Question

A body with moment of inertia I = 10 kgm² rotates uniformly with T = 2s. Find the moment of momentum of the body, relative to the axis of rotation.

Answer 1

By definition, the moment of momentum is:

$$L = I\omega$$

where $I = 10kg\cdot m^2$ is the moment of inertia, and $\omega$ is the angular speed. In turn, the angular speed is:

$$\omega = \dfrac{2\pi}{T}$$

where $T = 2s$ is the period of rotation. Thus, obtian:

$$L = \dfrac{2\pi I}{T}\\ L = \dfrac{2\pi \cdot 10}{2} = 10\pi\approx 31.4\space \dfrac{kg\cdot m^2}{s}$$

Answer. $31.4\space \dfrac{kg\cdot m^2}{s}$.