# Kinematics - Minimization of a period

1. Jun 30, 2009

### RoyalCat

1. The problem statement, all variables and given/known data
A mass is released from rest atop an incline angled at Θ relative to the horizon, from a certain distance up the incline (Not height), x_0. This point up the incline is denoted P.

The mass then travels along a horizontal path of length S.

The mass then goes up an incline angled at Θ relative to the horizon.

All surfaces are smooth, the layout looks something like this:

\______/

What is the exact value of x so that T, the period of the motion (The time it takes the mass to return to the point P), is minimal?

The text-book answer is $$\frac{S}{2}$$
I keep getting $$\frac{S}{4}$$

I've also double-checked the book's answer, and the period is shorter for the book's answer.

2. Relevant equations
Kinematic equations.

3. The attempt at a solution
One period of motion consists of the mass going down the incline, traveling along the distance S, going up the opposite incline, going down the opposite incline, traveling back along the distance S and going back up the first incline.

The two motions along the distance S are identical, as are all 4 of the motions of the mass going up/down the inclines (As energy is conserved).

Therefore, the period is:
$$T=4t_1+2t_2$$
Where $$t_1$$ is the time it takes the mass to travel down the incline, or up the incline, and $$t_2$$ is the time it takes the mass to travel the distance S.

For the motion down the incline, with x measured along the incline from the point P:
$$x(t)=\frac{g\sin{\theta}}{2}t_1^{2}$$
$$x(t_1)=x_0$$
$$x_0=\frac{g\sin{\theta}}{2}t_1^{2}$$
$$t_1=\sqrt{\frac{2x_0}{g\sin{\theta}}}$$

The speed at the bottom of the incline is:
$$v_s=\sqrt{2g\sin{\theta}x_0}$$
$$x(t)=v_st$$
$$x(t_2)=S$$
$$t_2=\frac{S}{v_s}=\frac{S}{\sqrt{2g\sin{\theta}x_0}}$$

$$T=4t_1+2t_2$$
$$T(x_0)=4\sqrt{\frac{2x_0}{g\sin{\theta}}}+2\frac{S}{\sqrt{2g\sin{\theta}x_0}}$$
$$z\equiv\sqrt{x_0}$$
$$T(z)=(4\sqrt{\frac{2}{g\sin{\theta}}})z+(2\frac{S}{\sqrt{2g\sin{\theta}}})\frac{1}{z}$$
$$T'(z)=4\sqrt{\frac{2}{g\sin{\theta}}}-(2\frac{S}{\sqrt{2g\sin{\theta}}})\frac{1}{z^{2}}$$
$$T'(z_{min})=0$$
$$4\sqrt{\frac{2}{g\sin{\theta}}}=(2\frac{S}{\sqrt{2g\sin{\theta}}})\frac{1}{z^{2}}$$
$$\frac{1}{z^{2}}=(4\sqrt{\frac{2}{g\sin{\theta}}})(\frac{\sqrt{2g\sin{\theta}}}{2S})$$
$$\frac{1}{z^{2}}=4*\frac{1}{2S}*2=\frac{4}{S}$$
$$x_{0_{min}}=z_{min}^{2}=\frac{S}{4}$$

Bleh, could someone please read through it and tell me where I went wrong?

Last edited: Jun 30, 2009
2. Jun 30, 2009

### KLoux

You dropped a factor of two when you solve for your minimum z. Your 4th-to-last line is correct, but you lose a two when you go to the 3rd-to-last line.

-Kerry

3. Jun 30, 2009

### LowlyPion

(1/z)' = -1/z2

4. Jun 30, 2009

### RoyalCat

Err, yeah, I just noticed it myself. Thing is though, that the factor of two in the fourth-to-last line, was a typo limited to that line, and the line before it. The last couple of lines leading to $$x_0=\frac{S}{4}$$ still stand. :\

I assume that's what LowlyPion was aiming at with his (1/z)'=-1/z² comment, as well.

I fixed the factor of two typo, but I'm still stumped.

Last edited: Jun 30, 2009
5. Jun 30, 2009

### LowlyPion

Fwiw, I get S/4 in working it out. Maybe the book is wrong?

6. Jun 30, 2009

### KLoux

Hmmm, when I substitute $$x_0=\frac{S}{4}$$ and solve for the period, I get something smaller than when I use $$x_0=\frac{S}{2}$$. So either your equation for the period is wrong, or the book's answer is wrong?

-Kerry

7. Jun 30, 2009

### RoyalCat

Yeah, I thought that might be the case, so I made the two substitutions $$x_0=\frac{S}{2}$$ and $$x_0=\frac{S}{4}$$

Drawing the curve using Graphmatica, I can see the minimum really is at $$\frac{S}{4}$$
So that would mean either the book is wrong, or my derivation for $$T(x_0)$$ is wrong.

Seeing how you got the same answer as I did, LowlyPion, I'm guessing the former is the reason. I guess I'll ask the writer then.

Thanks a bunch, everyone. :)