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 $3 H z$ $3 Hz$ to $4 H z$ $4 Hz$ , by how much does its angular momentum change?

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Answer 1 · 2017-06-14T07:27:15

The angular momentum changes by $= 0.52 k g m^{2} s^{- 1}$ $=0.52kgm^2s^-1$

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

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

Mass, $m = k g$ $m=kg$

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

So, $I = 8 \cdot \frac{({0.08}^{2} + {0.12}^{2})}{2} = 0.0832 k g m^{2}$ $I=8*((0.08^2+0.12^2))/2=0.0832kgm^2$

The change in angular momentum is

$DeltaL=IDelta omega$

The change in angular velocity is

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

The change in angular momentum is

$DeltaL=0.0832*2pi=0.52kgm^2s^-1$

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 3 Hz3Hz3 Hz to 4 Hz4Hz4 Hz, by how much does its angular momentum change?