Question

A disc of mass $2kg$ and radius $\text{0.2}m$ is rotating with an angular velocity of $30rad{s}^{-1}$ . The angular velocity of the disc, if a mass of $\text{0.25}kg$ is put on the periphery is

• A. $24rad{s}^{-1}$
• B. $36rad{s}^{-1}$
• C. $15rad{s}^{-1}$
• D. $26rad{s}^{-1}$

Answer 1

Answer: (A)

If no external torque acts on a system of particles then angular momentum of the system remains constant, that is,
$\tau =0$
$\Rightarrow$ $\frac{dL}{dt}=0$
$\Rightarrow$ $L=I\omega =$ constant
$\Rightarrow$ $I_1{\omega }_1=I_2{\omega }_2$
$\therefore$ $\frac{1}{2}{\left(Mr\right)}^2{\left(\omega \right)}_1=\frac{1}{2}\left(M+2m\right)r^2{\left(\omega \right)}_2$ ...(i)
Here, $M=2kg$ , $m=\text{0.25}kg$ and $r=\text{0.2}m$
${\omega }_1=30rad{s}^{-1}$
Using values in equation (i), we get
$\frac{1}{2}\times 2\times {\left(0.2\right)}^2\times 30=\frac{1}{2}\times \left(2+2\times 0.25\right){\left(0.2\right)}^2\times {\left(\omega \right)}_2$
$\Rightarrow$ $60=2.5{\omega }_2$
$\therefore$ ${\omega }_2=24rad{s}^{-1}$