# A cylinder has inner and outer radii of 2 cm and 3 cm, respectively, and a mass of 1 kg. If the cylinder's frequency of counterclockwise rotation about its center changes from 10 Hz to 15 Hz, by how much does its angular momentum change?

Apr 24, 2018

The change in angular momentum is $= 0.041 k g {m}^{2} {s}^{-} 1$

#### 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 $m = 1 k g$

The radii of the cylinder are ${r}_{1} = 0.02 m$ and ${r}_{2} = 0.03 m$

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

So, $I = 1 \cdot \frac{\left({0.02}^{2} + {0.03}^{2}\right)}{2} = 0.0013 k g {m}^{2}$

The change in angular velocity is

$\Delta \omega = \Delta f \cdot 2 \pi = \left(15 - 10\right) \times 2 \pi = 10 \pi r a {\mathrm{ds}}^{-} 1$

The change in angular momentum is

$\Delta L = I \Delta \omega = 0.0013 \times 10 \pi = 0.041 k g {m}^{2} {s}^{-} 1$