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Skippy · #2 07-10-2008, 10:42 PM

I think I've got this, but it's also been a while since I've done this type of calculation. Here goes:

I assumed 1 m was the center to center distance between the masses, and that the masses are rigidly coupled to the rod.

First, I found the center of mass for our uneven barbell:

I drew a free body diagram with the masses and rod, then assumed it to be stationary in a 1g envrionment due to being simply supported at the CG. I called the location of the 1.5 kg mass x=0, with the location of the 2.9 kg mass being x=1. Since it's a stationary rigid object, the sum of the moments about any point on the object will be zero. So I did a moment balance at x=0. At this point, we have a 2.9 kg*m (improper unit I know but it works here) torque in the clockwise direction balanced by an equal torque generated by the upward force of the simple support, which must be 4.4 (1.5+2.9) kg. So,

-2.9+4.4x=0

where x is the location of the CG. Solving the equality reveals x=.659.

Next I went back through some of my college books to find the formula for mass moment of inertia, which seems to be Im (I with an m subscript, just use your imagination)=mr^2.

I summed the moments of inertia of the two masses:

Im=(1.5x.659^2)+(2.9*.341^2)

which works out to .988 kg*m^2

Then I decided to apply conservation of energy so I found the kinetic energy of the uneven barbell

Ek=0.5*Im*w^2

where w (omega-use that imagination again) is 30 rpm or Pi rad/s to make the units come out right. This came to 4.88 joules (lots of things squared over lots of things squared comes out to joules-if I did this right).

Next I had to find a torque that if applied over five seconds would provide the same energy. Ek is also equal to the definate integral over time of the applied torque, which I assumed to be constant. I don't know how to type in the symbols and make it look pretty, but I came up with 5T=4.88, so the required torque should be 1.02 Nm.

Somebody care to check that?