## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

We can find the rod's moment of inertia. $I = \frac{1}{3}ML^2$ $I = \frac{1}{3}(0.75~kg)(0.50~m)^2$ $I = 0.0625~kg~m^2$ The angular impulse is equal to the change in angular momentum. We can find the angular impulse exerted by the force. $\Delta L = \tau ~t$ $\Delta L = (r\times F)~t$ $\Delta L = (0.25~m)(1000~N)(2.0\times 10^{-3}~s)$ $\Delta L = 0.50~N~s$ Since the rod started from rest, the angular momentum just after the blow from the force is $0.50~N~s$. We can find the angular velocity immediately after the blow. $L = I~\omega$ $\omega = \frac{L}{I}$ $\omega = \frac{0.50~N~s}{0.0625~kg~m^2}$ $\omega = 8.0~rad/s$ The angular velocity is 8.0 rad/s