# A cylinder has inner and outer radii of 8 cm and 12 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?

Apr 6, 2016

$\Delta L = \frac{9}{2} \left(8.0 \times {10}^{-} 3\right) = 0.432 \pi \approx 1.357 \frac{k g {m}^{2}}{s}$

#### Explanation:

Given:
1) A hollow cylinder radii: R_o=8cm; R_i=12cm
2) Mass, $m = 12 k g$
3) omega_1 =2pif_1; omega_2 =2pif_2; f_1=1Hz; f_2=7Hz

Required change in angular momentum:
$\Delta L = I \left({\omega}_{2} - {\omega}_{1}\right)$

Theorem, Definition and Principles:
Angular Momentum, $L = I \omega$ where
$I$ = Moment of Inertia
$\omega$ = Angular Velocity

Solution Strategy:
A) Compute the Moment of Inertia of Hollow cylinder
B) Convert frequency to radians: $\omega = 2 \pi f$
C) Compute the Angular momentum for ${\omega}_{1} \mathmr{and} {\omega}_{2}$

$\textcolor{b r o w n}{\text{A)}}$ Moment of Inertia of Hollow Cylinder (HC)
Let R_o=8cm; R_i=12cm be outer and inner radii of HC
then we can derive the moment if inertia, I of HC as follows:
$I = \int {r}^{2} \mathrm{dm}$ where $r$ is arbitrary radius in $r : r \in \left[{R}_{o} , {R}_{i}\right]$
Also let $\rho = \frac{M}{V}$ be the density and $l = \text{length of cylinder}$
then $\mathrm{dm} = \rho \textcolor{b l u e}{\mathrm{dv}} = \rho \cdot \textcolor{b l u e}{2 \pi \cdot r \cdot l \cdot \mathrm{dr}}$
Now because we are dealing with HC,
$\rho = \frac{M}{\pi \left({R}_{i}^{2} - {R}_{o}^{2}\right) l}$

Integrate over the inner and outer radiuses and inserting the expression for $\rho$
$I = \rho 2 \pi l {\int}_{{R}_{o}}^{{R}_{i}} {r}^{3} \mathrm{dr}$
$= \frac{M}{\pi \left({R}_{i}^{2} - {R}_{o}^{2}\right) l} 2 \pi l \frac{1}{4} \left[{R}_{i}^{4} - {R}_{o}^{4}\right]$

$= \frac{M}{\cancel{\pi \left({R}_{i}^{2} - {R}_{o}^{2}\right)} l} 2 l \frac{1}{4} \cancel{\pi \left[{R}_{i}^{2} - {R}_{o}^{2}\right]} \left[{R}_{i}^{2} + {R}_{o}^{2}\right]$
${I}_{H {C}_{0}} = \frac{1}{2} M \left[{R}_{i}^{2} + {R}_{o}^{2}\right]$

$\textcolor{b r o w n}{\text{B) Convert } \omega}$
omega_1= 2pi; omega_2=2pi*7=14pi

color(brown)("C) Compute the angular momentum before and after"
${L}_{o} = I {\omega}_{1}$ Before
${L}_{i} = I {\omega}_{2}$ After
Change $\Delta L = I \left({\omega}_{2} - {\omega}_{1}\right)$
$= \frac{9}{2} \left[1.44 \times {10}^{-} 2 - 6.4 \times {10}^{-} 3\right] 2 \pi \left(7 - 1\right)$
$\Delta L = \frac{9}{2} \left(8.0 \times {10}^{-} 3\right) = 0.432 \pi \approx 1.357 \frac{k g {m}^{2}}{s}$