(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A thin, uniform 3.60kg bar has two balls glued on either end, each with a weight of 2.5kg. The length of the rod is 90cm and is balanced in the center. When one ball falls off calculate the angular acceleration just after this happens. (e.g the rod is falling in a rotational motion)

2. Relevant equations

τ=I*α

τ=mg*r

I(rod)=1/12(mg*r)

I(ball)=mg*r

3. The attempt at a solution

I'm truly struggling with this, I began by finding the [itex]\Sigma[/itex]I=I(ball)+I(rod) which looked a little something like this: 1/12[(3.6*9.8)*.45]+(2.5*9.8)(.45) which resulted in a value of 5.5566.

Next I calculated the τ. ( i feel this is where i screwed something up) My logic was that the torque due to the bar wouldn't matter as it is a constant on either side regardless of ball or no ball. In lieu of this τ=m(ball)*9.8*.45=11.025. (this value absolutely has to be wrong)

After i got these values i subbed them into the initial τ=I*α to solve for α which I got to be 1.98m/s which is far from correct.

Anyone care to head me in the right direction?

Thanks,

Z

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# Homework Help: Dynamics of Rotational motion

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