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Next: Angular Momentum Up: Angular Momentum and Torque Previous: Angular Momentum and Torque


First think about what a force does. If you apply a force to an object, it'll have a tendency to go in that direction. Now you want to know what kind of thing you apply to twist an object. For example, if you have a merry-go-round, you can apply a force to the rim of it. If the force is towards the center, it won't move at all.


If it's tangent to the rim, it'll move a lot more easily.


Also if you apply a force close to the center, it won't move much.


Given this, we see that what causes a rotating object to twist is not just the force, it depends on where the force is being applied and it what direction. So now we define the torque to be


tex2html_wrap_inline664 is the defines the point that the force is being applied. Does this make sense? Well the dependence on r means that the larger r, the more effective the force will be in twisting the object. The cross product makes sense too. If tex2html_wrap_inline664 and tex2html_wrap_inline730 are in the same direction, you get no twist, if they're perpendicular, you get the maximum effect. The whole thing seems pretty sensible. Now the direction of tex2html_wrap_inline732 is a bit odd at first sight. It's perpendicular to both tex2html_wrap_inline664 and tex2html_wrap_inline736 . Think about appling torque with a wrenth to undue a screw. The direction of the torque is in the direction of the bolt. It defines the axis that you're attempting to twist the object.

But does our analogy we translational motion hold? If I is like the mass m, and tex2html_wrap_inline742 is like a, then the analog to f = ma should be tex2html_wrap_inline748 . Is this true?

Let's try to see if this works out. The strategy is as follows. We'll consider some object like a disk rotating about a fixed axis. We'll consider it to be two dimensional for simplicity. That way we don't have to worry too much about the vector nature of the problem. We'll divide it up into lots of tiny masses and understand each one individually. Then we'll sum up the effect from all of the.

First of all consider just one point mass circling at a distance r from an axis. Let's say we apply a force tex2html_wrap_inline736 to this particle. Both tex2html_wrap_inline664 and tex2html_wrap_inline736 are both in the x-y plane, so their cross product always points in the z direction.


Here tex2html_wrap_inline758 is the tangential component of the force.


But tex2html_wrap_inline760 So plugging this in we have


The moment of inertia for a single particle is just tex2html_wrap_inline762 , so we get tex2html_wrap_inline748 for this one particle. Now let's say you've got a jillion particles. To each one you apply a different torque tex2html_wrap_inline766 , then the sum of all these torques is just


Now we've understood the analogous quantity to the force, let's do the same for momentum.

next up previous
Next: Angular Momentum Up: Angular Momentum and Torque Previous: Angular Momentum and Torque

Joshua Deutsch
Sun Feb 23 15:54:50 PST 1997