## College Physics (4th Edition)

(a) The ratio of the angular momentum after the collapse to the angular momentum before the collapse is 1. (b) $\frac{\omega_f}{\omega_0} = 10^8$ (c) $\frac{KE_f}{KE_0} = 10^8$ (d) The period after the collapse is $0.10~s$
(a) Since there is no external net torque on the star, by conservation of momentum, the ratio of the angular momentum after the collapse to the angular momentum before the collapse is 1. (b) Let $R$ be the original radius. We can find the initial rotational inertia: $I_0 = \frac{2}{5}M~R^2$ We can find the final rotational inertia after the collapse: $I_f = \frac{2}{5}M~(R\times 10^{-4})^2$ $I_f = 10^{-8}\times \frac{2}{5}M~R^2$ $I_f = 10^{-8}\times I_0$ We can use conservation of angular momentum to find the ratio of the angular velocity after the collapse to the angular velocity before the collapse: $L_f = L_0$ $I_f~\omega_f = I_0~\omega_0$ $\frac{\omega_f}{\omega_0} = \frac{I_0}{I_f}$ $\frac{\omega_f}{\omega_0} = \frac{I_0}{10^{-8}\times I_0}$ $\frac{\omega_f}{\omega_0} = 10^8$ (c) We can find the ratio of the rotational kinetic energy after the collapse to the rotational kinetic energy before the collapse: $\frac{KE_f}{KE_0} = \frac{\frac{1}{2}I_f~\omega_f^2}{\frac{1}{2}I_0~\omega_0^2}$ $\frac{KE_f}{KE_0} = \frac{(10^{-8}\times I_0)~(10^8~\omega_0)^2}{I_0~\omega_0^2}$ $\frac{KE_f}{KE_0} = 10^8$ (d) We can write an expression for the period before the collapse: $P_0 = \frac{2\pi}{\omega_0}$ We can find the period after the collapse: $P_f = \frac{2\pi}{\omega_f}$ $P_f = \frac{2\pi}{10^8\omega_0}$ $P_f = 10^{-8}\times \frac{2\pi}{\omega_0}$ $P_f = 10^{-8}\times P_0$ $P_f = 10^{-8}\times (1.0\times 10^7~s)$ $P_f = 0.10~s$ The period after the collapse is $0.10~s$.