Conservation of Energy, Ball going down slope

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SUMMARY

The discussion focuses on calculating the angular speed of a solid ball weighing 162 N and rolling down a 5.9 m ramp inclined at 33°. The conservation of energy principle is applied, where the initial mechanical energy (MEi) equals the final mechanical energy (MEf). The relevant equations include potential energy (PEg = mgh) and kinetic energy (KE = 0.5mv² + 0.5Iω²), with the relationship ω = v/r for angular speed. The user initially attempted to solve for velocity but encountered difficulties, indicating a need for a clearer approach to the problem.

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Homework Statement


A solid 162 N ball with a radius of 0.350m rolls 5.9 m down a ramp tha is linclined at 33° with the horizontal. If the ball starts from rest at the top of the ramp what is the angular speed of the ball at the bottom of the ramp?


Homework Equations


MEi= MEf
PEg=KE + KErot
mgh= .5mv2 +.5Iω2
ω= v/r

The Attempt at a Solution


I solved the conservation of energy equation for velocity then divided by the radius of the ball, but it did not work, am I approaching the problem wrong?
 
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Show your work in detail, please.

ehild
 

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