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

AI Thread Summary
The discussion focuses on proving the moment of inertia of a uniform annulus with inner radius 1 meter and outer radius 3 meters. The area of the annulus is calculated, leading to the mass per unit area and the mass of a differential ring within the annulus. By applying integration techniques, the moment of inertia is derived using the formula I = (2M/(R^2 - r^2))∫x^3dx. After performing the integration and substituting the given values, it is confirmed that the moment of inertia equals 5m. The calculations and reasoning validate the initial claim regarding the moment of inertia of the annulus.
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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.
 
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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 $R$?
 
i don't really know how to continue could you show me.
 
markosheehan said:
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?
 
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