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- Thread starter ItDoesn'tMatter
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BvU

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Simple case to try out: two point masses at the ends of a massless stick.

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The kinetic energy of your rod -- or any rigid body -- is composed of one part having to do with the rotation of the body about some axis fixed in the body and one part having to do with the translational motion of a point on that axis. If the point with respect to which the rotation is defined is stationary, then translational kinetic energy does not factor into the calculation; if the point is moving then translational kinetic energy does factor into the calculation.

There are two ways of calculating the kinetic energy in your problem. One is to find the moment of inertia about the end point and treat the motion as a pure rotation (i.e. no translational part). This gives, as you say, $$T=\frac{1}{2}I\omega ^2 = \frac{ml^2}{6}\omega ^2.$$

The other way of doing it is to calculate the moment of inertia about the center of mass, which is moving, so that we must add in a translational part. The moment of inertia about the center of mass is ##\frac{1}{12}ml^2## so we have $$T=\frac{1}{2}I\omega^2 + \frac{1}{2} m v^2 = \frac{1}{24}ml^2\omega^2 + \frac{1}{2}m\bigg(\frac{l\omega}{2}\bigg)^2 = \frac{ml^2}{6}\omega^2.$$

Clearly they are the same. So what you are doing when you get the wrong answer is essentially you are calculating the energy due to translation of the center of mass and ignoring the fact that there is also a rotation about the center of mass.

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