A particle slides down a non-fixed hemisphere

  • #1
Jason Ko
21
6
Homework Statement
A uniform hemisphere of radius 𝑅𝑅 and mass 𝑀𝑀 is placed with its flat side on a
smooth surface. A particle of mass 𝑚𝑚 is at rest at the very top of the hemisphere in
unstable equilibrium condition. A very small disturbance causes the particle to
slide down the spherical surface from rest. At the same time the hemisphere will
be pushed by the particle to the left. You can assume no friction at any contact
points and surfaces. Find the angle theta where the particle lose contact with the hemisphere. The full description is shown below.
So far I work in hemisphere frame and wrote down equation of motion. I try to use the velocity obtained from conservation of energy and substitute it into the equation of centripetal force. However, I don't know how to get the expression of a in my work as shown below.
Relevant Equations
mgh=1/2mv^2
F=ma
3. 80-9.jpg
Screenshot 2024-09-17 at 11.34.22 PM.png
 
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  • #2
I do not see any point in equations involving the normal force from the ground.
I would find working in the frame of the hemisphere confusing here.
Your energy equation ignores the KE of the hemisphere.
Consider another conserved quantity.
 
  • #3
haruspex said:
I do not see any point in equations involving the normal force from the ground.
I would find working in the frame of the hemisphere confusing here.
Your energy equation ignores the KE of the hemisphere.
Consider another conserved quantity.
I now consider the conservation of momentum and get ##v^2 =\frac{gR(1-cos\theta)}{1+\frac{m}{M}}##. Working in the frame of hemisphere, I get an equation of ##N-mgcos\theta +masin\theta = \frac{mv^2}{R}## where ma is the pseudo force as the hemisphere is accelerating leftward. In case of the particle loses contact with the hemisphere, N=0 and ##v^2## can use the above mentioned expression. However, I am unable to find the expression of a and hence cannot solve this problem.
 
  • #4
Jason Ko said:
I now consider the conservation of momentum and get ##v^2 =\frac{gR(1-cos\theta)}{1+\frac{m}{M}}##.
Please show how you get that.
Jason Ko said:
Working in the frame of hemisphere, I get an equation of ##N-mgcos\theta +masin\theta = \frac{mv^2}{R}##
Check the signs. Which way are you taking as positive for ##a##?

What about the force balance on the hemisphere?
 

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