# A Difficult Mechanics Problem (Work-Energy?)

1. Nov 24, 2005

### chaose

The string is L = 1m long. Peg P is located d below the support point. When the ball is released from rest (as shown), it swings along the dashed path. WHen it catches on the peg, it swings upward but does not complete a circular path. Instead it leaves its path and strikes the peg.
Part A. What is the distance d below the support point at which the peg can be placed for this to occur?
Part B. If the Peg P is now located at an angle A below the horizontal (as in the picture). When the string catches on the peg, the ball strikes the peg. Express d as a function of A if this were to occur.

I'm thinking of approaching it by applying the law of conservation of energy and figuring the energy and velocity at various points, and get a relation in terms of energy. So far, I'm not getting anywhere useful. Can anyone please help (and explain?)
thanks .

Last edited: Nov 24, 2005
2. Nov 25, 2005

### ONJ's Noble Steed

I wrote up a solution to the first problem. Click on the thumbnail to see it. I do not claim that my solution is the most efficient one, nor do I even claim that it is correct.

Note: Apparently, I skipped a major step in that solution. Here's the corrected solution:

Last edited: Nov 25, 2005
3. Nov 25, 2005

### daniel_i_l

Overall it looks good, but if alpha is the angle over the horizontal then shouldn't the total force be: $$F = \frac{mv^2}{r} + mgsin(\alpha)$$

4. Nov 25, 2005

### ONJ's Noble Steed

Oh, well, if you say that g is negative, then you add, but I just assumed that it's positive. I guess it might be more correct to say that it's negative, because to calculate the angle, I chose a coordinate system with the x-axis going right and the y-axis going up. Since g is directed downward, it should be negative. But obviously this is of no consequence if you say beforehand that g is positive, which I forgot to do.

I also noticed that I misspelled "parabolic" as "parobolic."

Last edited: Nov 25, 2005
5. Nov 28, 2005

### vtmemo

Quite a nasty "introductory" problem.

This problem required you to make several fairly high-level observations about the situation; as an engineer, it took me a fair amount of effort to prove the assumptions I made.
The attached PDF shows all work, derivations, and assumptions used to reach the final conclusions.
For those of you playing the home-game, here's what I got:
$$d = L - L\frac{2}{2+\sqrt{3}}$$
(this solution should hold for any length, L)
... If L=1m, then
$$d = 1 - \frac{2}{2+\sqrt{3}}$$, whatever that comes out to. All I know is that it's less than 1.
By the way, if you're wondering what the ideal angle is offhand, I got
$$\tan{\theta} = \frac{1}{\sqrt{2}}$$
The second part was just a matter of doing some quick trig based on the previous part, finding d as a function of beta ($$d(\beta)$$). I was able to get beta as a function of r, the secondary swing radius. If you want it as a function of d, then substitute.
$$r(\beta) = \frac{L \sin{\beta}}{\frac{1}{\sqrt{3}}+\frac{1}{2\sqrt{3}}+\cos{\beta}}$$
I know it seems very complicated, but I left a lot of the numbers there as radical constants as opposed to evaluating / approximating them. I hope this helps, because it made my brain hurt.

- vtMeMo

P.S., THIS IS NOT AN INTRODUCTORY PROBLEM. Professional Engineers called this problem "mean" for even an advanced student, just by virtue of how much conditional thinking it requires. It took me almost 2.5 hours to go through it by hand (checking).

#### Attached Files:

• ###### pendulum_mmilo_11_28_05.pdf
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6. Nov 28, 2005

### daniel_i_l

Thats exactly what I got. It actually wasn't that hard if you use energy conservation to get the speed as a function of the angle and L, centripital force to get another speed-angle equation (the tension is 0 when the mass starts to fly to the middle) and the kinimatics equations to get an equation connecting the radius of the small circle and the speed. With those 3 equations it just takes algebra to find d as a function of L.