# A cylinder has inner and outer radii of 4 cm and 8 cm, respectively, and a mass of 6 kg. If the cylinder's frequency of counterclockwise rotation about its center changes from 7 Hz to 8 Hz, by how much does its angular momentum change?

Mar 27, 2017

The change in angular momentum is $= 0.15 k g {m}^{2} {s}^{-} 1$

#### Explanation:

The angular momentum is $L = I \omega$

where $I$ is the moment of inertia

Mass, $m = 6 k g$

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

So, $I = 6 \cdot \frac{\left({0.04}^{2} + {0.08}^{2}\right)}{2} = 0.024 k g {m}^{2}$

The change in angular momentum is

$\Delta L = I \Delta \omega$

The change in angular velocity is

$\Delta \omega = \left(8 - 7\right) \cdot 2 \pi = \left(2 \pi\right) r a {\mathrm{ds}}^{-} 1$

The change in angular momentum is

$\Delta L = 0.024 \cdot 2 \pi = 0.15 k g {m}^{2} {s}^{-} 1$

Mar 27, 2017

$L = 0.15 k g {m}^{2} {s}^{-} 1$

#### Explanation:

Angular momentum formula: $L = I \cdot \omega$
$L$ = Angular momentum
$I$ = Moment of Inertia
$\omega$ = Angular velocity

$I = \frac{1}{2} M {R}^{2}$
$M$ =Mass
$R$ =Distance of Axis
$I$ = Moment of Inertia

$I = \frac{1}{2} \cdot 6 k g \cdot \left({\left(0.04 m\right)}^{2} + {\left(0.08 m\right)}^{2}\right) = 0.024 k g {m}^{2}$

$\omega = 2 \pi \cdot F$
$F$ = Frequency
$\omega$ = Angular Vecloicty
$\omega = 2 \pi \left(8 H z - 7 h z\right) = 2 \pi H z \mathmr{and} 2 \pi {s}^{-} 1$

$L = 0.024 k g {m}^{2} \cdot 2 \pi {s}^{-} 1 = 0.15 k g {m}^{2} {s}^{-} 1$