A spool of thin wire rotates without friction about its axis. A man pulls down on a wire connected to an inner radius with force of 16.3 N. How long does it take to increase the angular velocity of the spool from 15.0 rad/s to 33.0 rad/s? Icm = 0.550 kgm2 Inner radius, r = 0.300 m Outer radius, R = 0.560 m. I got this part, the answer was 2.02 s. For the second part, it says, Consider the same spool as in the previous problem. Now, instead of pulling on the wire, the man attaches the wire to a mass of 0.900 kg. The man lets go of the mass, and the spool starts to turn. What is the speed of the mass after it has travelled downward a distance of 0.650 m? (Assume there is no energy lost to friction.) Here's what I did Conservation of energy: [tex] (1/2)Iw_f^2 - (1/2)Iw_i^2 = W [/tex] (1/2)(.550)(33.0)^2 - (1/2)(.550)(15)^2= 237.6 J [tex]W= \tau (\Theta_f - \Theta_i) [/tex] [tex]237.6 = \tau (.650) [/tex] [tex]\tau=109.5 [/tex] [tex]\tau= I \alpha[/tex] [tex]\alpha= 199.1 [/tex] I don't think I'm doing this the right way.. can someone please help me?