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At what angle does an object fall off a sphere?

  • Thread starter Inkage
  • Start date
1. The problem statement, all variables and given/known data
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.


2. Relevant equations
None


3. 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 =(
 
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, [tex]\theta[/tex] and find for what angle [tex]\theta[/tex] it holds that [tex]N=0[/tex]

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 =.=
 
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 [tex]\theta[/tex] from the horizontal, in which case you are correct!

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

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