# Physics Question

1. Nov 15, 2004

### T@P

I am taking a beginning physics class, and I have some questions about the moment of inertia. First of all, what is the momet of inertia of a uniform disc of mass M and two objects of mass m on either side of the disc (diametrically opposite).

Also, my physics book (Giancoli) has a list of formulas for the moments of inertia of many objects, but no explanation of how they got it. Im guessing its through calculus (which I know). Unfortunately the course I am taking does not "have" calculus in it, so I was wondering if someone could tell me generally how its done? Thanks

2. Nov 15, 2004

### Nylex

The moment of inertia of a system of point masses is given by $$I = \sum_i m_ir_i^2$$, where m is the mass of a certain point mass, r is the distance from the axis about which you're taking the moment of inertia.

If you want to calculate the moment of inertia of a continous body, you need to use integration.

Last edited: Nov 15, 2004
3. Nov 15, 2004

### T@P

I am aware that you use integration to calculate the moment of inertia, but as in the case i stated above, what is the integral that calculates it? also could you at least give the integral that would let me calculate the moment of inertia of a simple object (sphere?)

4. Nov 15, 2004

### arildno

You are to sum over/integrate over all particles making up the body B.
We may represent this with the integral:
$$I=\int_{B}r^{2}dm$$
where it is understood that each particle has some position $$\vec{x}$$ and some mass dm (specific to that particle).
By knowing where the axis is in space, we may find the correct "r"-value for each particle (by use of its position vector $$\vec{x}$$).

The above integral is rewritten in terms of the density distribution of the body, that is, the (infinitesemal) mass of each particle fullfills:
$$dm=\rho(\vec{x})dV$$

Hence, we may rewrite our integral as:
$$I=\int_{V}\rho(\vec{x})r^{2}(\vec{x})dV$$

Let us consider the case of the sphere with constant density $$\rho_{0}$$, radius $$\mathcal{R}$$, and let the axis we are considering go through the center of the sphere, which we also set as the origin in spherical coordinates.

We let the angle the position vector to some particle from the origin makes with the rotation axis be $$\phi$$

Consider the particle situated at the spherecial coordinate $$(R,\theta,\phi)$$
($$\theta$$ being the planar angle)

For that particle, we have:
$$dV=R^{2}\sin\phi{dR}d\theta{d}\phi$$
$$r=R\sin\phi$$

Hence, our integral becomes:
$$I=\int_{0}^{\mathcal{R}}\int_{0}^{2\pi}\int_{0}^{\pi}\rho_{0}R^{4}\sin^{3}\phi{d\phi}d\theta{dR}$$

Using $$\sin^{3}\phi=\sin\phi(1-\cos^{2}\phi)$$
we find the antiderivative: $$-(\cos\phi-\frac{1}{3}\cos^{3}\phi)$$

You ought to end up with the answer $$I=\frac{2}{5}MR^{2}$$
where M is the mass of the sphere.

5. Nov 15, 2004

### Nylex

Sorry, I thought you said you didn't need calculus, so I didn't include the integral. For a sphere, you can also calculate the moment of inertia using a single integral, by considering the sphere to be made up of spherical shells (for which I = (2/3)MR^2).

6. Nov 15, 2004

### T@P

thank you
would the moment of inertia change if i place two equal masses on a flat disc? (in diametrically opposite places)?

7. Nov 15, 2004

### Staff: Mentor

Of course. You would have to add the rotational inertia of the two masses.

8. Nov 15, 2004