Question

A disc of moment of inertia $\frac{9.8}{{\pi }^2}{\text{kgm}}^2$ is rotating at $600\text{rpm}$. If the frequency of rotation changes from $600\text{rpm}$ to $300\text{rpm}$, then what is the work done.

• A. $1467\text{J}$
• B. $1452\text{J}$
• C. $1567\text{J}$
• D. $1632\text{J}$

Answer 1

The correct option is A $1467\text{J}$

According to Work energy theorem

Workdone $=$ Change in kinetic energy

$W=\frac{1}{2}I\times ({\omega }_1^2-{\omega }_2^2)$

$=\frac{1}{2}I\times 4{\pi }^2({\nu }_1^2-{\nu }_2^2)$

$=\frac{1}{2}\times \frac{9.8}{{\pi }^2}\times 4{\pi }^2({10}^2-5^2)$

$=9.8\times 2\times 75=1470\text{J}$

Hence, option $\left(A\right)$ is correct.