# 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?

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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

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

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