# Newton's second law with linear and angular momentum

1. Jan 10, 2010

### simbil

If a force is applied to an object, will the object always experience the same linear momentum change regardless of where the force is applied, or will applying the force away from the objects centre of mass cause lesser linear momentum to develop but with the addition of angular momentum?

For example, a non-accelerating rod in space, applying a force of the same magnitude to its COM compared to applying it perpendicularly to one of its ends.
Will both produce the same magnitude of linear momentum with the second case causing angular momentum in addition to the linear momentum?
Or, will the second case see a smaller magnitude of linear momentum plus angular momentum with the total momentum being the same as the first case?
Or, something else I didn't predict?

The background to the question is a conversation about accelerated running and the correction of angular momentum of the runner (the rod in space is just an easy to discuss example), in case anyone is interested.

2. Jan 10, 2010

### Staff: Mentor

Yes. The linear acceleration of the object's center of mass does not depend on where the force is applied.
No. But a force applied so as to exert a torque about the center of mass will create an angular acceleration about the center of mass as well as a linear acceleration.

3. Jan 10, 2010

### Gerenuk

Some time ago I posted a derivation
https://www.physicsforums.com/showpost.php?p=2493708&postcount=3
Indeed the acceleration of the center of mass is always equal to the sum of external forces irrespectively of their points of action.
Also the change in angular momentum about the center of mass is always the torque from external forces about the center of mass. No matter if the object is free in space or a part in a physical device.

4. Jan 11, 2010

### simbil

That's very helpful - thank you both.

5. Jan 11, 2010

### rcgldr

Although the force is the same if applied to the center or end of a rod, the average speed of that force over time is faster if applied to the end of the rod. The work done is equal to the force times the distance of the path that the force is applied, and more work is done when the force is applied at a faster average speed to the end of the rod over a longer path, and this difference in work done accounts for the additional angular kinetic energy of the rod.

I did the math comparing linear only versus linear + angular acceleration of a sphere on a treadmill in post #7 of this thread.

Last edited: Jan 11, 2010
6. Jan 11, 2010

### Gerenuk

I'm not sure if I understood this incorrectly or if it is written misleadingly.
Work does not depend on velocity. In your threadmill example the difference is that in one case the sphere acquires some rotational energy but not in the other case.

Or maybe one difference to the threadmill example is that there the speed of the threadmill is given, but the actual force acting on the sphere is a complicating subject. Instead with the rod example the force is given a definite magnitude and creates a well-defined position and velocity independent change in linear momentum.

7. Jan 11, 2010

### rcgldr

If a force is applied at a higher velocity, then per unit time, that force is applied over a longer distance, and the amount of work done (force x distance) is greater. I was just explaining how the higher velocity affected the result: power = force x speed, and work = power x time = force x distance.

In my treadmill example, the force was the same in both cases, 1 Newton. The difference in the two cases was in the treadmills rate of acceleration, velocity, and distance involved (except at time=0).

8. Jan 11, 2010

### Gerenuk

Oh OK. Maybe you have an alternative view with power and time.

It still want to make clear at with the rolling sphere on the threadmill the force transfer is not the same as with the glued sphere. If the force were the same then both would accelerate at the same rate.

9. Jan 11, 2010

### rcgldr

In my example of that treadmill, the force is the same in both cases, 1 Newton, and the linear acceleration of the spheres is the same in both cases, 1 meter/second^2. The difference is in the rate of acceleration of the treadmill surface, 1 meter/second^2 for the glued sphere case, 3.5 meters/second^2 for the rolling sphere case.

10. Jan 11, 2010

### Gerenuk

Ah ok. Sorry, I didn't go through your derivation.