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Gravitational potential energy and rotational motion

  1. Oct 29, 2011 #1
    1. The problem statement, all variables and given/known data
    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?

    2. Relevant equations
    PE = mgh
    KE = 1/2mv²
    ΔE = ΔKE + ΔPE
    F = ma
    ac = v²/r


    3. 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.
     
  2. jcsd
  3. Oct 29, 2011 #2

    Simon Bridge

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    well - from your answer to 3, where does this happen? Can you make an equation relating the normal force to the angle?
     
  4. Oct 30, 2011 #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.
     
  5. Oct 30, 2011 #4

    Simon Bridge

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    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:

    v2=2gR.sin(A)

    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?
     
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