Final angular velocity of a mass on a pulley

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SUMMARY

The discussion focuses on calculating the final angular velocity of a spool in a pulley system using Newton's 2nd Law and kinematic equations. The user initially calculated the angular acceleration (α) as 20 rad/s² but encountered discrepancies when determining the final angular velocity (ωf). The correct approach involves analyzing the forces acting on both the bucket and the spool, specifically the tension in the string, rather than treating the system as a whole. The conservation of energy method provided a more accurate ωf of 11.01 rad/sec.

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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.
1dzor5.png



Homework Equations


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?
 
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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.
 

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