How Does Angular Velocity Change as a Particle Slides Down a Sphere?

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SUMMARY

The discussion focuses on calculating the angular velocity, \(\omega\), of a particle sliding down a frictionless sphere using the conservation of energy principle. The initial attempt incorrectly expressed the total energy as \(E_{tot} = \frac{1}{2}m\omega^2 - mg\theta\). The correct approach involves expressing total energy in terms of linear speed and height, then converting these into angular terms. The correct formula for angular velocity derived from the conservation of energy is \(\omega = \sqrt{2g\theta}\).

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Homework Statement



A particle sits at the top of a fixed sphere of radius a. The particle is given a tiny nudge so that it begins to slide down the frictionless surface of the sphere. Consider the point at which the particle is still in contact with the sphere and its position is indicated by the value of the angle [tex]\theta[/tex].

Use the conservation of energy principle to calculate the angular velocity, [tex]\omega[/tex], as a function of theta.



Homework Equations



Etotal = Ek + Ep



The Attempt at a Solution



Etot = 1/2*m*([tex]\omega[/tex])^2 -m*g*[tex]\theta[/tex]

[tex]\omega[/tex] = (2*g*[tex]\theta[/tex])^1/2


Please can anyone verify if I have completed this part correctly?
 
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James01 said:
Etot = 1/2*m*([tex]\omega[/tex])^2 -m*g*[tex]\theta[/tex]
This is incorrect. Instead, write the total energy in terms of the usual variables of speed and height and then translate that into angular terms. (For a given linear speed, what's the angular speed? For a given height along the sphere, what angle does it make?)
 

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