AP EAPCET 23-Aug-2021 Shift 2 Paper

Given, mass of disc, $M=4\text{ kg}$ Radius of disc, $R=0.4\text{ m}$ Initial angular velocity of disc, $\omega_{1}=30\text{ rad s}^{-1}$ Mass of point-masses, $m=0.25\text{ kg}$ Now, moment of inertia of disc, $I_{3}=\frac{1}{2}MR^{2}$ Moment of inertia of disc and two point mass system, $I_{2}=\frac{1}{2}MR^{2}+2mR^{2}$ Let $\omega_{2}$ be the final angular velocity. According to conservation of angular momentum, $I_{1}\omega_{1}=I_{2}\omega_{2}\Rightarrow\omega_{2}=\frac{I_{1}\omega_{1}}{I_{2}}$ Substituting the values, we get $\omega_{2}=\frac{\frac{1}{2}MR^{2}(30)}{\frac{1}{2}MR^{2}+2mR^{2}}$ $=\frac{\frac{1}{2}\times4\times(0.4)^{2}\times30}{\frac{1}{2}\times4\times(0.4)^{2}+2\times0.25\times(0.4)^{2}}$ $\omega_{2}=24\text{ rad/s}$ Hence, the final angular velocity of system is $24\text{ rad/s}$.