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

In summary, we derived the formula for the moment of inertia of an annulus about an axis through its center, using the given values of its inner and outer radii and mass. The moment of inertia was found to be equal to 5 times the mass of the annulus.
  • #1
markosheehan
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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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  • #2
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:

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

The mass per unit area is:

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

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

\(\displaystyle M_R=\frac{2M}{R^2-r^2}x\,dx\)

We know the moment of inertia for this ring is:

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

And so the moment of inertia for the annulus is:

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

Can you proceed to carry out the integration and derive the formula into which you can put the given values for $r$ and $R$?
 
  • #3
i don't really know how to continue could you show me.
 
  • #4
markosheehan said:
i don't really know how to continue could you show me.

Okay, we have:

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

So, we may apply the power rule:

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

And the anti-derivative form of the FTOC:

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

And we obtain:

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

Now, we are given:

\(\displaystyle r=1,\,R=3,\,M=m\)

And so:

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

Does that make sense?
 

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

What is the formula for calculating moment of inertia for an annulus?

The formula for calculating moment of inertia for an annulus is I = (π/4)(r4o - r4i), where ro and ri are the outer and inner radii, respectively.

Can the moment of inertia of an annulus with radii 3m and 1m be proven?

Yes, the moment of inertia of an annulus can be proven using the formula mentioned above and substituting the given values for ro and ri.

What is the unit of measurement for moment of inertia?

The unit of measurement for moment of inertia is kilogram square meters (kg·m2).

How does the moment of inertia change if the outer radius is increased?

If the outer radius of an annulus is increased, the moment of inertia will increase as well. This is because the increase in radius results in a larger distance from the axis of rotation, which increases the moment of inertia.

Is the moment of inertia of an annulus with radii 3m and 1m affected by the mass distribution?

Yes, the moment of inertia of an annulus can be affected by the mass distribution, specifically the distribution of mass in relation to the axis of rotation. This is because the moment of inertia takes into account not only the mass of an object, but also the distribution of that mass in relation to the axis of rotation.

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