Why is my calculation for the speed of a falling rod off by sqrt(2)?

[faculaganymede]

A thin uniform rod of mass M and length L is positioned vertically above an anchored frictionless pivot point and then allowed to fall to the ground. With what speed does the free end of the rod strike the ground?
My answer is off by sqrt(2) from the correct answer, sqrt(3gL), and I don't understand why. Please help!

 

[andrewchang]

the gpe is being transformed into KE rotational. you would use the height as L/2 since the center of mass is changing that much distance.
the moment of inertia of a stick rotating about the end is 1/3mL^2

$$mg(0.5L) = 0.5I\omega^2$$

 

[marlon]

A thin uniform rod of mass M and length L is positioned vertically above an anchored frictionless pivot point and then allowed to fall to the ground. With what speed does the free end of the rod strike the ground?
My answer is off by sqrt(2) from the correct answer, sqrt(3gL), and I don't understand why. Please help!

You should not have posted this here. Did you not read https://www.physicsforums.com/showthread.php?t=89899 thread ?


regards
marlon

 

[faculaganymede]

sorry, that was my first post. didn't know better.

 

[faculaganymede]

Thanks Andrew!