# 1/2 m v^2 in a circle?

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1. Feb 12, 2017

### JiggaMan

1. The problem statement, all variables and given/known data
in circular motion (e.g. a pendulum) is the kinetic energy still 1/2 m v ^2 or is it a different equation?

2. Relevant equations
1/2 m v ^2

3. The attempt at a solution

2. Feb 12, 2017

### Staff: Mentor

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".

3. Feb 17, 2017

### 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.