# A cylinder has inner and outer radii of 8 cm and 12 cm, respectively, and a mass of 8 kg. If the cylinder's frequency of rotation about its center changes from 3 Hz to 2 Hz, by how much does its angular momentum change?

Jan 26, 2017

The answer is $= 0.523 k g m {s}^{- 1}$

#### Explanation:

The angular momentum is $L = I \omega$

and $\Delta L = I \Delta \omega$

where $I$ is the moment of inertia

For a cylinder, $I = \frac{m}{2} \left({r}_{1}^{2} + {r}_{2}^{2}\right)$

So, $I = \frac{8}{2} \cdot \left({0.08}^{2} + {0.12}^{2}\right) = 4 \cdot 0.0208 = 0.0832 k g {m}^{2}$

$\Delta \omega = \left(3 - 2\right) \cdot 2 \pi = 2 \pi r a {\mathrm{ds}}^{-} 1$

$\Delta L = 0.0832 \cdot 2 \pi = 0.523 k g m {s}^{- 1}$