# Calculating the velocity of a hanging object attached to a disc

1. Oct 19, 2014

### EzequielSeattle

1. The problem statement, all variables and given/known data
A length of thread is wrapped many times around a steel disc of mass M and radius R, which is free to rotate around a fixed, frictionless, horizontal axle. The end of the thread is connected to a small object of mass m. If this small mass is held at rest and then released, how fast is it moving after it has fallen through a vertical distance h? Express in terms of M, R, m, g, and h. Make sure this makes sense in extreme limits of M >> m and M << m.

2. Relevant equations
V = mgh
K_rot = (1/2)Iω^2
V + K = 0

3. The attempt at a solution
Energy is conserved, so K = -V.
The object has a potential energy at the initial time of m*g*h. This will all be converted into rotational kinetic energy. Therefore, the final rotational energy will be equal to -mgh.
(1/2)Iω^2=-mgh
The moment of inertia for a disc is (1/2)MR^2.
Some basic algebra to isolate the angular velocity gives
ω=√(MmghR^2)

Am I correct? I guess I was unsure about whether the moment of inertia selected was correct. I also wasn't sure about the extreme conditions. Thanks in advance!

2. Oct 19, 2014

### voko

The heavier the disk is, the faster it rotates. That is what your final formula means. How can that be true?

3. Oct 20, 2014

### EzequielSeattle

Oh. Fair. I hadn't considered that. The question asked for it to be put in terms of M, and I'm not sure that the mass of the disc will matter at all now. How can I reconcile this?

4. Oct 20, 2014

### EzequielSeattle

OK, actually I now realize that the mass of the disc will matter, because it will rotate as the ball falls. The moment of inertia should then be... the same? Or shouldn't it be greater, taking the attached ball into account? Shouldn't I need the know the length of the string to calculate a new moment of inertia?

5. Oct 21, 2014

### voko

Since your result is so obviously wrong, you should review your method. You started from conservation of energy. It is most certainly a valid starting point. So you may have made errors in the application of this principle to this particular system, or errors in the subsequent algebraic manipulation.

6. Oct 21, 2014

### haruspex

Perhaps not that basic. Check all your steps. If you still get that, post them.
You are also overlooking that the small mass will have KE too. The extreme condition M << m implies you are not supposed to ignore that.

7. Oct 21, 2014

### EzequielSeattle

OK, so I checked my algebra, and yeah, I messed that up.

My new value is

K_rot_M + K_rot_m = mgh
(1/2)(1/2)Mω^2 + mω^2 = mgh
Dividing all by m gives
(1/4)(M/m)ω^2 + ω^2 = gh

Solving for ω gives me

√(gh/(1/4(M/m)+1))

Am I on the right track? For M >> m, the velocity will end up quite small. For m >> M, the velocity will end up quite large. Seems intuitive. Am I on the right track?

8. Oct 22, 2014

### voko

What is K_rot_m? Does the small mass rotate?

9. Oct 22, 2014

### NTW

Don't complicate things unnecessarily... Think in the potential energy of the mass with respect to the floor, at a height h, before the it starts falling. One instant before the mass reaches floor level, that energy would have been converted in KE of two sorts...

You have just to equate the initial energy with the final energy and solve for v...

10. Oct 22, 2014

### haruspex

What happened to R? And why no 1/2 on the KE for m?