Prove Moment of Inertia of Annulus w/ Radii 3m & 1m: 5m

[markosheehan]

a uniform annulus consists of disc of radius 3 meters with a disc of radius 1 meters removed from its center. the mass of the annulus is m. Prove that the moment of inertia of the annulus about an axis through its center is 5m.

 

[MarkFL]

Consider an annulus whose inner radius is $r$ and outer radius is $R$, and has mass $M$. The area of the face of the annulus is:

$$A=\pi\left(R^2-r^2\right)$$

The mass per unit area is:

$$M_A=\frac{M}{A}=\frac{M}{\pi\left(R^2-r^2\right)}$$

Next, consider a ring within the annulus having inner radius $x$ and outer radius $x+dx$. The mass of this ring is:

$$M_R=\frac{2M}{R^2-r^2}x\,dx$$

We know the moment of inertia for this ring is:

$$I_R=\frac{2M}{R^2-r^2}x\cdot x^2\,dx$$

And so the moment of inertia for the annulus is:

$$I=\frac{2M}{R^2-r^2}\int_r^R x^3\,dx$$

Can you proceed to carry out the integration and derive the formula into which you can put the given values for $r$ and $R$?

 

[markosheehan]

i don't really know how to continue could you show me.

 

[MarkFL]

i don't really know how to continue could you show me.

Okay, we have:

$$I=\frac{2M}{R^2-r^2}\int_r^R x^3\,dx$$

So, we may apply the power rule:

$$\int u^r\,du=\frac{u^{r+1}}{r+1}+C$$ where $$r\ne-1$$

And the anti-derivative form of the FTOC:

$$\int_a^b f(u)\,du=F(b)-F(a)$$

And we obtain:

$$I=\frac{2M}{R^2-r^2}\left[\frac{x^4}{4}\right]_r^R=\frac{M}{2\left(R^2-r^2\right)}\left[x^4\right]_r^R=\frac{M}{2\left(R^2-r^2\right)}\left(R^4-r^4\right)=\frac{M}{2}\left(R^2+r^2\right)$$

Now, we are given:

$$r=1,\,R=3,\,M=m$$

And so:

$$I=\frac{m}{2}\left(3^2+1^2\right)=5m$$

Does that make sense?