Question

A cylinder has inner and outer radii of `8 cm` and `12 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

`DeltaL= 9/2(8.0xx10^-3)=0.432pi~~1.357(kgm^2)/s`

Explanation:

Given:
1) A hollow cylinder radii: `R_o=8cm; R_i=12cm`
2) Mass, `m=12 kg`
3) `omega_1 =2pif_1; omega_2 =2pif_2; f_1=1Hz; f_2=7Hz`

Required change in angular momentum:
`DeltaL= I(omega_2-omega_1)`

Theorem, Definition and Principles:
Angular Momentum, `L=Iomega` where
`I` = Moment of Inertia
`omega` = Angular Velocity

Solution Strategy:
A) Compute the Moment of Inertia of Hollow cylinder
B) Convert frequency to radians: `omega =2pif`
C) Compute the Angular momentum for `omega_1 and omega_2`

`color(brown)("A)")` Moment of Inertia of Hollow Cylinder (HC)
Let `R_o=8cm; R_i=12cm` be outer and inner radii of HC
then we can derive the moment if inertia, I of HC as follows:
`I= intr^2dm` where `r` is arbitrary radius in `r:r in[R_o,R_i]`
Also let `rho=M/V` be the density and `l="length of cylinder"`
then `dm=rhocolor(blue)(dv)=rho*color(blue)(2pi*r*l*dr)`
Now because we are dealing with HC,
`rho = M/(pi(R_i^2-R_o^2)l)`

Integrate over the inner and outer radiuses and inserting the expression for `rho`
`I=rho2pil int_(R_o)^(R_i) r^3dr`
`= M/(pi(R_i^2-R_o^2)l) 2pil 1/4[R_i^4-R_o^4 ]`

`= M/(cancel(pi(R_i^2-R_o^2))l) 2l 1/4cancel(pi[R_i^2-R_o^2 ])[R_i^2+R_o^2]`
`I_(HC_0)=1/2M [R_i^2+R_o^2]`

`color(brown)("B) Convert "omega)`
`omega_1= 2pi; omega_2=2pi*7=14pi`

`color(brown)("C) Compute the angular momentum before and after"`
`L_o=Iomega_1` Before
`L_i=Iomega_2` After
Change `DeltaL= I(omega_2-omega_1)`
`=9/2[1.44xx10^-2-6.4xx10^-3]2pi(7-1)`
`DeltaL= 9/2(8.0xx10^-3)=0.432pi~~1.357(kgm^2)/s`