# A cylinder has inner and outer radii of 12 cm and 18 cm, respectively, and a mass of 6 kg. If the cylinder's frequency of rotation about its center changes from 4 Hz to 9 Hz, by how much does its angular momentum change?

Feb 8, 2017

The answer is $= 1.47 k g {m}^{2} {s}^{- 1}$

#### Explanation:

The angular momentum is $L = I \omega$

where $I$ is he [moment of inertia

The change in angular momentum is

$\Delta L = I \Delta \omega$

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

So, $I = 6 \cdot \frac{{0.12}^{2} + {0.18}^{2}}{2} = 0.0468 k g {m}^{2}$

$\Delta \omega = \left(9 - 4\right) \cdot 2 \pi = 10 \pi r a {\mathrm{ds}}^{-} 1$

$\Delta L = 0.0468 \cdot 10 \pi = 1.47 k g {m}^{2} {s}^{- 1}$