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

The angular momentum is $2.1~kg~m^2/s$
We can express the angular velocity in units of rad/s; $\omega = (120~rpm)(\frac{2\pi~rad}{1~rev})(\frac{1~min}{60~s})$ $\omega = 4\pi~rad/s$ We can find the moment of inertia of the bar as; $I = \frac{1}{12}MR^2$ $I = \frac{1}{12}(0.50~kg)(2.0~m)^2$ $I = 0.167~kg~m^2$ We can find the angular momentum about the axle: $L = I~\omega$ $L = (0.167~kg~m^2)(4\pi~rad/s)$ $L = 2.1~kg~m^2/s$ The angular momentum is $2.1~kg~m^2/s$.