Torque, beam, angular acceleration problem

[CricK0es]

Homework Statement

[/B]
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.

3. The Attempt at a Solution

Torque = I.α
The torque here is provided by the single ball = 0.4 x 2.5g

The moment of inertia for the rod and the ball, I = (1/12 M L^2) + (2.5 x 0.4^2)

Solving for α, I get the wrong answer. The correct answer is 16.3 rad s^-2

I would appreciate some guidance on this, and where I may have gone wrong. Many thanks

 

[BvU]

Check your calculations. Typing error on the calculator ?
Can you post your steps ?

 

[Ian Baughman]

The torque here is provided by the single ball = 0.4 x 2.5g

Remember that Torque = F_{perpendicular} × r
What is the force exerted by the weight of the ball?

 

[CricK0es]

1.) The one ball falls off and so it is no longer balanced. Therefore, the torque is provided by the distance (from pivot) multiplied by the force
Torque = Weight of ball x distance = 0.4 x 2.5g = g

2.) The moment of inertia has to be determined for the rod AND the ball, I , = I = (1/12 M L^2) + (2.5 x 0.4^2)

3.) Torque = I.α ... Therefore α = T / I

g / [ (1/12) . (3.8) . (0.8)^2 + (2/5)... which I now have working xD = 16.3

Sorry, I must have rearranged things incorrectly the first time, or as you said put it in the calculator wrong. Oh well... I have another question, regarding centre of percussion (where I barely even know where to start), but I'll post a different thread tomorrow. But still, many thanks

 

[CWatters]

#1 says the balls are 2.5kg not 2.5g

Edit: Oh wait I see you mean g as in 9.8m/s/s not grams