Simple(ish) mechanics problem, conservation of energy

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SUMMARY

The discussion focuses on applying the conservation of energy principle to calculate the angular velocity \( \frac{d\theta}{dt} \) of a particle sliding down a frictionless sphere of radius \( a \). The total energy at the top is established as \( E_{total} = 2mga \), equating gravitational potential energy and kinetic energy at any point. The participant derived the expression \( v^2 = 2g(2a - h) \) and expressed height \( h \) in terms of \( a \) and \( \theta \) as \( h = a \cos(\theta) + a \). The final angular frequency is calculated as \( \frac{d\theta}{dt} = \frac{v}{a} \).

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Homework Statement


A particle sits at the top of a sphere of fixed radius a. It is given a tiny nudge and begins to slide down the frictionless surface of the sphere. My attachment won't work so I'll try and explain. Theta, [tex]\theta[/tex], is the angle which the particle makes with the vertical axis of the sphere. When the particle is at angle theta, it is still in contact with the sphere.

Part (c): Use the conservation of energy principle to calculate the angular velocity d[tex]\theta[/tex]/dt as a function of [tex]\theta[/tex].


Homework Equations


I'm assuming I'll need the kinetic and potential energy equations...


The Attempt at a Solution


I've had a go, but I think its wrong :confused:

I start with saying the total energy of the system is equal to the gravitational energy of the particle when it is at the top of the sphere, at height 2a. So E(total) = 2mga.
Now, at any other moment in time, the kinetic and potential of the particle must equal this. So: 2mga = 0.5mv2 + mgh. A bit of rearranging and I canceled m and got this: v2 = 2g(2a-h).

Now, I put h in terms of a and theta like this: h = cos([tex]\theta[/tex]) + a. So I have an expression for v in terms of a and theta.
I put this into the formula: d[tex]\theta[/tex]/dt = v/r = v/a to get the final answer for angular frequency in terms of theta!

It seems a little hefty for just 4 marks, which is why I think I'm wrong or there might be an easier method.
Any advice would be fantastic!
 
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That looks basically fine, except I get h=a*cos(theta)+a (typo, I hope). I can't really think of any easier way to do it.
 
Dick said:
That looks basically fine, except I get h=a*cos(theta)+a (typo, I hope). I can't really think of any easier way to do it.

Thanks very much! It was a typo, thankfully.
:-p
 

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