Question

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

• A. $24 \, r a d \, s^{- 1}$
• B. $36 \, r a d \, s^{- 1}$
• C. $15 \, r a d \, s^{- 1}$
• D. $26 \, r a d \, s^{- 1}$

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 $ $L=Iω=$ constant
$\Rightarrow $ $I_{1}\omega _{1}=I_{2}\omega _{2}$
$\therefore \frac{1}{2} Mr ^{2} \omega_{1}=\frac{1}{2}( M +2 m ) r ^{2} \omega_{2} \ldots$ (i) ...(i)
Here, $M=2kg$ , $m=\text{0.25}kg$ and $r=\text{0.2}m$
$\omega _{1}=30 \, rads^{- 1}$
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 $ $\omega _{2}=24 \, rads^{- 1}$