A cylinder has inner and outer radii of $8 c m$ $8 cm$ and $12 c m$ $12 cm$ , respectively, and a mass of $8 k g$ $8 kg$ . If the cylinder's frequency of rotation about its center changes from $1 H z$ $1 Hz$ to $5 H z$ $5 Hz$ , by how much does its angular momentum change?

Question

1346 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-16T16:27:05

The change in angular momentum is $= 2.09 k g m^{2} s^{- 1}$ $=2.09kgm^2s^-1$

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

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

and $ω$ $omega$ is the angular velocity

The mass of the cylinder is $m = 8 k g$ $m=8kg$

The radii of the cylinder are $r_{1} = 0.08 m$ $r_1=0.08m$ and $r_{2} = 0.12 m$ $r_2=0.12m$

For the cylinder, the moment of inertia is $I = m \frac{(r_{1}^{2} + r_{2}^{2})}{2}$ $I=m((r_1^2+r_2^2))/2$

So, $I = 8 \cdot \frac{({0.08}^{2} + {0.12}^{2})}{2} = 0.08322 k g m^{2}$ $I=8*((0.08^2+0.12^2))/2=0.08322kgm^2$

The change in angular velocity is

$Delta omega=Deltaf*2pi=(5-1) xx2pi=8pirads^-1$

The change in angular momentum is

$DeltaL=IDelta omega=0.0832 xx8pi=2.09kgm^2s^-1$

A cylinder has inner and outer radii of 8 cm8cm8 cm and 12 cm12cm12 cm, respectively, and a mass of 8 kg8kg8 kg. If the cylinder's frequency of rotation about its center changes from 1 Hz1Hz1 Hz to 5 Hz5Hz5 Hz, by how much does its angular momentum change?