}⟹2al=(M-m)g{R}^2-(M+m)g{R}^2⟹2a(2I+M{R}^2+m{R}^2)=(M-m)g{R}^2⟹a=\frac{(M-m)g{R}^2}{2I}+M{R}^2+m{R}^2⟹a=\frac{(M-m)g{R}^2}{M+m}+\frac{2I}{R^2}$

Hence the acceleration of the block at M is

$\frac{(M-m)g}{M+m}+\frac{2I}{R^2}$