Question

The pulley shown in figure has moment of inertia $I=5{\text{kg-m}}^2$ about its axis and radius $R=1\text{m}$. Find the acceleration of the blocks if the masses of the blocks are $M=3\text{kg}$ and $m=2\text{kg}$. Assume that the string is light and does not slip on the pulley and $g=10{\text{m/s}}^2$.

• A. $3{\text{m/s}}^2$
• B. $1{\text{m/s}}^2$
• C. $2{\text{m/s}}^2$
• D. $\frac{1}{2}{\text{m/s}}^2$

Answer 1

The correct option is B $1{\text{m/s}}^2$


From F.B.D of block $M$
$Mg-T_1=Ma$ ... (i)
From F.B.D of block $m$
$T_2-mg=ma$ ... (ii)
Torque equation about point $\left(\text{O}\right)$ of pulley.
$T_{1}R-T_{2}R=I\alpha$
As string does not slip over the rim of the pulley,
$\alpha =\frac{a}{R}$
$\Rightarrow T_{1}R-T_{2}R=I\alpha =\frac{Ia}{R}$ ... (iii)
Substituting $T_1$ and $T_2$ from equation (i) and (ii)
$\left[M\right(g-a)-m(g+a\left)\right]R=\frac{Ia}{R}$
Solving
$a=\frac{(M-m)g{R}^2}{I+(M+m)R^2}$
$=\frac{(3-2)\times 10\times 1^2}{5+(3+2)\times 1^2}$
$=\frac{10}{10}=1{\text{m/s}}^2$