1. Dec 19, 2006

### dkgojackets

1. The problem statement, all variables and given/known data

A bead slides without friction around a loop-the-loop. The bead is released at height y from the bottom of the loop, which has radius r. What is the instantaneous kinetic energy K at the top of the loop so that the bead would press the track with an upward force F=mg/2?

Express the height y in terms of K, m, and r.

2. Relevant equations

K=.5mv^2
cent acceleration=(v^2)/r

3. The attempt at a solution

I drew a FBD with weight mg down and F mg/2 up. The sum was mg/2 down, which I set equal to m(v^2)/r. I arranged the equation to get v^2=gr/2, and plugged that into the kinetic energy equation, getting mgr/4 as the incorrect solution.

I was able to get the second part through energy conservation, setting mgy=2mgr + K and solving.

2. Dec 19, 2006

### Staff: Mentor

Hmm. I get mgr/4 = KE at the top as well. We must be missing something.

3. Dec 19, 2006

### Saketh

According to the problem, the bead is pressing up. In other words, the bead overcomes its weight to push up on the wire.

According to your analysis, the bead is pressing down. If the bead is pressing up, then the weight force must already be canceled out by some force, with a resulting net force of $\frac{mg}{2}$. It just so happens, however, that your net force and the actual net force are equivalent. This is just a coincidence.

However, I do not understand why your answer is incorrect, because I am getting the same answer that you did, although with different reasoning. (EDIT: Never mind. See my later post for the error in my reasoning.)

Last edited: Dec 19, 2006
4. Dec 19, 2006

### dkgojackets

So a total force of 3mg/2? That would make kinetic energy 3mgr/4, which seems right.

5. Dec 19, 2006

### Saketh

The total force is not that.

The problem is asking for the kinetic energy at the top of the loop.

$$KE_i + PE_i = KE_f$$

Your analysis finds the instantaneous kinetic energy at the bottom of the loop. Now you need to solve for the kinetic energy at the top of the loop.

Here's the problem - when you found the net force of the bead, you had it going down. It should be going up. The problem tells you that the net force of the bead is upwards. Therefore, centripetal force is equal to mg/2. Your reasoning was slightly incorrect.

6. Dec 19, 2006

### OlderDan

Not so. The bead is pushing up on the track with a force of mg/2, so the track is pushing down on the bead with a force of mg/2 (Newton 3). This is added to the weight of the bead for a total downward force of 3mg/2, and that is the centripetal force at the top of the loop.

That looks right to me.

Last edited: Dec 19, 2006