
#1
Apr705, 12:29 AM

P: 6

I'm kinda having trouble figuring out this problem... i was thinking of doing a system of equations to relate the present angular and new angular speed but that didn't really get me anywhere because i wasn't really including all the variables in the problem... that and i couldn't simplify it down to one variable to solve for. If anyone could help me out i'd appreciate it a lot.
A student sits on a freely rotating stool holding two weights, each of mass 2.92 kg. When his arms are extended horizontally, the weights are 1.09 m from the axis of rotation and he rotates with an angular speed of 0.746 rad/s. The moment of inertia of the student plus stool is 2.92 kg · m2 and is assumed to be constant. The student pulls the weights in horizontally to a position 0.307 m from the rotation axis. Question: Find the new angular speed and find the kinetic energy before and after he pulls the weight inward. 



#2
Apr705, 01:48 AM

P: 2,223

Whats changing here is the moment of inertia of the weights. They are originally at distance 1.09m and are contracted to 0.307m. Find this new moment of inertia.
Then since momentum is conserved in the system, you can find the new angular speed by: [tex] I_i\omega_i = I_f\omega_f [/tex] solved for [tex] \omega_f [/tex] 



#3
Apr705, 02:05 AM

P: 289

Just to concur with whozum:
You need to use conservation of angular momentum: [tex] L_i = L_f [/tex] Where: [tex] L = p\times r [/tex] 


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