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

[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 can't I do it like I did it? Is it wrong to use the formula for kinetic energy (1/2)mv^2?

 

[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.

 

[stockzahn]

You can't use the formula 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 infinite small slices of mass along the rod.

 

[Warlic]

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.

But if mgL=1/2*mv^2, then I can just cross out m on both sides, and it shouldn't have anything to say?

 

[stockzahn]

But if mgL=1/2*mv^2, then I can just cross out m on both sides, and it shouldn't have anything to say?

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.

 

[phyzguy]

But if mgL=1/2*mv^2, then I can just cross out m on both sides, and it shouldn't have anything to say?

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.