A cylinder has inner and outer radii of $12 c m$ $12 cm$ and $18 c m$ $18 cm$ , respectively, and a mass of $6 k g$ $6 kg$ . If the cylinder's frequency of rotation about its center changes from $9 H z$ $9 Hz$ to $8 H z$ $8 Hz$ , by how much does its angular momentum change?

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Answer 1 · 2017-05-29T06:12:22

The angular momentum changes by $= 0.88 k g m^{2} s^{- 1}$ $=0.88kgm^2s^-1$

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

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

Mass, $m = 6 k g$ $m=6kg$

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

So, $I = 6 \cdot \frac{({0.12}^{2} + {0.18}^{2})}{2} = 0.1404 k g m^{2}$ $I=6*((0.12^2+0.18^2))/2=0.1404kgm^2$

The change in angular momentum is

$DeltaL=IDelta omega$

The change in angular velocity is

$Delta omega=(9-8)*2pi=(2pi)rads^-1$

The change in angular momentum is

$DeltaL=0.1404*2pi=0.88kgm^2s^-1$

A cylinder has inner and outer radii of 12 cm12cm12 cm and 18 cm18cm18 cm, respectively, and a mass of 6 kg6kg6 kg. If the cylinder's frequency of rotation about its center changes from 9 Hz9Hz9 Hz to 8 Hz8Hz8 Hz, by how much does its angular momentum change?