# Apparent paradox with angular and linear motion

1. Jul 25, 2012

### cybie

A friend raised an observation which I thought would be easy to explain. But it seem I have a gap in my physics knowledge...

An object (rigidbody) is stationary in zero gravity empty space. In the first situation we apply an impulse to its COG and its linear momentum changes, and no angular momentum change. Correct?

In the second situation, same initial condition, but the same impulse is apply off center, inducing a change in angular velocity, as well as the same change in linear momentum as in the first situation.

In the first situation, the impulse resulted in the object having an linear kinetic energy. In the second situation, the impulse caused the object to have the same linear kinetic energy, but also an additional rotational kinetic energy.

Why aren't the total kinetic energy after the impulse the same in both situation?

2. Jul 25, 2012

### mikeph

Energy must be conserved, so the energy from the impulse is divided between linear and angular kinetic energy. The ratio of the division depends on the magnitude of the components of the impulse along the directions parallel and perpendicular to the direction from the point of impact to the body's centre of mass, and also the mass and moment of inertia about the axis of induced rotation.

3. Jul 25, 2012

### jbriggs444

One way to understand this is to realize that the work performed by this impulse is proportional to the distance moved by the point of application of the force.

If you apply the impulse at the center of mass, the only resulting motion is linear.

If you apply the impulse off-center, there is both linear motion and rotational motion. The point at which the force is applied moves further under application of the same force.

So even though the force * time might be the same for the two situations, force * distance can be different.

4. Jul 25, 2012

### cybie

>So even though the force * time might be the same for the two situations, force * distance can be different.

I want narrow to just impulse and not force (I am assuming it is theoretically possible), because I don't want to deal with how the application position of the force change over time.

In all the text I read about change in momentum, I didn't find one which change in linear momentum is a function of 'offset' from the center where the impulse is apply. That why I assume the change in momentum is just a function of mass and the magnitude of the impulse.

If that's the case, then by offsetting the impulse, I will get same change in linear momentum plus an addition angular momentum, resulting in a bigger total kinetic energy. So that doesn't sound right.

What did I miss?

5. Jul 25, 2012

### Staff: Mentor

That is correct.

That is also correct.

You missed that impulse ≠ energy. The center and offset impulses have the same impulse but different energies. You cannot match both except in special cases (like both on center, or both offcenter by same amount).

6. Jul 25, 2012

### 256bits

The second part of that statement is untrue.
While a body rotaing and moving linerally at the same time can be analyzed for the velocities of certain points by suposition of translation and rotation, the same cannot be said for an inpulse applied off centre to the centre of gravity - ie you do not calculate first the angular momentum as if the impulse was applied off some distance to the CG and then calculate a linear momentum as if the same impulse was at the CG.

If you do think about what you have stated is that
1. Apply impulse off from the CG so that the rigid body aquires an angular momentum.
2. At an oportune moment, apply the same impulse at the CG so that the body aquires a linear momentum.

I will explain it in terms of collisions in regards to conservation of momentum. Note that when two bodies do collide they are acting upon each other with impulse.

You should look up coefficient of restitution, e.
If e = 0 , a collision is non elastic
If e = 1 , a collision is elastic and energy is conserved
For e=0 to e<1, energy is not conserved.

For a linear collision, two bodies with collinear velocities of the CG V1a and V1b with the point of contact will collide and separate with CG velocities V2a and V2b. And one can calculate the coeficient of restitution from the values of velocity.

You may find this surprising, but rigid bodies have the same coeficient of restitution whether the contact is colliner with the CG or off centre. What that means is that the point(s) of contact of the bodies respectively having velocities V1a and V1b before contact will have velocites V2a and V2b after contact. Other parts of the rigid bodies will have different velocities. ( The velocities mentioned here are vectors ).

Note that it follows that the CG of the rigid body(s) when hit off centre cannot have the same V1b or V2b as the point of contact or there would not be any rotaion.

7. Jul 25, 2012

### cybie

Thanks for all the replies.

Then how do you calculate the change in linear and angular momentum due to an off-center impulse?

8. Jul 25, 2012

### AlephZero

I don't think it is very helpful to bring collisions and coefficients of restitution into this. Applying a known impulse to a body doesn't necessarily have anything to do with collisions. It is equally relevant to analysing "steady state" situations like rockets or jet engines.

Dalespam has it right: The kinetic energy is different in the two situations because impulse ≠ energy - they are measured in different units, so they can't be "the same thing".

9. Jul 25, 2012

### AlephZero

You seem to have done the right things in your OP, but the problem was you didn't believe the answer.

Impulse = change of linear momentum.
Moment of the impulse about the CG = change of angular momentum.

10. Jul 25, 2012

### cybie

Noting that energy does not equal impulse.

But the difficulty I think I am having is the that 1st situation (on center) and 2nd situation (off center), both with the same impulse, give different total kinetic energy. How do I explain that the 2nd case have higher total kinetic energy.

Let me elaborate: Let's say the situation we are talking about is coin flipping. Assuming my thumb is asserting an impulse on the coin. In situation 1 is my thumb impact the coin on-center. 2nd situation my thumb impact the coin off-center. Assuming I use the same amount of energy to flick my thumb, how come the 2nd case resulted in higher total kinetic energy?

Last edited: Jul 25, 2012
11. Jul 25, 2012

### Staff: Mentor

This is incorrect, the statement was true. An off-center impulse imparts the same linear momentum as an on-center impulse.

12. Jul 25, 2012

### Staff: Mentor

You can work it out mathematically by the process jbriggs444 suggested above. Then, if you want to go to an instantaneous impulse you can simply fix the impulse and take the limit as the time goes to 0. You will see that the work done is not the same, and so therefore by the work energy theorem the KE is not the same either.

13. Jul 25, 2012

### cybie

Thanks for continuing to answer to help me get my head around this.

I stated that the same amount of energy was use to flick my thumb to impact the coin. It must follow that the kinetic energy of the coin be the same whether I hit it on or off center. If I hit it on center, the linear velocity should be less than the linear velocity if I hit it off center (because now it also has angular momentum). But that seem to contradict the statement that the change in linear momentum does not take into account of where the impulse was apply.

14. Jul 25, 2012

### Staff: Mentor

An on center force and an off center force may have the same impulse, or they may do the same work, not both. If you state as a fact that the work is the same then the off-center impulse must be less.

15. Jul 25, 2012

### rcgldr

Maybe it would help to know that in order to generate the same impulse off center, the point of application of that force will be accelerating faster, resulting in the force being applied at a faster speed over time. Since power = force x speed, a greater amount of power is involved when the same impulse is applied off center.

16. Jul 26, 2012

### Darwin123

"So that doesn't sound right."
What you missed is that "sounding right" is not a law of physics.

Law of physics: The impulse is equal to the total change in linear momentum of the system. This is a law of physics.
Statement that sounds right:The impulse determines the total change of kinetic energy of the system.

The "law of physics" is correct. The "statement that sounds right" is incorrect.

The impulse is equal to the change in linear momentum of the system. Therefore,
F_Ext Δt=M_Total ΔV_CM
where "F_Ext" is the average force over the time interval "Δt", "M"_Total is the total mass of the system and "ΔV_CM" is the change in velocity of the center of mass of the system. I couldn't write the vector the vector caps on the force or velocity, but they are there,
You can't determine the total kinetic energy of the system from "F_Ext Δt". As you pointed out, there are other degrees of freedom that are not "linear". The system has angular degrees of freedom, too.

17. Jul 26, 2012

### jbriggs444

Not a very accurate experiment -- much energy is lost and little measurement is done in that scenario.

If you were able to look very closely, you would see that in order to apply the same impulse in an off-center impact as in an on-center impact, the thumb has to hit harder.

That's because in an off-center impact, the coin rebounds more quickly. This means that the duration of the impact is reduced. And since impulse is the product of force * duration, that means that the force has to be increased in order to impart the same impulse.

18. Jul 26, 2012

### cybie

Thanks rcgldr! I think this is what I am looking. In order to explain the off-center case resulting in greater total kinetic energy, the energy it take to produce the impulse is greater than the on-center case, even though magnitude of both impulse are the same. Hence conservation of energy is not violated. Right?

Thanks

19. Jul 26, 2012

### schaefera

Try thinking about it this way:

In both cases, the impulse is the same. We agree on this.

But, for the ON CENTER case, your force (or the energy you supply) is only "fighting" linear inertia (that is, the mass).
In the OFF CENTER case, your force (or the energy you supply) is "fighting" both the linear inertia (mass) and the angular inertia (its moment of inertia about the CoG). So, in order for you to supply the same impulse to both cases, you MUST use more energy in the off-center case, since you're "fighting" more inertia, and using more energy is exactly what allows the total KE at the end to be greater because it is rotating.

Thus, impulse= same in both cases, but KE=different.

20. Jul 26, 2012

### rcgldr

You can think of the energy added as power x time or as force x distance, in the first case, the power is greater because of the greater speed, in the second case, the distance is greater because of the greater speed (while the force and time remain the same).

In the off center case, there's less resistance to the force, because the object is free to rotate as well as translate (move linearly), so in order to generate the same force, the acceleration needs to be greater.

Last edited: Jul 27, 2012