Kinetic Energy in Circular Motion: Is 1/2 m v^2 Still Applicable?

[JiggaMan]

Homework Statement

in circular motion (e.g. a pendulum) is the kinetic energy still 1/2 m v ^2 or is it a different equation?

Homework Equations

1/2 m v ^2

The Attempt at a Solution

 

[gneill]

It depends on what the object is. For a point mass at the end of a massless string you might use 1/2 m v^2. For a more complicated object that can't be viewed as a point mass you'll want to look at the moment of inertia and rotational kinetic energy. Investigate: "Physical Pendulum".

 

[James R]

Just to add: for a point mass, the kinetic energy depends on the speed ($|\vec{v}|^2$) rather than the (vector) velocity, so the direction of travel (e.g. circular motion) doesn't matter.