Question

A circular disc of radius ${}^{\prime }{r}^{\prime }$ and mass ${}^{\prime }{m}^{\prime }$ is rotating about an axis passing through its COM perpendicular to its plane with an angular speed $\omega$. If the radius of the disc is halved and the angular speed is doubled, which of the following options is correct?

• A. $K.E_1=\frac{K.E_2}{2}$
• B. $K.E_1=2K.E_2$
• C. $K.E_1=K.E_2$
• D. $K.E_1=\frac{K.E_2}{4}$

Answer 1

The correct option is C $K.E_1=K.E_2$

Given,
$r_1=r,m_1=m,w_1=w$
$r_2=\frac{r}{2};m_2=m;w_2=2w$
Rotational kinetic energy for $1st$ case: $(K.E_1)=\frac{1}{2}{I}_1{w}_1^2$
$=\frac{1}{2}\left(\frac{m_1{r}_1^2}{2}\right)w_1^2$
Rotational kinetic energy for $2nd$ case: $(K.E_2)=\frac{1}{2}{I}_2{w}_2^2$
$=\frac{1}{2}\left(\frac{m_2{r}_2^2}{2}\right)w_2^2$

$\therefore \frac{K.E_1}{K.E_2}=\frac{r_1^2}{r_2^2}\times \frac{w_1^2}{w_2^2}$
$\frac{K.E_1}{K.E_2}={\left(\frac{2r}{r}\right)}^2{\left(\frac{w}{2w}\right)}^2=1$
$\Rightarrow K.E_1=K.E_2$