# A cylinder has inner and outer radii of 2 cm and 5 cm, respectively, and a mass of 5 kg. If the cylinder's frequency of counterclockwise rotation about its center changes from 12 Hz to 9 Hz, by how much does its angular momentum change?

Apr 28, 2018

Let's assume this cylinder is uniform (which is very important).

Recall for a solid cylinder,

$I = \frac{1}{2} m {r}^{2}$

Moreover, recall angular momentum,

$L = I \omega$

Now, let's derive the moment of inertia of this circular body,

$I = \frac{1}{2} m {r}^{2} \approx 5 {\text{kg"*"m}}^{2}$

which is generally constant in our calculations.

Now, recall,

$\omega = 2 \pi f$

Hence,

I_(1)2pif_1 approx (75"kg"*"m"^2)/"s" , and

I_(2)2pif_2 approx (57"kg"*"m"^2)/"s"

therefore DeltaL = L_2 - L_1 approx (18"kg"*"m"^2)/"s"

is the change in angular momentum given your data.

Apr 29, 2018

The change in angular momentum is $= 0.137 k g {m}^{2} {s}^{-} 1$

#### 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 $m = 5 k g$

The radii of the cylinder are ${r}_{1} = 0.02 m$ and ${r}_{2} = 0.05 m$

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

So, $I = 5 \cdot \frac{\left({0.02}^{2} + {0.05}^{2}\right)}{2} = 0.00725 k g {m}^{2}$

The change in angular velocity is

$\Delta \omega = \Delta f \cdot 2 \pi = \left(12 - 9\right) \times 2 \pi = 6 \pi r a {\mathrm{ds}}^{-} 1$

The change in angular momentum is

$\Delta L = I \Delta \omega = 0.00725 \times 6 \pi = 0.137 k g {m}^{2} {s}^{-} 1$