Question

1

Answer 1

What you have for a) is right, and what you have for the moment of inertia of the hoop is right, but I'm not sure about your answer for the moment of inertia of the bracket. If the mass is negligible i.e. taken to =0, and the moment of inertia of a rod is mL^2 /12, then doesn't that mean the moment of inertia is zero?

Answer 2

I think the perpendicular axis theorem might be relevant to this question.

Answer 3

Okay, I've thought about it a bit more and I think your answer to the moment of inertia of the hoop isn't what the question is looking for either. Here's why.

The moment of inertia of an object depends on what axis of rotation you're talking about. Your answer of M3R3^2 for I_hoop assumes we're talking about the hoop rotating about an axis which is perpendicular to its own plane, passing through its centre (so it spins like Saturn's rings spin around saturn). But for part c), we're going to have to use the same axis of rotation for both this hoop and the disk below. So let's take our axis to be the one you used in a), which passed orthogonally through the centre of the bottom disk. Let I_h be the moment of inertia (let's say m.o.i. from now on of the hoop with respect to this axis.

Then we can use the parallel axis theorem, which states that the m.o.i. of an object about one axis is the same as its m.o.i. about another axis which is parallel to the first one, PLUS the moment of inertia of a point particle with the mass of the object, about the first axis. Look it up for clearer explanations – it makes sense if you think about it. So using that:

\[I_h=M3L^2+I_h'\]

where I'_h is the m.o.i. of the hoop about the vertical axis which passes through the centre of the hoop. This axis is parallel to the one you used in part a), which is why we're allowed to use the parallel axis theorem.

Now we have to find I'_h, and we can do that with the perpendicular axis theorem. This is less easy to explain and I'll let you look it up yourself, but what it tells us is that

\[I_h'=\frac{ 1 }{ 2 }I_h''\]

where I''_h is the m.o.i. of the hoop about the axis perpendicular to the hoop which passes through its centre, so the hoop is spinning like Saturn's rings. But we already know what that is – it's M3R3^2, like you said already.

\[I_h''=M3R3^2\]
so
\[I_h'=\frac{ 1 }{ 2 }M3R3^2\]
and
\[I_h=M3L^2+\frac{ 1 }{ 2 }M3R3^2=M3(L^2+\frac{R3^2}{2})\]

Then the angular momentum of the hoop is I_h times its angular velocity, which we're told is w0. So

\[L_h=M3\omega_0(L^2+\frac{R3^2}{2})\]
But angular momentum is a vector, like velocity. It has both a size (or amplitude) and a direction). What we've just found is the amplitude of L_h. You can use the right hand rule to find that the direction of L_h is pointing along the bracket, towards the centre of the rotating system.

So that's part b). Sorry it was so long. And never mind if you've already solved the problem in the last 13 hours...

If not, what you're going to have to do for part c) is write down two equations, which are: 'total angular momentum before = total angular momentum after', and 'total rotational kinetic energy before = total rotational kinetic energy after'. Since the two people (and therefore their disk) are initially stationary, what we've done in part b) is worked out 'total angular momentum before'.

Sorry for the essay, hope some of it was helpful! I'd really recommend reading the chapters in the Feynman lectures on physics (like the bible of physics) which are about rotation – I found them really helpful. It's all available for free online here: http://www.feynmanlectures.caltech.edu/I_toc.html

Answer 4

Argh sorry, I got the perpendicular axis theorem wrong. Actually I'_h = 2*I''_h, rather than half of it. So if you carry that mistake forward, the final answer for the amplitude of the angular momentum should be

\[L_h=M3\omega_0(L^2+2R3^2)\]