What is the moment of inertia about an axis through its center of a sphere of radius R, mass M and density varying with radius as Ar.

Can anyone explain this? This is a Conquering physics book problem and I did not get this solution.

## Moment of inertia problem

### Re: Moment of inertia problem

I assume you are referring to Q2 in the 1.4.5 problem set from "Conquering the Physics GRE".

I also found this to be a tricky problem. What I would do is first take the volume integral of the density and set it equal to the mass. Then I solve for A. I believe this is shown pretty well in the solution, giving us $$M = 4 \pi \int \rho (r^2 dr) = \pi A R^4 \rightarrow A = \dfrac{M}{\pi R^4}$$. (Note that the $$r^2$$ is because it's a volume integral in spherical coordinates and that the $$4 \pi$$ is from integrating over $$\theta$$ and $$\phi$$.

Once you have this you integrate $$I = \int s^2 \pho dV$$, where s is not the same as r. This is the really tricky part of this problem. s, as is explained in the solution, is the perpendicular distance from the axis of rotation to a point on the sphere. Moments of inertia are always computed using this, which is also known as a Moment Arm. Thus we can define $$s = r \sin(\theta)$$Thus, your fully integral would be $$\int \left( r \sin(\theta) \right)^2 \rho r^2 \sin(\theta) dr d \theta d \phi$$.

I hope that this was helpful. If you have any questions please feel free to ask.

Thanks.

I also found this to be a tricky problem. What I would do is first take the volume integral of the density and set it equal to the mass. Then I solve for A. I believe this is shown pretty well in the solution, giving us $$M = 4 \pi \int \rho (r^2 dr) = \pi A R^4 \rightarrow A = \dfrac{M}{\pi R^4}$$. (Note that the $$r^2$$ is because it's a volume integral in spherical coordinates and that the $$4 \pi$$ is from integrating over $$\theta$$ and $$\phi$$.

Once you have this you integrate $$I = \int s^2 \pho dV$$, where s is not the same as r. This is the really tricky part of this problem. s, as is explained in the solution, is the perpendicular distance from the axis of rotation to a point on the sphere. Moments of inertia are always computed using this, which is also known as a Moment Arm. Thus we can define $$s = r \sin(\theta)$$Thus, your fully integral would be $$\int \left( r \sin(\theta) \right)^2 \rho r^2 \sin(\theta) dr d \theta d \phi$$.

I hope that this was helpful. If you have any questions please feel free to ask.

Thanks.

### Re: Moment of inertia problem

So how did find that

s = r sin(theta)

thanks.

t

s = r sin(theta)

thanks.

t

### Re: Moment of inertia problem

If you draw it you'll see immediately why.tazi wrote:So how did find that

s = r sin(theta)

thanks.

t