Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Ball in a bowl

  1. Aug 3, 2006 #1
    This problem has been driving me crazy. There is spherical ball of mass M and radius R. It is in a sphererical bowl with radius 10R. What is its frequency if pure rolling motion is assumed?

    I figured it out if there is no friction so it slips.

    Sum of Torque = I ddz/ddt = Mg10RsinZ
    I = 1/2 MR^2 + M(9R)^2
    W^2 = MG10R/(163/2MR^2) = (20 G)/(163 R)

    I am at a loss in figuring it out if the ball rolls.
  2. jcsd
  3. Aug 3, 2006 #2


    User Avatar

    Staff: Mentor

    What force is causing the rolling motion? You need a force to move the ball and get some "frequency".
  4. Aug 3, 2006 #3
    Gravity pulls the ball down and we assume that it starts a small angle Z. We assume pure rolling so friction keeps it from slipping down the bowl.
    Assume no damping.
  5. Aug 3, 2006 #4
    dont forget the ball's moment of inertia
  6. Aug 3, 2006 #5


    User Avatar

    Staff: Mentor

    Oh, so the ball starts at the waist of the inside of the sphere, and then rolls down and back up, kind of like a modified pendulum. Got it. As jasc15 points out, the main difference for this system versus a pendulum is the energy that goes into the rolling inertia of the ball. Write the equation for the energy of the ball as a function of height from the bottom of the arc, and include terms for the linear PE, linear KE and the energy in the rotation. That should start you on the way to working out the motion as a function of time, which gives you the natural frequency of the system.
  7. Aug 4, 2006 #6
    Ok. Here are mathmatical symbols for the system.

    Ftangential = m a = mgsinZ - F(friction)
    Fcentripetal = ma = N- mg cosZ
    Torque = I ddZ/ddt = 9Rmg sinZ - 10R F(friction)

    E= -mgrcosZ = (1/2)mv^2 + 1/2I w^2 - mgR
    mgR(1-cosZ) = (1/2)mv^2 + (1/2)(2/5)mR^2(v/R)^2
    gR(1-cosZ) = 7/10v^2

    I really don't know what to do. This isn't for a class, I just study in my spare time. Taking the derivative twice doesn't seem like it would help. Further more the W in the equation refers to angular velocity of spinning and not the angular velocity of the ball going up and down.

    Maybe something along the lines of:
    v^2 = 10/7gR(1-cosZ)
    ma = m v^2/9R = mgsinz- F
    F =mgsinz-mv^2/9R = mgsinz-(10/63)mg(1-cosZ)
    Torque= I ddz/ddt = 9Rmg sinZ - 10Rmgsinz-(10/63)g(1-cosZ)
    w = (RMg/I)^(1/2) = (RMg/(2/5*MR^2 + M(9R)^2))^(1/2)
    w = (g/(R(2/5+81))^(1/2)

    Does this solution seem correct? Can we say a centipetal = v^2/R if it isn't a point mass?
  8. Aug 4, 2006 #7


    User Avatar

    Staff: Mentor

    The first energy equation is almost correct, but I would write it more in the form of total energy TE = PE + KE(linear) + KE(rotational).

    Then write an equation for the rotational velocity w of the ball as a function of position around its arc in the bowl -- you have the TE and PE (just related to the vertical height, right?), so you can figure out the w from the TE equation. Once you have the w, that gives you the linear velocity v as a function of position around the bowl arc, and that will give you the position (angular displacement in the bowl) as a function of time. That gets you to the frequency number that you want to derive.
  9. Aug 7, 2006 #8
    Ei =-mg9Rcos(Zmax)
    Ef = -mg9R cos(Z) + 1/2mv^2 + 1/2 I w^2 = -mg9R cos Z + 7/10Mv^2
    v = [10/7*gR(cosZ - cosZmax)]^(1/2)

    I don't see how to make the velocity s function of time. The accelleration in not consant so kinematics won't work.
  10. Aug 7, 2006 #9


    User Avatar

    Staff: Mentor

    I'm having trouble tracking your equations, so I'll try a couple in latex. I've sketched a semicircular bowl bottom of radius 10R, and I show the ball in two positions: (1) in the starting position at the top left of the bowl where the ball has only PE, call that position theta = +90 degrees, measured from the vertical, and (2) in the bottom position where the ball has all KE, and theta = 0 degrees.

    [tex]PE(1) = 9MgR[/tex]

    [tex]KE(2) = \frac{1}{2} mv^2 + \frac{1}{2} Iw^2[/tex]

    [tex]KE(2) = \frac{1}{2} mv^2 + \frac{1}{2} * \frac{2}{5} mR^2 w^2[/tex]

    [tex]v = wR [/tex]

    [tex]KE(2) = \frac{1}{2} mv^2 + \frac{2}{10} mR^2 ({\frac{v}{R}})^2[/tex]

    Set PE(1) = KE(2) and solve for v at (2). That puts bounds on v(theta). Now write a more general equation for the KE as a function of theta (since you know what the PE is as a function of theta -- just the height above the bottom of the bowl).

    EDIT -- Latex makes that last term look like v^2/R, but it's obviously (v/R)^2. I'll try adding explicit parens there.... ahh, good. That worked.
    Last edited: Aug 7, 2006
  11. Aug 10, 2006 #10
    I am still trying!

    Even for a simple pendulem it is only harmonic for small angles of Zmax.
    I just want to do simple harmonic motion; my differetial equations is limited.

    Part I:

    Energy at highest point:
    E = - mg(9R)cos(Zmax)

    Energy at lowest point:
    E = .5m(Vmax)^2 + .5Iw^2 -mg9R = .5m(Vmax)^2 + .5(2/5)mR^2(v/R)^2 -mg9R
    = (7/10)*m(Vmax)^2 - mg9R

    Setting the energies equal:
    E= - mg(9R)cos(Zmax) = (7/10)*m(Vmax)^2 - mg9R

    Solving energy equality for the velocity max:
    (Vmax)^2 = G9R*(1- cos(Zmax)) *(10/7)

    Part II:
    This is where I think I am making a mistake.

    S= (9R)*(Z) = (9R)*(Zmax sinwt)

    This starts it off at the bottum at time = 0.

    v = ds/dt = 9R*Zmax*w*coswt
    Vmax = 9R*Zmax*w

    Part III: combining the previous parts
    (Vmax)^2 = G9R*(1- cos(Zmax)) *(10/7)

    (9R*Zmax*w)^2 = G9R*(1- cos(Zmax)) *(10/7)

    Taylor series and small angles lead us to:
    cos(Zmax) = 1 - (1/2)*(Zmax)^2

    Substituting taylor series into eqn
    9R*(Zmax)^2*w^2 = G[1 - (1- (1/2)*(Zmax)^2)]*(10/7)
    = G*(Zmax)^2)*(10/7)

    Solving for w^2:

    w^2 =g/9R*(10/7) =(10 g)/(63R)

    The thing is i used a variation of a problem from the book and when i solved it with the same method I got a different answer. Does anyone see any fudamental mistakes? Also, what is Latex? I am on a public library computer and can't download programs. Sorry my math looks so mess. I wish i could draw pictures and write it neatly.
  12. Aug 10, 2006 #11


    User Avatar

    Staff: Mentor

    Why do you insist on calling the PE at the highest point negative? And why to you then subtract that energy out of the KE at the bottom of the bowl?
  13. Aug 10, 2006 #12
    Thank you very much for your help. I got it to work out. The method was right I just made a algebra mistake. Anyways, I just use the center of the circle or bowl as height zero and that is why they are negative. The difference of the potential energy is the same where ever you call height zero.
  14. Feb 2, 2011 #13
    Why do you not need to take into account the translational kinetic energy?
  15. Feb 3, 2011 #14


    User Avatar

    Staff: Mentor

    My post was from over 4 years ago. I don't remember that far back.... :wink:
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook