Question

Each pulley in the figure has radius $r$ and moment of inertia $I$. The acceleration of the block is

• A. $\frac{(M-m)g}{\left(M+m+\frac{2I}{r^2}\right)}$
• B. $\frac{(M-m)g}{\left(M+m-\frac{2I}{r^2}\right)}$
• C. $\frac{(M-m)g}{\left(M+m+\frac{I}{r^2}\right)}$
• D. $\frac{(M-m)g}{\left(M+m-\frac{I}{r^2}\right)}$

Answer 1

Answer: (A)

$T_1-mg=ma...(i)$
$r\left(T_2-T_2\right)=I\alpha ...(ii)$
$Mg-T_3=Ma...(iii)$
$r\left(T_3-T_2\right)=I\alpha ...(iv)$
and $a=R\alpha$
From Eqs. (ii) and (iv), we get
$T_3-T_1=\frac{2Ia}{R^2}$
From Eqs. (i) and (iii), we get
$(M-m)g=(M+m)a+T_3-T_1$
$(M-m)g=(M-m)a+\frac{2Ia}{r^2}$
$\Rightarrow a=\frac{(M-m)g}{\left(M+m+\frac{2I}{r^2}\right)}$