Question

011 (part 1 of 2) 10.0 points solid cylindrical spool with a radius of 0.01 m and a mass of 0.3 kg. A 5 kg mass is then attached to the free end of the string, causing the string to unwind from the spool. 012 (part 2 of 2) 10.0 points How fast will the spool be rotating after all of the string has unwound? Answer in units of rad/s

How fast will the spool be rotating after all the string has unwound?

Answer 1

By Conservation of energy

Mgh = (1/2)mV²+(1/2)IW²

Since Moment of inertia of spool

I=(1/2)mr²=(1/2)*0.3*0.01² =1.5*10^-5 Kg-m²

and V=rW

=>Mgh =(1/2)M(rW)²+(1/2)IW²

(Mr²+I)W² =2Mgh

W=sqrt[2Mgh/(I+Mr²)]

W=sqrt[2*5*9.81*4.5/(1.5*10^-5+5*0.01²)]

W=925.84 rad/s