# A cylinder has inner and outer radii of 12 cm and 15 cm, respectively, and a mass of 4 kg. If the cylinder's frequency of rotation about its center changes from 5 Hz to 8 Hz, by how much does its angular momentum change?

Mar 28, 2016

$\text{change of angular momentum=} 0 , 4428 \cdot \pi$

#### Explanation:

$\text{change of angular momentum=} I \cdot \Delta \omega$
$I : \text{moment of inertia}$
$\Delta \omega : \text{change of angular velocity}$
$I = \frac{1}{2} \cdot m \cdot \left({a}^{2} + {b}^{2}\right) \text{cylinder axis at center}$
$m : 4 k g$
$a = 12 c m = 0 , 12 m$
$b = 15 c m = 0 , 15 m$
$I = \frac{1}{\cancel{2}} \cdot \cancel{4} \left({\left(0 , 12\right)}^{2} + {\left(0 , 15\right)}^{2}\right)$
$I = 2 \cdot \left(0 , 0144 + 0 , 0225\right)$
$I = 2 \cdot 0 , 0369$
$I = 0 , 0738$
$\Delta \omega = {\omega}_{2} - {\omega}_{1}$
$\Delta \omega = 2 \cdot \pi \cdot {f}_{2} - 2 \cdot \pi \cdot {f}_{1}$
$\Delta \omega = 2 \cdot \pi \left({f}_{2} - {f}_{1}\right)$
${f}_{1} = 5 H z \text{ } {f}_{2} = 8 H z$
$\Delta \omega = 2 \cdot \pi \left(8 - 5\right)$
$\Delta \omega = 6 \cdot \pi$

$\text{change of angular momentum=} 0 , 0738 \cdot 6 \cdot \pi$
$\text{change of angular momentum=} 0 , 4428 \cdot \pi$