A cylinder has inner and outer radii of $12 c m$ $12 cm$ and $15 c m$ $15 cm$ , respectively, and a mass of $6 k g$ $6 kg$ . If the cylinder's frequency of rotation about its center changes from $7 H z$ $7 Hz$ to $3 H z$ $3 Hz$ , by how much does its angular momentum change?

Question

1449 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-17T14:48:46

The angular momentum changes by $0.93 k g m^{2} s^{- 1}$ $0.93kgm^2s^-1$

The angular momentum is $L = I ω$ $L=Iomega$

where $I$ $I$ is the moment of inertia

For a cylinder, $I = m \frac{r_{1}^{2} + r_{2}^{2}}{2}$ $I=m(r_1^2+r_2^2)/2$

So, $I = 6 \cdot \frac{({0.12}^{2} + {0.15}^{2})}{2} = 0.0369 k g m^{2}$ $I=6*((0.12^2+0.15^2))/2=0.0369kgm^2$

The change in angular momentum is

$DeltaL=IDelta omega$

The change in angular velocity is

$Delta omega=(7-3)*2pi=(8pi)rads^-1$

The change in angular momentum is

$DeltaL=0.0369*8pi=0.93kgm^2s^-1$

A cylinder has inner and outer radii of 12 cm12cm12 cm and 15 cm15cm15 cm, respectively, and a mass of 6 kg6kg6 kg. If the cylinder's frequency of rotation about its center changes from 7 Hz7Hz7 Hz to 3 Hz3Hz3 Hz, by how much does its angular momentum change?