At what angle does an object fall off a sphere?

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A mass placed on top of a smooth hemisphere begins to slide down after receiving a small impulse. The discussion focuses on determining the angle at which the mass will fly off the hemisphere, specifically when its vertical height decreases by a third of the radius. It is established that this occurs at an angle of 48 degrees, but confusion arises regarding the conditions for the mass to leave the surface, particularly when the normal force becomes zero. Participants suggest analyzing the forces acting on the mass, including centripetal acceleration and gravitational components, while also recommending the use of energy considerations to find the necessary centripetal acceleration. The conversation emphasizes the importance of correctly applying trigonometric principles to solve the problem.
Inkage
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Homework Statement


A mass m is placed on the top of a smooth hemisphere of radius a such that theta = pi/2 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.

Show that the mass flies off the surface of the hemisphere when its vertical height has decreased by a/3.

Homework Equations


None

The Attempt at a Solution



I managed to work out that when the vertical height is reduced by a/3, theta = 48 degrees. But I have no idea how to show that the mass flies off the hemisphere. I suppose when N = 0 the object will fall off, but surely that would be when cos(theta) = 0? Which would mean... theta = 90 degrees... Help please >_> I feel a bit retarded for not being able to do this question =(
 
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As long as the mass is on the hemisphere (This is circular motion!), what are the forces acting on it? (There are 3)
Find these three forces as dependent on the angle relative to the vertical, \theta and find for what angle \theta it holds that N=0

If you're familiar with accelerated reference frames you should move your observer to the rotating frame of reference.
 
The three forces would be... (I think)

1. Centripetal acceleration (towards centre)
2. mg [mgcos(theta)]
3. Normal contact force? mgsin(theta)?

Im so confused =.=
 
Inkage said:
The three forces would be... (I think)

1. Centripetal acceleration (towards centre)
2. mg [mgcos(theta)]
3. Normal contact force? mgsin(theta)?

Im so confused =.=

Check your trig. :) Unless you were measuring your \theta from the horizontal, in which case you are correct!

Now use energy considerations to find the centripetal acceleration and you're good to go.
F_{centripetal} = \frac{mv^2}{R}
 
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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