# Homework Help: Conceptual Center of Mass acceleration problem

1. Nov 12, 2008

### bocobuff

1. The problem statement, all variables and given/known data
3 identical rectangular blocks on frictionless surface have equal magnitudes of force and in the same direction are exerted on them. The CoM is the same for all the blocks and in the center of the rectangle. For block 1, the force is exerted to the left of the CoM, and halfway to the edge of the rectangle. Block 2, the force is exerted at the CoM and for block 3, the force is exerted on the far right edge of the rectangle. What is the direction of the acceleration of the block's CoM? Rank the magnitudes of CoM accelerations.

The attempt at a solution
I am thinking that block 1 will move cw, block 2 will go straight, and block 3 will go ccw. Also I think the magnitude of Block 3 is greater than blcok 1 because the distance between CoM and where the force is applied for block 1 is half that of block 3.

I can envision the whole movement but I can't think about the direction of the CoM. Anything would help.

2. Nov 12, 2008

### bocobuff

3. Nov 12, 2008

### LowlyPion

The distance from the C of M will determine which portion of the force will go into torque and which portion will go into translating linear motion won't it?

2 will go straight as you suggest with nothing to introduce torque..

The only problem you have is determining which of the other 2 will give more linear motion and to which direction the center of mass will fall off onto.

4. Nov 12, 2008

### bocobuff

So less force is going to torque in block 1 because the point of force is closer to the CoM? Which would mean that it has a greater linear acceleration? So am I supposed to compare the magnitudes of the angular acceleration of CoM or linear?

5. Nov 12, 2008

### LowlyPion

I hope not. Only rank them looks like to me.

Cheers

6. Nov 12, 2008

### bocobuff

I must be retarded then. Still don't get it. Wouldn't the force exerted be the tangential component that causes torque?

7. Nov 12, 2008

### borgwal

The CoM motion is determined only by the net force on the block, and so is equal for all 3 blocks.

There is no "portion of the force" that goes into translating linear motion.

8. Nov 12, 2008

### LowlyPion

Since it is frictionless and the blocks do have mass there will be varying degrees of linear momentum and angular momentum imparted. In 2) there is no net torque and so it is all linear. In 1) there will be a force couple between mass toward the outside edge, and the mass toward the center of mass. In 3) the force couple will be potentially minimal, hence the center of mass will will drift more quickly away than 1)as the edge is engaged.

9. Nov 12, 2008

### LowlyPion

I think you need to read the problem more carefully. Only one of the forces is acting directly through the center of mass of a block. That force will produce only a change in linear momentum. The other 2 are applied in an unbalanced way.

10. Nov 12, 2008

### borgwal

The accelerations are all three the same. The amounts of work done by the identical forces are different, and that's how some blocks will rotate in addition to their *identical* CoM motions.

11. Nov 12, 2008

### borgwal

Check your textbook, and see what causes the CoM motion.
You sum all external forces, independent of where they act.

12. Nov 12, 2008

### LowlyPion

Borgwal, my apologies. You are exactly right. It was my muddy thinking. Thank you for straightening out my faulty understanding.

So bocobuff, please disregard my earlier direction. That was incorrect on my part. The forces will act on the CoM as Borgwal has indicated.

13. Nov 12, 2008

### borgwal

Okay, no problem!