Question

The moment of inertia of the pulley as shown in the figure is $4{\text{kg-m}}^2$. Its radius is $1\text{m}$. The system is released from rest. Find the linear velocity of the block, when it has descended through $40\text{cm}$.

• A. $3\sqrt{\frac{6}{7}}\text{m/s}$
• B. $2\sqrt{\frac{6}{7}}\text{m/s}$
• C. $4\sqrt{\frac{6}{7}}\text{m/s}$
• D. $5\sqrt{\frac{6}{7}}\text{m/s}$

Answer 1

The correct option is B $2\sqrt{\frac{6}{7}}\text{m/s}$

Work Done by gravity,

$w_g=mgh=3\times 10\times 0.4=12\text{J}$

According to work - energy theorem,

$W_{ext}=\Delta KE$

$\Rightarrow w_g=\frac{1}{2}I{\omega }^2+\frac{1}{2}m{v}^2$

$\Rightarrow 12=\frac{1}{2}\left(4\right){\omega }^2+\frac{1}{2}\left(3\right)v^2$

Substituting, $\omega =\frac{v}{r}$ and further simplifying, we get,

$12=\frac{7}{2}{v}^2$

$\therefore v=\sqrt{\frac{24}{7}}=2\sqrt{\frac{6}{7}}\text{m/s}$