# Conservation of Energy and Final Tangential Speed

1. Oct 22, 2007

### Nghi

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

After doing some exercises on the floor, you are lying on your back with one leg pointing straight up. If you allow your leg to fall freely until it hits the floor, what is the tangential speed of your foot just before it lands? Assume the leg can be treated as a uniform rod 0.83 m long that pivots freely about the hip.

2. Relevant equations

KE = 0.5mv^2(1 + I/(mr^2))
PE = mgh
I = (1/3)mr^2

3. The attempt at a solution

This problem is driving me crazy! I've tried so many different ways, and I can't seem to get it. Since the problem gives us that it's a uniform rod that pivots around the hip, I equated that to the rod being rotated on an axis at an endpoint. Hence me getting I = (1/3)mr^2.

From there, I basically plugged and chugged. Since it's starting at a height, the initial PE = mgh, and initial KE = 0. The final KE = the long equation in 2, and the final PE = 0 since it lands on the floor.

initial PE + initial KE = final PE + final KE
mgh + 0 = 0 + 0.5mv^2(1 + I/(mr^2))
mgh = 0.5mv^2(1 + I/(mr^2))

The two masses then cancelled out to give the following:

gh = 0.5v^2(1 + I/(mr^2))

Since I = (constant)mr^2, the two terms cancel out, so all that's left is:

gh = 0.5v^2(1 + (constant))
(9.81)(0.83) = 0.5v^2(1 + (1/3))
v = [2(9.81)(0.83)/(4/3)]^0.5

And when I solved for v, I got 3.49 m/s. But this isn't the right answer. I'm not sure why it isn't, and I'm not too sure what to do now. I thought maybe that the leg was only doing rotational kinetic energy since it wasn't really doing any linear movement, but that sort of got me confused, so I abandoned that trail of thought.

Some help would be very appreciated. Thanks!

2. Oct 22, 2007

### Staff: Mentor

I have no idea where you got this equation from.
These are OK.

(When calculating the change in PE as the leg falls, remember that it is a uniform rod.)

So far, so good.

....
Time to get back on that trail! Since the leg is in pure rotation (it pivots) just compute the increase in rotational KE. Use that to find the angular speed and then the tangential speed of the foot.

3. Oct 22, 2007

### Nghi

The KE I got is the equation used for rolling motion without slipping. But I guess that wouldn't make sense, since the leg isn't really doing any translational kinetic energy. But about the rotational energy thing: I also did that, I believe, and didn't get the right answer. :(

initial PE + initial KE = final PE + final KE
mgh + 0 = 0.5Iw^2 + 0
mgh = 0.5Iw^2
mgh = 0.5((1/3)mr^2)w^2
gh = (1/6)(r^2)w^2

I substituted in x = 0.83 m for both height 'h' and radius 'r' to get the following:

(9.81)(0.83) = (1/6)(0.83^2)w^2
w = 8.421
v/r = 8.421
v/(0.83) = 8.421
v = 6.9895 m/s

This answer wasn't also the right one, either. Gahhhf;sdlfisf

4. Oct 22, 2007

### Staff: Mentor

This time your method is correct but you're using the wrong value for 'h'. While the foot falls a distance equal to 0.83 m, other parts of the leg do not. Hint: Follow the center of mass.

5. Oct 22, 2007

### Nghi

So the foot falls in a quarter of a circle instead. So... that mean I have to use C = 0.5(pi)r to find height h?

6. Oct 22, 2007

### Nghi

Wait, I just reread what you wrote. If I'm suppoesd to follow its center of mass, and it's a uniform rod, does that mean the center of mass is located at the geometric center? And would that geometric center be located at the knees? Because if so, then that would make the height h = 0.83/2, or 0.415. Would that be the actual height the leg falls?

7. Oct 23, 2007

### Staff: Mentor

Yes, the center of mass of a uniform rod is in the middle. In calculating the change in gravitational PE of an extended object, what counts is the change in height of the center of mass. Obviously, being a rotating stick, some parts move more that others. But the average change in height (and the change in height of the c.o.m.) is 0.83/2 m. That's what you need to use.

8. Oct 23, 2007

### Nghi

Ahh, that makes sense. So a few small after-questions, then:

1. does this center of mass come into play for objects that rotate? Because all of the other problems I've done concerning PE and KE didn't include center of mass as a factor.

2. If it was a rotating sphere instead of a stick, would you still have to use center of mass?

Sorry if I'm clogging up your inbox. I have a test on Thursday, and the whole 'center of mass' being introduced is creating a little bit of an obstacle for me. :/

9. Oct 23, 2007

### Staff: Mentor

If you need to find the change in gravitational PE associated with a rotating or moving object, then sure you'd use the center of mass. Depending on the exact problem, you might not need it.

Realize that this is equivalent to asking for the work done by gravity. And where does gravity act? Effectively at the center of mass.

I don't understand the situation you have in mind.

Feel free to keep asking until it's clear.

Perhaps a specific example might help clarify things. Find one that bugs you and we can discuss it.

10. Oct 23, 2007

### Nghi

I guess what I don't understand is WHEN to use center of mass. Is it used exclusively for gravitational PE? It just seems really random for the problem to pull out center of mass. There really isn't any specific problem in the book that I've encountered (sorry! :(), but it's just in general, I guess. How would I know when the problem calls for COM?

Oh, and as for the rotating sphere/rod thing. You know how in the example you have the uniform rod rotating at an endpoint? Well, you had to go and find the COM in order to be able to find the "h" height that the rod really fell through. Well, if it was a uniform sphere that was rotating around an endpoint instead, would you still have to find the COM to find the height "h"?