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Consider an imaginary vertical plane. Now say, a body is falling freely (under earth's attraction). If you consider any axis that is perpendicular to that plane. We get a non-zero value for torque. Then why is it that body is not moving in a circular path about that axis?

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diagram? And a little more detail in the question, it doesn't make much sense right now. – Nic Aug 28 '11 at 21:21
up vote 2 down vote accepted

When you exert a torque on something, its angular momentum changes. This doesn't necessarily mean things are moving in a circle, or even rotating. You can have torque and changing angular momentum even when things are going in a straight line.

Most of the time, when we discuss torque and angular momentum in basic physics, we are talking about things like spinning bicycle wheels, precessing tops, or ice skaters doing a fancy maneuver, but that's only because torque and angular momentum are easiest to exemplify and visualize in those situations. The simple facts about torque and angular momentum, such as $\vec{\tau} = \dot{\vec{L}}$, hold generally.

To make your example precise, suppose a particle's trajectory is

$$\vec{r} = x_0\hat{x} - (y_0 - \frac{1}{2}gt^2)\hat{y}$$

Then the force on it is

$$\vec{F} = m\ddot{\vec{r}} = -mg\hat{y}$$

The torque about the origin is

$$\vec{\tau} = \vec{r}\times\vec{F} = -x_0mg\hat{z}$$

The angular momentum about the origin is

$$\vec{L} = \vec{r}\times\vec{p} = \vec{r} \times (-mgt\hat{y}) = -x_0mgt\hat{z}$$

Finally, we see that the time derivative of the angular momentum is the torque.

$$\dot{\vec{L}} = -x_0mg\hat{z} = \vec{\tau}$$

So we can indeed analyze the torque and angular momentum of an object falling in a straight line and find a consistent and accurate description.

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