Question

A disc of moment of inertia $\frac{9.8}{{\pi }^2}{\text{kg m}}^2$ 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 approximately?

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

Answer 1

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

${\omega }_i=600\times \frac{2\pi }{60}=20\pi rad/s$
Similarly ${\omega }_f=10\pi rad/s$
Using work energy theorem,
$\begin{array}{cc}WD& =\Delta K{E}_{\text{rot}}\\ & =\frac{1}{2}I{\omega }_f^2-\frac{1}{2}I{\omega }_i^2\\ & =\frac{1}{2}I\left(\right(10\pi {)}^2-\left(20\pi {)}^2\right)\end{array}$

$=\frac{1}{2}\times \frac{9.8}{{\pi }^2}\times (-300{\pi }^2)=-1470J$
For detailed solution watch the next video.