In summary, a marble is placed at the top of an inverted hemispherical bowl of radius R = 0.30 m and slides down without friction. At an angular position of θ = 16.6°, the radial acceleration is 0.817 m/s² and the tangential acceleration is 2.80 m/s². The normal force is 0 at this point. The marble loses contact with the bowl at an angle of θ = π and a free body diagram at this point would not have a normal force and would show a centripetal force pointing towards the center of the bowl.
  • #1

Homework Statement

A marble is placed at the top of an inverted hemispherical bowl of radius R = 0.30 m. It starts from rest and slides down the bowl without friction. Draw a free body diagram when the marble reaches an angular position θ = 16.6°. From your FBD, sketch the approximate direction of the acceleration.
1.Calculate the radial component of the acceleration (assuming that the radius of the marble is negligible). (Hint: First find the velocity at θ = 16.6°.)
2. What is the tangential component of the acceleration when θ = 16.6°? (Hint: Use Newton's second law.)
3. What is the magnitude of the normal force when the marble loses contact with the bowl?
4. Draw a free body diagram for the angular position where the marble loses contact with the bowl. Draw the direction of the acceleration next to your FBD. What is the angle θ when the marble loses contact with the bowl?

Homework Equations

PE = mgh
KE = 1/2mv²
F = ma
ac = v²/r

The Attempt at a Solution

1. I found the radial acceleration to be 0.817 m/s² by using ΔE = ΔKE + ΔPE and found v final. I then used ac = v²/r
2. Tangential acceleration is 2.80 m/s² which I found by using ƩFx= Nx -> ma = mgsin(16.6)
3. Normal force is 0
4. I am not sure how to find this value. I tired to find the velocity when the marble loses contact and sub that into ΔE = ΔKE + ΔPE to find the final height but I wasn't sure how to find the velocity or if this is even the correct method.
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  • #2
well - from your answer to 3, where does this happen? Can you make an equation relating the normal force to the angle?
  • #3
I'm not sure how I would set that up. I had thought that perhaps the angle might be 90 since that would make the x-component of the normal force equal to zero (nx = mgsinθ). But the answer was not correct. As to what happens at this point I thought that at the point the ball leaves the acceleration will be equal to gravity, but now I am unsure.
  • #4
Reality check: (check my working)
The normal force is always normal to the surface of the bowl.
Gravity always points down.
Measure angle A from the top of the bowl.

for A=0, N=0, F=mg
for 0 < A < pi, F2+N2=(mg)2

so: F=mg.cos(A) and N=mg.sin(A) for the magnitudes of the tangential and normal forces. The motion is circular but not uniform - there is an angular acceleration - it's not even kinematic.

By conservation of energy, the kinetic energy will be:

T=mgR.sin(A) where R is the radius of the bowl, which puts the tangential speed at:


centripetal acceleration is, therefore, ac=2gsin(A), giving a centripetal force of:

Fc = 2mgsin(A)

this will be the force pointing towards the center ... which is also what the normal force does! This is twice the normal force - how come?

Anyway - from your own analysis: you'd want to say that the first time there was no contact was at A=0, the location of maximum push is at A=pi/2, so the ball loses contact with the wall at A=pi. (all angles in radiens)

So - draw a free body diagram which does not have a normal force ... what does it look like? Would this be what happens at A = pi?

Suggested for: Gravitational potential energy and rotational motion