What is the angular acceleration of a falling rod with two attached balls?

[zarentina]

Homework Statement

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)

Homework Equations

τ=I*α
τ=mg*r
I(rod)=1/12(mg*r)
I(ball)=mg*r

The Attempt at a Solution

I'm truly struggling with this, I began by finding the $\Sigma$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

 

[azizlwl]

I think you have to check again the formula for Moment of Inertia.

 

[zarentina]

Well I found one mistake in my I(bar) I was using R instead of L. So my new calculated inertia is 7.34. Yet this still leaves me in square one, any other insight?

 

[PhanthomJay]

Looks like your torque is correct, but your I's are not. Correct for the proper moment of inertia which has units of kg-m^2.

 

[gneill]

Moments of inertia usually involve a mass multiplied by the square of a length. Units: kg-m². There's no gravitational constant in there.

 

[zarentina]

*sigh, if there has even been a /facedesk moment* My new value for the I= 1/12(3.6(.9^2))+(2.5(.45^2)) --> .243+.50625 = .74925. Setting this into the initial equation my values resulted in 11.025=.74925a solving for a yields a=14.72. Does this look correct or have a goofed up else where?

Thanks!

 

[PhanthomJay]

looks OK don't forget units for α, what are they?

 

[zarentina]

rads/sec^2 correct?

 

[PhanthomJay]

rads/sec^2 correct?

^^ Yes!