# Can't use formula for kinetic energy 1/2*mv^2?

1. Nov 19, 2015

### Warlic

Question number 1; find the angular speed.

I imagine a point mass at the very top, it has potential energy mgL, this is converted to kinetic energy 1/2*mv^2.
I put them equal to each other and say that v=ωL. This gives me ω=sqrt(2g/L), the answer is supposed to be ω=sqrt(3g/L).

Why cant I do it like I did it? Is it wrong to use the formula for kinetic energy (1/2)mv^2?

2. Nov 19, 2015

### phyzguy

Your solution assumes all of the mass is at the top of the bar. Actually the mass is distributed along the bar. You could do it your way by assuming a string of point masses, each of length dL and mass m/L dL, then integrating along the length of the bar.

3. Nov 19, 2015

### stockzahn

You can't use the formular for translational kinetic energy, because every point of the rod has a different velocity depending on its position. You would have to integrate the kinetic energy of all the infinte small slices of mass along the rod.

4. Nov 19, 2015

### Warlic

But if mgL=1/2*mv^2, then I can just cross out m on both sides, and it shouldnt have anything to say?

5. Nov 19, 2015

### stockzahn

No, that's not possible, because in the formula for translational kinetic energy the entire mass (every mass point) is supposed to have the same velocity.

6. Nov 19, 2015

### phyzguy

You are right that the m cancels out, but each point along the rod has a different L and v. Think of a point 1/2 way out. It is at a distance L/2 from the pivot.

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