Moment of Inertia of a Thin Rod about its Center

[PsychonautQQ]

Homework Statement

A thin, uniform, 3.8kg bar, 80cm long has two 2.5 kg balls glued on at either end. It is supported horizontally by a thin, horizontal, frictionless axle passing through its centers and perpendicular to the bar. Suddenly the right hand ball becomes detached and falls off, but the other ball remains glued to the bar. Find the angular acceleration of the bar just after the ball falls off.

The Attempt at a Solution

So the net torque will just be m_ball*g*r because each side of the bar will cancel each other out. So if we set
m_ball*g*r = I(alpha) = m_ball*(r^2)*(alpha)

so
g/r = (alpha)
(9.8)/4 = (alpha)
and I get 24.5 which is wrong by a factor of 2/3 (correct answer is 16.333)

The question is asking for the bars angular acceleration and I'm giving It the balls, but wouldn't they be equal?

 

[voko]

Isn't it strange that the mass of the bar does not enter the equations? Think for a second that the bar is very massive or is very light; would they behave in exactly the same way under equal torques?

 

[ehild]

Homework Statement

A thin, uniform, 3.8kg bar, 80cm long has two 2.5 kg balls glued on at either end. It is supported horizontally by a thin, horizontal, frictionless axle passing through its centers and perpendicular to the bar. Suddenly the right hand ball becomes detached and falls off, but the other ball remains glued to the bar. Find the angular acceleration of the bar just after the ball falls off.

The Attempt at a Solution

So the net torque will just be m_ball*g*r because each side of the bar will cancel each other out.

That is correct, the net torque is that of the weight of the ball.

So if we set
m_ball*g*r = I(alpha) = m_ball*(r^2)*(alpha)

The moment of inertia includes also the contribution from the rod. What is it about the centre?


ehild