Question

A cylinder has inner and outer radii of `9 cm` and `14 cm`, respectively, and a mass of `9 kg`. If the cylinder's frequency of counterclockwise rotation about its center changes from `1 Hz` to `7 Hz`, by how much does its angular momentum change?

Answer 1

Change in angular momentum = `(I delta omega)`

now,change in angular velocity= `delta omega = 2pi(7-1) (rad)/s` i.e `12 pi (rad)/s`

`I` = moment of inertia of of a cylindrical shell of given outer and inner radii will be `9/2((9/100)^2 + (14/100)^2) Kg.m^2` i.e `0.124 Kg.m^2` (formula is `M/2((r_1)^2+(r_2)^2)`)

So,change in angular momentum is `(0.124 * 12 pi) N.m.s=4.674 N.m.s`

Answer 2

The change in angular momentum is `=4.7kgm^2s^-1`

Explanation:

The angular momentum is `L=Iomega`

where `I` is the moment of inertia

and `omega` is the angular velocity

The mass of the cylinder is `m=9kg`

The radii of the cylinder are `r_1=0.09m` and `r_2=0.14m`

For the cylinder, the moment of inertia is `I=m((r_1^2+r_2^2))/2`

So, `I=9*((0.09^2+0.14^2))/2=0.12465kgm^2`

The change in angular velocity is

`Delta omega=Deltaf*2pi=(7-1) xx2pi=12pirads^-1`

The change in angular momentum is

`DeltaL=IDelta omega=0.12465xx12pi=4.7kgm^2s^-1`