# Quick question on the conservation of angular momentum

1. Dec 2, 2012

### LiftHeavy13

okay, i realize that the angular momentum of a moving point mass could be looked at about any point, and that angular momentum is conserved as long as no torque is acted on that point mass. but, something i don't understand is how, then, the angular velocity could increase if the moment of inertia decreases.

Here me out:

We have a string moving a point mass on a horizontal table at a constant speed in a radius of R1. The angular momentum of the point mass about the center of the circle is L1. Now, we pull the spring in, decreasing the distance between the point mass and the center to R2. Technically, since the force always acted parallel/antiparallel to the radial vector from the center to the point mass, no torque was done on the point mass about the center, and angular momentum is conserved... But we decreased the distance between them, and therefore the moment of inertia of the point mass about the center. Hence, the angular momentum of the point mass about the center increased... But how is this possible if there was no torque done on the point mass about the center and therefore no angular acceleration?

2. Dec 2, 2012

### Rap

The angular velocity of the particle increases as you draw it inward, the moment of inertia decreases, and their product - the angular momentum, remains unchanged.

3. Dec 3, 2012

### Kevin Willis

Correct me if I am wrong people :)

There was an acceleration. The speed increased and the mass stayed the same. The inertia concept can be demonstrated with in a experiment that has two wheels of identical weight and diameter on a incline. One wheel is solid and the other is ring shaped without the center. The solid wheel accelerates faster because its inertia is less even though they weigh the same. I cant explain this concept much further but this might help.

4. Dec 3, 2012

### dev70

hello, Liftheavy 13, i think you should think about the conservation of energy which is 1/2 I w^2, where I=moment of inertia and w= angular velocity..
i think this should explain..

5. Dec 3, 2012

### Rap

No, energy is not conserved, the particle is being subjected to a force.

6. Dec 3, 2012

### dev70

ok..i understand..but as per liftheavy angular momentum should increase? why? as no external torque is applied angular momentum is always conserved. so, IW remains constant. and the angular velocity changes as per the above equation.

7. Dec 3, 2012

### Khashishi

The classic example is the figure skater doing twirls. When one pulls one's limbs in, it decreases the moment of inertia, so the angular velocity increases. The angular momentum stays the same.

At a microscopic level, I suppose you could picture the molecules moving on average about the center of mass with some linear velocity perpendicular to the direction to the center of mass. If these molecules are pulled in toward the center of mass, the linear velocity doesn't change (conservation of linear momentum), but the same linear velocity becomes a larger angular velocity because it's closer.

(edit: Oops ignore the second statement.)

Last edited: Dec 3, 2012
8. Dec 3, 2012

### jbriggs444

So far, so good.

1. The skater has to exert significant force to draw his or her arms in. That force does work on the arms, increasing their kinetic energy. If their kinetic energy increases, it follows that their linear velocity has increased.

2. Moment of inertia scales as the square of radius. If you scale down the radius by a factor of two and conserve angular momentum, it follows that angular velocity increases by a factor of four. It is then clear that linear velocity has increased by a factor of two.

3. As you reel in an object on a string the only force is radial -- there is no torque. But the radial direction and the tangential direction are not at right angles as an object is pulled in on a spiral trajectory. There is a net tangential acceleration. So the object ends up moving faster than it started.

9. Dec 3, 2012

### LiftHeavy13

okay... that still didn't answer the question. how can angular velocity increase if there is no net torque and therefore no angular acceleration?

Last edited: Dec 3, 2012
10. Dec 3, 2012

### bp_psy

It doesn't.

11. Dec 3, 2012

### LiftHeavy13

yeah, sorry made a quick edit. no one has answered it yet, not even my teacher. lol, it makes no sense at all

12. Dec 3, 2012

### Rap

Why is this so hard? L=I w where L is angular momentum, I is moment of inertia, w is angular velocity. As you draw the object inward, I goes down, w goes up, L stays the same. Angular momentum does not increase. Moment of inertia goes down. Angular velocity goes up.

13. Dec 3, 2012

### LiftHeavy13

hey, maybe you should read what I'm asking. I know that angular momentum is conserved; I even stated that in the OP. My question is, once again, how angular velocity can increase if the net toque is zero and therefore angular acceleration is zero.

14. Dec 3, 2012

### rcgldr

The force from the string on the point mass is not perpendicular to the spiral path of the mass that occurs when the string is pulled in or released out. There's a non-zero component of force in the direction of the mass, which speeds up (or slows down if tension reduced) the point mass.

Take a look at post #3 of this thread, which also includes the math to show that the (internal) work done equals the change in kinetic energy. Post #4 of this thread shows the case where the speed of the point mass is kept constant by winding (or unwinding) a string around a pole, but in this case angular momentum is not conserved unless you include the effect of the torque exerted on the central pole.

Last edited: Dec 4, 2012
15. Dec 4, 2012

### Rap

Because net torque of zero does not imply that angular acceleration is zero. Torque is not equal to the moment of inertia times the angular acceleration. This is only true when the moment of inertia is constant. The complete definition of torque is that it is the time rate of change of angular momentum. $\tau=dL/dt$ where $L=I\omega$, where $\tau$ is torque, $L$ is angular momentum, $I$ is moment of inertia and $\omega$ is angular velocity. This means that $$\tau=I \frac{d\omega}{dt}+\omega\frac{dI}{dt}$$ $\tau=I \frac{d\omega}{dt}$ is NOT true in this case. Sorry, I did not understand what it was that you did not understand.

Last edited: Dec 4, 2012
16. Dec 4, 2012

### cosmic dust

Actually, you could set Ï„=0 to get:

$\frac{d\omega}{dt}=-\frac{\omega}{I}\frac{dI}{dt}\ne 0$