# Final angular velocity of a mass on a pulley

## Homework Statement

Use Newton's 2nd Law, sum of torques, and kinematic equations to determine the angular speed of the spool shown in the figure below. Assume the string has a negligible mass, and it turns without slipping. Use g=10 m/s2 for acceleration due to gravity. Ʃτ=rxF=Iα
F=mA
ωf2i2+2αΔθ

## The Attempt at a Solution

Sum of torques:
Ʃτ=rxf
=(.6m)(3kg*10m/s2)(sin 90°)=Iα
--> 18 kgm2/s2=Iα
α=18/[.5(5)(.62)]
α=20 rad/s2

But plugging this into the kinematic equation to solve for ωf doesn't give me the right answer (I know this because I calculated ωf based on conservation of energy first (ωf=11.01 rad/sec), so I think I may have messed up somewhere up until this point. Any suggestions?

## Answers and Replies

Doc Al
Mentor
Do not assume that the force exerted on the spool equals the weight of the bucket. You need to figure out the tension in the string. (Analyze forces on the bucket as well as on the spool.)

Thank you! I knew it was something having to do with the tension. And I wasn't thinking of doing forces on both objects, more of just doing it as a whole.