I = 3kgm²
m = 3kg
d1 = 1m
d2 = 0.3m
w1 = 0.75rad/s
w2 = ??
The total inertia is the sum of the individual inertias.
(I + md1² + md1²)w1 = (I + md2² + md2²)w2
w2 = (I + 2md1²)w1/(I + 2md2²) = 1.9rad/s
A student sits on a freely rotating stool holding two weights, each of which has a mass of 3.00 kg. When his arms are extended horizontally, the weights are 1.00 m from the axis of rotation and he rotates with an angular speed of 0.750 rad/s. The moment of inertia of the student plus stool is 3.00 kgm2 and is assumed to be constant. The student pulls the weights inward horizontally to a position 0.300 m from the rotation axis.
(a) Find the new angular speed of the student.
(b) Find the kinetic energy of the rotating system before and after he pulls the weights inward.
Momentum is conserved, Figure the initial momentum. Ignore arm mass, just use the distance tot he weight and the length. Notice the radius is .5, and the mass is 6kg. Figure the moment of inertia for that.
set that equal to the final angular momentum, and solve for w.
wouldnt the radius be 1 m because the weights are 1m from the axis of rotation?
so I did
1/2(6 kg(1m)^2 and that gives me a Intertia of 3.0. I multiplied that by the angular speed given in the problem; .750
then I set that equal to 1/2(6 kg)(.300m)^2 and solved for w!
ooops, you are correct, the radisu is 1m.
im just not getting the correct answer
what could i be doing wrong?
use your weights to help you find your total inertia. so 2(3)(1)^2 +3
Which your total inertia will be 9. Now use your 9 kg * M^2 in your kinetic inertia initial. (1/2)(9)(.750)^2 which will be 2.53125. Figure out your inertia final use that to help figure out the kinetic final. don't forget Inertia inital * dist=inertia final * distance.
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