Physics - Rotation Question with Solution | TestHub

The moment of inertia of disk A is $I_A=\frac{1}{2}M{R}^2$. The moment of inertia of disk B is $I_B=\frac{1}{2}(\frac{M}{2})(\frac{R}{2}{)}^2=\frac{1}{16}M{R}^2$. The combined moment of inertia is $P=I_{total}=I_A+I_B=\frac{1}{2}M{R}^2+\frac{1}{16}M{R}^2=\frac{9}{16}M{R}^2$, so P-(1). By conservation of angular momentum, $I_A{\omega }_0=(I_A+I_B){\omega }_f$. So, $\frac{1}{2}M{R}^2{\omega }_0=\frac{9}{16}M{R}^2{\omega }_f\Rightarrow Q={\omega }_f=\frac{8}{9}{\omega }_0$, so Q-(2). The initial kinetic energy is $K_i=\frac{1}{2}{I}_A{\omega }_0^2=\frac{1}{4}M{R}^2{\omega }_0^2$. The final kinetic energy is $K_f=\frac{1}{2}{I}_{total}{\omega }_f^2=\frac{1}{2}(\frac{9}{16}M{R}^2)(\frac{8}{9}{\omega }_0{)}^2=\frac{2}{9}M{R}^2{\omega }_0^2$. The loss in kinetic energy is $R=K_i-K_f=(\frac{1}{4}-\frac{2}{9})M{R}^2{\omega }_0^2=\frac{1}{36}M{R}^2{\omega }_0^2$, so R-(3). The angular impulse on disk B is its change in angular momentum: $S=I_B{\omega }_f-0=(\frac{1}{16}M{R}^2)(\frac{8}{9}{\omega }_0)=\frac{1}{18}M{R}^2{\omega }_0$, so S-(4).