Solving Mechanics Problem: Forces, Eqns of Motion, Reactions

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Homework Statement


A mass m is placed on top of a smooth hemisphere of radius a such that \vartheta=\pi/2 (so it is basically on the top of the semicircle, with \vartheta being the angle between it and the horizontal).

It is given a very small impulse and as a result begins to slide down one side of the hemisphere under the influence of the gravitational acceleration g.

State the forces acting on the mass, giving their directions, and write down its radial and angular equations of motion in polar coordinates as long as it remains sliding on the sphere.

Find the reaction force between the mass and the surface of the hemisphere as a function of the angle \vartheta, and hence show the mass flies off the surface of the hemisphere when its vertical height has decreased by a/3.

Homework Equations


I guess that
a= -r\omega^2 r^ + r \delta\omega/\deltat \vartheta^

The Attempt at a Solution


Only just started this section of the course and so struggling to get my head around a lot of the material. The forces acting are gravity and the normal force, and so I suppose the equation of motion would be:

ma= -mgsin\vartheta + N r^ - mgcos\vartheta \vartheta^

Not sure how to express the normal force, but would I be right in saying it flies off where -mgsin\vartheta + N < -mr\omega^2

also when it's at height a/3 sin\vartheta=1/3

But yeah basically I'm just pretty confused with this topic so far so any help would be appreciated.
 
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You are almost there for the acceleration. Think of the sphere as an incline with continuously changing angle theta. Perpendicular to the "incline" is the radial direction and parallel to the incline is the "theta" direction. What are the components of the weight along these directions?

The mass flies off when the sphere can no longer exert a force on the mass in which case the mass is in free fall. The only force that the sphere can exert on the mass is N. So what do you think the value of N must be when the mass flies off?
 
right so when the normal force is 0 it flies off.

the component of gravity acting towards the centre is -mgsin\vartheta I think.

So the overall centripetal force which is -mr\omega^2 which equals -mgsin\vartheta\vartheta + N and so

N = -mr\omega^2 + mgsin\vartheta and so when N = 0

r\omega^2 = gsin\vartheta


not sure where to go now. I guess working out \omega as a function of \vartheta? Although not sure how I'd do that exactly.
 
Use energy conservation and v = ωR.
 
ah I see, got it now! Thanks :D
 
hw excatly did u do it??
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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