# A cylinder has inner and outer radii of 9 cm and 14 cm, respectively, and a mass of 9 kg. If the cylinder's frequency of counterclockwise rotation about its center changes from 1 Hz to 7 Hz, by how much does its angular momentum change?

Feb 3, 2018

Change in angular momentum = $\left(I \delta \omega\right)$

now,change in angular velocity= $\delta \omega = 2 \pi \left(7 - 1\right) \frac{r a d}{s}$ i.e $12 \pi \frac{r a d}{s}$

$I$ = moment of inertia of of a cylindrical shell of given outer and inner radii will be $\frac{9}{2} \left({\left(\frac{9}{100}\right)}^{2} + {\left(\frac{14}{100}\right)}^{2}\right) K g . {m}^{2}$ i.e $0.124 K g . {m}^{2}$ (formula is $\frac{M}{2} \left({\left({r}_{1}\right)}^{2} + {\left({r}_{2}\right)}^{2}\right)$)

So,change in angular momentum is $\left(0.124 \cdot 12 \pi\right) N . m . s = 4.674 N . m . s$

Feb 3, 2018

The change in angular momentum is $= 4.7 k g {m}^{2} {s}^{-} 1$

#### 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 $m = 9 k g$

The radii of the cylinder are ${r}_{1} = 0.09 m$ and ${r}_{2} = 0.14 m$

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

So, $I = 9 \cdot \frac{\left({0.09}^{2} + {0.14}^{2}\right)}{2} = 0.12465 k g {m}^{2}$

The change in angular velocity is

$\Delta \omega = \Delta f \cdot 2 \pi = \left(7 - 1\right) \times 2 \pi = 12 \pi r a {\mathrm{ds}}^{-} 1$

The change in angular momentum is

$\Delta L = I \Delta \omega = 0.12465 \times 12 \pi = 4.7 k g {m}^{2} {s}^{-} 1$