## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

We can convert the angular velocity to units of rad/s. $\omega = (100~rpm)(2\pi~rad/rev)(\frac{1~min}{60~s})$ $\omega = 10.5~rad/s$ We can find the rotational inertia of the baton. $I = \frac{1}{12}ML^2$ $I = \frac{1}{12}(0.40~kg)(0.96~m)^2$ $I = 0.0307~kg~m^2$ We can find the rotational kinetic energy. $KE_{rot} = \frac{1}{2}I~\omega^2$ $KE_{rot} = \frac{1}{2}(0.0307~kg~m^2)(10.5~rad/s)^2$ $KE_{rot} = 1.7~J$ The rotational kinetic energy is 1.7 J