Question

The moment of inertia of pulley is $I$ and its radius is $R$. The string doesn't slips on the pulley as system is released then, the acceleration of the system is:

• A. $\frac{g}{3+\frac{I}{m{R}^2}}$
• B. $\frac{g}{1+\frac{I}{m{R}^2}}$
• C. $\frac{3g}{1+\frac{I}{m{R}^2}}$
• D. $\frac{3g}{2+\frac{I}{m{R}^2}}$

Answer 1

Answer: (A)

$\frac{g}{3+\frac{I}{m{R}^2}}$

Given,

Tension in both cables is $T_{1}and{T}_2$

The equation of forces on both masses

$T_1-mg=ma\Rightarrow T_1=(ma+mg)......\left(1\right)$

$2mg-T_2=2ma\Rightarrow T_2=2(mg-ma)......\left(2\right)$

$torqueonpulley,(T_2-T_1)R=I\alpha ......\left(3\right)$

From (1), (2) and (3)

$=I\alpha =[2(mg-ma)-(ma+mg)]R$

$\Rightarrow I\alpha R=(mg-3ma)R^2$

$\Rightarrow Ia+3ma{R}^2=mg{R}^2$

$\Rightarrow a=\frac{mg{R}^2}{I+3m{R}^2}$

Acceleration of system is $\frac{mg{R}^2}{I+3m{R}^2}$