# Levers physics problem

1. Dec 1, 2008

### saltine

1. The problem statement, all variables and given/known data
Force F is applied to a rod resting on a frictionless surface. What is the acceleration at B?

2. Relevant equations
The length of the rod is L. Its mass is m.
Icenter = mL2/12

3. The attempt at a solution
The rod is supposed to spin about its center of mass. The acceleration at B is the same as that at A in polar coordinate center at the center of mass.

τ = (L/2) F = Icenter α => α = 6F/mL
a = (L/2) α = 3F/m eθ

where <b>eθ</b> is a unit vector in the counter-clockwise tangent direction relative to the center of rotation.

Is this correct? Where did the centripetal force go?

There is no centripetal force because there is no actual rotation at the moment (ω=0 rad/s).

- Thanks

2. Dec 1, 2008

### LowlyPion

Re: Levers

Since there is no friction, the "a" should be F/m shouldn't it?

3. Dec 1, 2008

### glueball8

Re: Levers

shouldn't it be just Fnet=ma?

4. Dec 1, 2008

### saltine

Re: Levers

That is my question: Should it??

Do these situations give the same movement at B?

When I tried it using a pen on a table, it is obvious that point B moves up in the top diagram and it moves down in the bottom diagram. The F is not applied to the center of mass in the original question. Does that make the difference?

If it was just F = ma, wouldn't it imply that point B would be moving downward? Which contradicts the reality?

5. Dec 1, 2008

### glueball8

Re: Levers

I know that the center mass will move in a line. There's was a MIT physics physics lecture on something very close to this. Can't seem to find where..

6. Dec 1, 2008

7. Dec 1, 2008

### saltine

Re: Levers

I understand it but I don't that the problem is the lecture is similar to this one. In the example in the video at 7:38 where there is a spinning rod, the rod is spinning about point P, but the center of mass of the rod is at point C. The professor was saying that spinning the rod like this generates a force along the rod, because the centripetal forces do not balance out.

The part that is almost the same as my problem comes at 15:22, it shows that the center of mass would move parallel to force F, and the rod will spin at the same time.

The rod it self would have an acceleration of acm = F/m (moving downward). The point B of the rod would have an additional component from the rotation 3F/m which moves up. So if I add the two together, point B would have the acceleration:

aB = 3F/m - F/m = 2F/m this says that the point B would move up momentarily.

If there was an object just above point B, when the force F is applied, point B will push against that object.

Last edited: Dec 1, 2008