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## 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

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 and 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 is a bit odd at first sight. It's perpendicular to both and . 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 is like a, then the analog to f = ma should be . 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 to this particle. Both and are both in the x-y plane, so their cross product always points in the z direction.

Here is the tangential component of the force.

But So plugging this in we have

The moment of inertia for a single particle is just , so we get for this one particle. Now let's say you've got a jillion particles. To each one you apply a different torque , 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: Angular Momentum Up: Angular Momentum and Torque Previous: Angular Momentum and Torque

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