Question

A 20-kg hoop travels around a 0.50-m radius circle with an angular velocity of 10 rad/s. The magnitude of its angular momentum about the center of the circle is:

Answer 1

The angular momentum of a particle about the center of rotation is given by the following formula:

$$L = \omega mr^2$$

where $m = 20kg$ is the mass, $r = 0.5m$ is the radius of the circle and $\omega = 10\space rad/s$ is the angular velocity. Thus, obtain:

$$L = \omega mr^2 = 10\cdot 20\cdot 0.5^2 = 50\space m^2\cdot kg/s$$

Answer. $50\space m^2\cdot kg/s$.