# A cylinder has inner and outer radii of 3 cm and 9 cm, respectively, and a mass of 4 kg. If the cylinder's frequency of counterclockwise rotation about its center changes from 9 Hz to 6 Hz, by how much does its angular momentum change?

Feb 22, 2018

The change in angular momentum is $= 0.34 k g {m}^{2} {s}^{-} 1$

#### Explanation:

The angular momentum is $L = I \omega$

where $I$ is the moment of inertia

and $\omega$ is the angular velocity

The mass of the cylinder is $m = 4 k g$

The radii of the cylinder are ${r}_{1} = 0.03 m$ and ${r}_{2} = 0.09 m$

For the cylinder, the moment of inertia is $I = m \frac{\left({r}_{1}^{2} + {r}_{2}^{2}\right)}{2}$

So, $I = 4 \cdot \frac{\left({0.03}^{2} + {0.09}^{2}\right)}{2} = 0.018 k g {m}^{2}$

The change in angular velocity is

$\Delta \omega = \Delta f \cdot 2 \pi = \left(9 - 6\right) \times 2 \pi = 6 \pi r a {\mathrm{ds}}^{-} 1$

The change in angular momentum is

$\Delta L = I \Delta \omega = 0.018 \times 6 \pi = 0.34 k g {m}^{2} {s}^{-} 1$