Question

A disc of mass $2 \, kg$ and radius $0.2m$ is rotating with an angular velocity $30 \, rads^{- 1}$ , about an axis passing through its centre and perpendicular to its plane. If a mass of $0.25 \, kg$ is stuck on the periphery of the disc, then its angular velocity will become

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

Answer 1

Answer: (A)

When there is no external torque then the angular momentum remains constant.
i.e. $\tau=0\Rightarrow \frac{d L}{d t}=0$
$\therefore \, \, \, I_{1}\omega _{1}=I_{2}\omega _{2}$ ... (i)
Here $M=2 \, kg, \, m=0.25 \, kg, \, r=0.2m,$
$\omega _{1}=30 \, rad \, s^{- 1}$
Hence, putting the value of $\omega _{1}$ in equation (i), we get
$\Rightarrow \, \, \, \frac{1}{2}\times 2\times \left(0.2\right)^{2}\times 30=\frac{1}{2}\times \left(2 + 2 \times 0.25\right)\left(0.2\right)^{2}\times \left(\omega \right)_{2}$
$\Rightarrow \, \, \, 1.2=0.05\omega _{2}$
$\Rightarrow \, \, \, \omega _{2}=24 \, rad \, s^{- 1}$