# Forces exerted somewhere other than CM

1. Feb 7, 2016

### Moazin Khatri

Let's say I have a pen up in space in a zero g environment. And I carefully give the pen a nudge at the very tip of it. That's pretty away from the center of the mass of the pen. I know that will exert a torque and the pen will spin but I want to know whether its center of mass (along with the whole body) will move in a forward direction or not? And If yes how am I going to calculate how much it will move forward?

2. Feb 7, 2016

### Orodruin

Staff Emeritus
You can compute it just the same way as you would normally, by computing the total momentum of the pen - which is going to depend on the impulse you give it.

3. Feb 7, 2016

### Moazin Khatri

Then is it true that whether I exert a force at the CM or at the very tip of the pen it doesn't make a difference? Will it still go with the same velocity?

4. Feb 7, 2016

### Orodruin

Staff Emeritus
Of course it will make a difference, the pen will start rotating in one case and not in the other. For the overall movement of the CoM, it will not matter.

5. Feb 7, 2016

### Moazin Khatri

Yes I was talking about the over all CoM's movement.
But in the rotating case my nudge has given the pen a spin and some energy consequently. So won't this be a violation of the conservation of energy? Because if, in both cases, the pens have equal linear momentum thus equal kinetic energy due to their forward velocity but only in the rotatory case it also has a kinetic energy due to the spin...

6. Feb 7, 2016

### PeroK

The pen's initial kinetic energy is 0 in both cases, so energy isn't conserved in either case.

7. Feb 7, 2016

### Moazin Khatri

Obviously the energy is given by my nudge. But suppose I give equal nudges. Or same force for the same period of time.

8. Feb 7, 2016

### PeroK

Force x time = impulse = change in linear momentum. Force x time does not equal change in Kinetic Energy.

(I think this is a really good question and worth thinking about. Here at PF, we like you to work it out for yourself as far as possible.)

9. Feb 7, 2016

### Prannu

Wait a minute, if the pen is in a zero-g environment, there won't be any torque on the pen (you need two forces that are acting at different parts of the pen to actually start the rotation).

10. Feb 7, 2016

### PeroK

That's not the case at all. Just think of something on a flat surface.

11. Feb 7, 2016

### PeroK

Here's a hint. If you think about an impulse as a instantaneous spike, it's difficult to see what's happening. Instead, imagine you apply a modest force for 1-2 seconds. Now, compare the two cases where you apply the force to the CM and to one end. You will notice significant differences in what happens. For example, in the second case, you'll have to make a decision about how you are going to continue to apply the force once the pen starts moving (and rotating).

12. Feb 7, 2016

### Prannu

Going with your first example PeroK, if a pen was on a flat surface, then the only reason why it rotates is because friction from the ground (acting at the CM) couples with the flick of your hand at a place that is not the CM. So if the pen was on a surface with 0 friction, the torque would be zero (unless I'm forgetting something here).

13. Feb 7, 2016

### Moazin Khatri

Thanks for your hint and advice. Yes I am already seeing it. I'll do a complete analysis and post it here for you to check it :) Thanks alot

14. Feb 7, 2016

### PeroK

An object will rotate as a result of a single torque. Any intro to mechanics will cover that. Just google "torque". Moreover, the friction at the CM exerts no torque on the object, so is certainly not required for rotation.

15. Feb 7, 2016

### Prannu

All right, thank you!

16. Feb 7, 2016

### Moazin Khatri

@PeroK
Case 1 => Force applied at the center
I apply a force at the CoM of the pen. The impulse lasts about 1 second and the pen goes off in a straight line with no spin at all.
Case 2 => Force at tip
I'm assuming that newton's laws can be applied to any object with any shape or size and at any location.
I might apply the same force in this case but the impulse time will be short. The reason is my finger's direction will be a straight line and the pen will rotate and eventually miss my finger in about half of a second or maybe even less. If I apply F=dP/dt here I can easily see that in this case the pen's final velocity is going to be very low because the impulse time was very less compared to the first case.
I also exerted a variable torque in this case and due to that torque the pen will have a spin as well. Variable because of changing angle.
I did this experiment on a table and also noticed that the closer you nudge to the CoM the more linear momentum and the less angular momentum you give. The reason is because the closer you are the more impulse time you get and so you transfer more linear momentum. While the torque becomes less.
So basically both cases aren't identical at all and due to that my first question becomes meaningless.
Please correct me wherever I am wrong. Thanks

17. Feb 7, 2016

### Orodruin

Staff Emeritus
You cannot do this experiment reliably on a pen on a table. The imparted energy and the imparted momentum are very different concepts. The answers you have gotten have been based on your initial assumption of constant force. If you have a constant force for a fixed amount of time the imparted momentum is going to be the same, but the imparted energy is not. The work done is proportional to the force times the displacement, not force times time, and the displacement in the different cases is going to be different. When you do your pen experiment you will have serious trouble controlling that you are actually giving the same force during the same time.

18. Feb 7, 2016

### Moazin Khatri

Yes. I got it. Thanks

19. Feb 7, 2016

### andrewkirk

Perhaps an easier way to get insight into this situation might be to consider a case where the impulse is delivered by a small object striking the end of the rod (let's say a rod because pens are not uniform, which makes the maths more difficult) in an elastic collision (so that kinetic energy is conserved).

The reason that might be easier is that you can then use the equations, for conservation of kinetic energy, linear momentum and angular momentum to solve for the final pattern of motion.

This contrasts with the 'nudge' scenario in which, as PeroK pointed out in post 11, there will be (potentially considerable) additional complexity in how the force is delivered over the non-zero time in which contact is made. We cannot use any conservation laws because the nudge is applied by an external body.

In the collision scenario, if we treat the colliding object as a point mass (so it has no rotation), we have five unknowns:
- final x velocity of rod
- final y velocity of rod
- final rotation rate of rod
- final x velocity of point mass
- final y velocity of point mass

And we have the following equations:
- conservation of linear x momentum
- conservation of linear y momentum
- conservation of kinetic energy
- conservation of angular momentum

We need one more equation to solve the problem. I'm having a bit of a dumbo morning and can't see right now what it should be.