# Not sure about my rotations problem

1. Jun 7, 2015

### toesockshoe

1. The problem statement, all variables and given/known data
A rod of length L and mass M is balanced in a vertical position at rest. The rod tips over and rotates to the ground with its bottom attachment to the ground never slipping. Find the velocity of the center of mass of the stick just before it hits the ground and also find the velocity of the tip just before it hits.

Can someone check my work to see if im correct?

2. Relevant equations
W=Delta E

3. The attempt at a solution
$W=\Delta E$
$0 = \Delta RKE + \Delta GPE$
$mg\frac{L}{2} = \frac{1}{2}Iw_f^2$
$mg\frac{L}{2} = \frac{1}{2} \frac{1}{3}mL^2w_f^2$
$w_f = \sqrt{\frac{3g}{L}}$

2. Jun 7, 2015

### Orodruin

Staff Emeritus
You are doing fine so far. You now need to relate this angular frequency to the velocities the question asks for.

3. Jun 7, 2015

### toesockshoe

oh yeah, so just multiply it by r correct? so it owuld be: $v_f = \frac{L}{2}\sqrt{\frac{3g}{l}}$

4. Jun 7, 2015

### Orodruin

Staff Emeritus
This is for the center of mass, you should also do it for the tip (although the relation is trivial). I also suggest including the L/2 inside the square root (of course, it then becomes L^2/4) to simplify the expression slightly.

5. Jun 7, 2015

### toesockshoe

would you just multiply it by 2?

6. Jun 7, 2015

### Orodruin

Staff Emeritus
Yes, the distance to the center of rotation is twice as large.

7. Jun 7, 2015

### toesockshoe

how would you prove this on a test, and what if they wanted the velocity of some point of b distnace away from the center? how would you find velocity then? multiply by b*2? Also, can you explain conceptually why its twice as fast at the tip? I just read it somewhere in my book so I said multiply by 2.

8. Jun 7, 2015

### Orodruin

Staff Emeritus
Any point in a rotational motion has the velocity $v = \omega r$, where r is the distance to the center of rotation. That the tip is twice as far away follows from the fact that the center of mass is in the middle of the rod.

9. Jun 7, 2015

### toesockshoe

ok so why is the center of rotation at the bottom of the wheel? doesnt the wheel rotate around the middle of the wheel? a bicycle tire certainly looks like it spins around the center.

10. Jun 7, 2015

### Orodruin

Staff Emeritus
This depends on the frame you are using to look at the wheel. In the current problem, it is pretty clear what the center of rotation is.

11. Jun 7, 2015

### toesockshoe

ok yes, in this problem the center of rotation is the bottom of the stick. but imagine a bicycle wheel rolling..... woudlnt it seem like the center of rotation in that case is the center of the bike?

12. Jun 7, 2015

### Orodruin

Staff Emeritus
If you use a system moving with the bike, yes. If you use a system fixed at the ground, no.

13. Jun 7, 2015

### toesockshoe

but it would be impossible to work with with a frame moving with the bike right? becuase wouldnt that be a noninertial frame?

14. Jun 7, 2015

### Orodruin

Staff Emeritus
As long as the bike is not accelerating, the frame is inertial. And there is nothing preventing you from doing computations in non-inertial frames, it is just that you will have to take additional things into account.

15. Jun 7, 2015

### toesockshoe

ok assuming the bike is accelerating and I'm to dumb to use pseudo-forces, does the origin have to be the road?

16. Jun 7, 2015

### toesockshoe

17. Jun 7, 2015

### Orodruin

Staff Emeritus
If you are interested in the velocity of a point on the wheel relative to the road, it is the most convenient choice.

18. Jun 7, 2015

### toesockshoe

ok fine, but the center of rotation is still the middle of the wheel correct? we're just letting the ground be the orgin because its earier to solve problems?

19. Jun 7, 2015

### Orodruin

Staff Emeritus
No. The (momentary) center of the rotation in the ground frame is the contact point (this is the point on the wheel with zero velocity).

20. Jun 7, 2015

### toesockshoe

ok that doesn't seem to make sense to me. can you explain what exactly a center of rotation is? i thought the center of rotation is the point where the object rotates around.... a bicycle tire doesnt rotate around the ground or else it would also have to go under the ground right?