A cylinder has inner and outer radii of $12 c m$ $12 cm$ and $15 c m$ $15 cm$ , respectively, and a mass of $4 k g$ $4 kg$ . If the cylinder's frequency of rotation about its center changes from $5 H z$ $5 Hz$ to $3 H z$ $3 Hz$ , by how much does its angular momentum change?

Question

1610 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-29T06:10:26

The change in angular momentum is $= 0.93 k g m^{2} s^{- 1}$ $=0.93kgm^2s^-1$

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

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

The mass, $m = 4 k g$ $m=4kg$

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

So, $I = 4 \cdot \frac{({0.12}^{2} + {0.15}^{2})}{2} = 0.0738 k g m^{2}$ $I=4*((0.12^2+0.15^2))/2=0.0738kgm^2$

The change in angular momentum is

$DeltaL=IDelta omega$

The change in angular velocity is

$Delta omega=(5-3)*2pi=(4pi)rads^-1$

The change in angular momentum is

$DeltaL=0.0738*4pi=0.93kgm^2s^-1$

A cylinder has inner and outer radii of 12 cm12cm12 cm and 15 cm15cm15 cm, respectively, and a mass of 4 kg4kg4 kg. If the cylinder's frequency of rotation about its center changes from 5 Hz5Hz5 Hz to 3 Hz3Hz3 Hz, by how much does its angular momentum change?