Question

The pulleys in figure (10 - E6) are identical, each having a radius R and moment of inertia I. Find the acceleration of the block M.

Answer 1

According to the equation,

$Mg-T_1=Ma...\left(1\right)$

$(T_1-T_2)R=I\frac{a}{R}$

$\Rightarrow (T_1-T_2)=I\frac{a}{R^2}...\left(2\right)$

$\Rightarrow (T_2-T_3)R=I\frac{a}{R^2}...\left(3\right)$

$\Rightarrow T_3-mg=ma...\left(4\right)$

$\text{By adding equation}\left(2\right)\text{and}\left(3\right),\text{we will get,}$

$\Rightarrow (T_1-T_3)=\frac{2Ia}{R^2}...\left(5\right)$

$\text{By adding equation}\left(1\right)\text{and}\left(4\right),\text{we will get,}$

$-mg+Mg+(T_3-T_1)=Ma+ma...\left(6\right)$

$\text{Substituting the value of}{T}_3-T_1\text{we will get,}$

$\Rightarrow Mg-mg=Ma+ma+\frac{2Ia}{R^2}$

$\Rightarrow a=\frac{(M-m)G}{(M+m+2I/R^2)}$