# Angular momentum and torque

1. Mar 30, 2010

### usljoo

i guess my question is actually really really simple.

lets consider rotation in the xy plane so that L points in the direction of the z axis.
the angular momentum is defined as L=rxp where r is the position vector and p the linear momentum.

suppose that p is constant in magnitude and perpendicular to r so that we have a constant rotation with constant angular velocity.

my problem is now that when one makes r=2r (makes r twice as large) then for the angular momentum to stay constant the linear momentum needs to drop by a factor of 2.
now one can also write L=rmv for the rotation described above now the problem comes with the torque T=rma witch means that if T is supposed to stay constant when r=2r then a=a/2
but this would mean that the force has dropped just because r did rise.
now im wondering what kind of physical meaning this would have, and i know that it must have some because of the lever.

now before i bite my thong explaining things, could someone please point to my mistakes thx

2. Mar 30, 2010

### tiny-tim

Welcome to PF!

Hi usljoo! Welcome to PF!

I'm confused

if the angular velocity is constant, then the angular acceleration and the torque (T) are both zero.

Can you explain your set-up?

3. Mar 30, 2010

### usljoo

yes thats true if the angular velocity is constant then the force and torque are zero but i only mentioned constant angular velocity in the context of angular momentum and not of torque and force.

that is to say that if there is an rice in linear momentum then there is of course a force.
and thats exactly the thing, newtons second law says that F=ma so if one doubles the mass one gets only half the acceleration if F is supposed to stay constant or to be the same.
now if the concept of torque is made to follow this idea that is T=I$$\alpha$$ whitch means i T is the same and one raises I then $$\alpha$$ must drop by the same factor.
but I=(r^2)m and $$\alpha$$=a/r so if r=2r then I=4R so T=I$$\alpha$$=4I$$\alpha$$/2 so that T=2T so that a must drop one more time by r.

i hope you get the idea of what troubles me if not please tell me

4. Mar 30, 2010

### tiny-tim

ok, let's ignore the angular velocity completely.

Suppose you have a turntable which is being turned by a belt applying a torque T.

Suppose the turntable is very light (so we can ignore its moment of inertia and it mass), and the only mass is an object of mass m at distance r (or 2r) on the turntable, and the applied torque is the same for both positions of the mass.

Then the angular acceleration of the system is T/mr2 or T/m(2r)2.

5. Mar 30, 2010

### usljoo

yes thats correct but im not concerned with numbers here, you see newton made his second law because of experiments, he applied the same force first to one mass and then to a different one and noticed that acceleration and force are proportional right, but as torque is defined it gives results witch should not exist.

for example the lever, the same torque is suposed to be aplied on both endpoints but one torque has a lover linear force and the other a higher but how can this be explained in terms of angular acceleration and linear acceleration of the endpoints

6. Mar 30, 2010

### tiny-tim

I'm sorry, I don't understand at all.

I gave a specific mechanical device … I think you agree with the analysis of it.

Can you please illustrate the point that you're making by also describing a specific mechanical device?

7. Mar 30, 2010

### usljoo

allright ill discribe the lever.
(hope everyone is familiar with it if not, http://en.wikipedia.org/wiki/Lever)

i allways thought that it is easier to raise the weight on the longer side from the pivot than it is from the shorter one, because there is more weight coming from the longer rod but thats not it as one can see on the picture from wikipedia)

now the explanation there says that this is so because the torques are equal on both sides but each torque is differently calculated as you can see.

as i see it, the angular accelerations on both endpoints must be the same.
but if that is so then because a=$$\alpha$$r than the linear acceleration on the longer endpoint must be greater than the one on the shorter so that a larger net force is acting on the endpoint that is longer. on the other side when considering torque it follows that a smaller force is acting there.

so how to combine these two views.

8. Mar 30, 2010

### tiny-tim

I'm looking at the picture, but I'm not seeing what you're describing.
Again, I don't see that explanation, or anything like it, on that page.

I'm sorry, but I still don't understand.​

9. Mar 30, 2010

### closet mathemetician

Maybe its the same principle as for the angular momentum, that is, in order for the angular momentum of the system to be the same all over, the points on the outside edge must travel at a higher velocity than the points closer to the center, because the outer points have more distance to travel (greater circumference) in the same amount of time. Seems to me if the system was constantly accelerating, you'd have the same issue with the acceleration.

10. Mar 30, 2010

### Naty1

I wonder if the poster is confused between Newtons F = ma which is a force applied directly to a mass (at its center of mass) which causes a linear acceleration and torque which represents a force applied at a distance r from the center of mass...and tends to rotate a mass about an axis....in other words torque measures the results of a force applied as a moment....and linear (straight line) KE of 1/2mv2 (from v =wr becomes) 1/2m(wr)2....

In any case, torque has nothing to do with acceleration....

11. Mar 31, 2010

### usljoo

yes thats exactly what im thinking too naty, torque should not have anything to do with linear acceleration right but T=r$$\times$$F and lets raise r then for T to stay the same F needs to drop.

or to put it in another light lets suppose that we have two particles rotating in the xy plane about the z axis but one of them having twice the radius of rotation than the other.
now for these two particles to have the same torque the linear force on the longer needs to be half the force on the shorter, by definition of torque, or that the longer particle has less linear acceleration if we suppose that both particles have the same mass.

now in newtons law we have F=ma so our situation above is analogues to the situation when we raise m twice so a must drop twice, but this is a fact of experience.

its really hard for me to explain too so i wanna thank everybody for responding at all.

but why should F drop if r raises and what kind of measure for rotation is this anyway, i mean torque.

so essentially what would follow from the above is that is we would apply the same torque to two particles witch rotate on different radius then the linear force on the longer would be less, but then again how can one apply torque without the linear force.

12. Mar 31, 2010

### sganesh88

Torque is indeed a messy non-intuitive concept. Why should a lesser force cause the same torque just because its applied at a greater lever arm tortured me too initially.Just get hold of some Mechanics book-I'd prefer Kleppner. Study the derivation of the expression between Torque and angular momentum of a system of particles.
The limitation having a greater torque using the mechanical advantage of a lever is the greater linear distance to cover the same angle. The same reason why lower gears give you a large force but poor speed.

13. Mar 31, 2010

### rcgldr

Perhaps part of the problem here is that angular_acceleration = torque / angular_inertia, not torque / angular_momentum.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook