moment of inertia

logistic_guy

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Find the moment of inertia I\displaystyle I of a uniform hollow cylinder of inner radius R1\displaystyle R_1, outer radius R2\displaystyle R_2, and mass M\displaystyle M if the rotation axis is through its center.

Hint: I=R2 dm\displaystyle I = \int R^2 \ dm
 
Hint: I=R2 dm\displaystyle I = \int R^2 \ dm
Using the hint. I will rotate around the z\displaystyle z axis.

Iz=R2 dm=R2ρ dV\displaystyle I_z = \int R^2 \ dm = \int R^2 \rho \ dV

If I take the volume of a thin cylindrical shell, I get:

Iz=R2ρ dV=R2ρ 2πRh dR=2πρhR3 dR\displaystyle I_z = \int R^2 \rho \ dV = \int R^2 \rho \ 2\pi Rh \ dR = 2\pi \rho h\int R^3 \ dR

I am ready to integrate😍

Iz=2πρhR1R2R3 dR=2πρhR44R1R2=2πρh(R244R144)\displaystyle I_z = 2\pi \rho h\int_{R_1}^{R_2} R^3 \ dR = 2\pi \rho h \frac{R^4}{4}\bigg|_{R_1}^{R_2} = 2\pi \rho h\left(\frac{R^4_2}{4} - \frac{R^4_1}{4}\right)


=πρh(R24R142)=πρh(R22+R12)(R24R12)2\displaystyle = \pi \rho h\left(\frac{R^4_2 - R^4_1}{2}\right) = \pi \rho h\frac{\left(R^2_2 + R^2_1\right)\left(R^4_2 - R^2_1\right)}{2}

We know that ρ\displaystyle \rho is the volume density, then

ρ=MV=Mhπ(R22R12)\displaystyle \rho = \frac{M}{V} = \frac{M}{h\pi\left(R^2_2 - R^2_1\right)}

This gives:

Iz=πMVh(R22+R12)(R24R12)2=πMhπ(R22R12)h(R22+R12)(R24R12)2\displaystyle I_z = \pi \frac{M}{V} h\frac{\left(R^2_2 + R^2_1\right)\left(R^4_2 - R^2_1\right)}{2} = \pi \frac{M}{h\pi\left(R^2_2 - R^2_1\right)} h\frac{\left(R^2_2 + R^2_1\right)\left(R^4_2 - R^2_1\right)}{2}

With a little simplification, finally we get🤩

Iz=12M(R22+R12)\displaystyle I_z = \textcolor{blue}{\frac{1}{2}M\left(R^2_2 + R^2_1\right)}
 
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