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 $7 H z$ $7 Hz$ to $4 H z$ $4 Hz$ , by how much does its angular momentum change?

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Answer 1 · 2017-01-29T08:25:17

The answer is $= 1.57 k g m s^{- 1}$ $=1.57kgms^(-1)$

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

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

The change of angular momentum is

$DeltaL=I Delta omega$

For a cylinder, $I=(m(r_1^2+r_2^2))/2$

So, $I=8*(0.08^2+0.12^2)/2=0.0832kgm^2$

$Delta omega=(7-4)*2pi =6pirads^-1$

$L=0.0832*6pi=1.57kgms^(-1)$

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 7 Hz7Hz7 Hz to 4 Hz4Hz4 Hz, by how much does its angular momentum change?