Rotational Kinematics - Angular velocity

[nahanksh]

Homework Statement

http://online.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?courses/phys211/oldexams/exam3/fa07/fig19.gif
A dumbbell consists of a slender rod of negligible mass and length L = 1 m and small steel balls attached to each end with mass 1 kg and 2 kg, respectively. It is pivoted at its center about a horizontal frictionless axle and initially held in place horizontally. The dumbbell is then released. What is the angular velocity of the system when the rod is vertical?

Homework Equations

The Attempt at a Solution

I have figured out the angular acceleration upon release is 6.5rad/s^2 and the angular acceleration is zero at vertical (because of torque = 0 at vertical)
But from there, i got stuck and i don't know how to link this acceleration into the velocity..


Please could someone help me out here?

 

[ehild]

It is easier to apply conservation of energy. When the rod is vertical, the centre of gravity is lower than in the horizontal position, the difference is equal to the kinetic energy.

ehild

 

[nahanksh]

It is easier to apply conservation of energy. When the rod is vertical, the centre of gravity is lower than in the horizontal position, the difference is equal to the kinetic energy.

I have tried to use the hint you have given..

Iw^2/2 = mgh

h = (2/3 - 1/2) ; length of center of gravity from pivot when it's at vertical

And i took moment of inertia I as 3*(2/3-1/2)^2

But when i calculated this, it was wrong...

I think RHS of the eqn is correct...

Is something wrong in LHS of the equation?

Please help me out...:(

I would really appreciate it .