Force in relation to angular momentum

by valvan1
Tags: angular, force, momentum, relation
 P: 3 1. The problem statement, all variables and given/known data so a student is sitting on a spinning stool and has a 2kg dumbbell in each hand angular velocity is 3rad/sec arms stretched out is a radius of 80cm and he pulls in his arms to 20cm. for this problem your ignoring the students weight. from other problems i have figured out so angular velocity to start with is 3 rad/sec angular velocity final is 48 rad/sec kinetic energy initial is 11.52( dont know what this dimensions is so im guessing J) kinetic energy final is 184.32 angular momentum = 7.68 and i need to find the force required to pull one of the dumbbells in at a constant speed is equal to F=((initial angular momentum of 1 block)^2)/(4*Mass of one weight*Radius^3)) 2. Relevant equations I=$$\sum$$mr^2 k=1/2I$$\omega$$^2 $$T$$=I$$\alpha$$ L=I$$\omega$$ 3. The attempt at a solution do i say Torque=F*d and Torque=I*alpha and then go Force = Ia/d? or is there another way to solve this that im not seeing ? or do i use the change in kinetic energy equation 1/2*I*$$\omega$$^2Final-1/2*I*$$\omega$$^2initial = F*d