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Stuck on 1 problem and part of another problem

  1. Jul 7, 2004 #1
    Well I'm back. I've done most of everything, except for the last part of one problem and one complete problem that has me stumped.

    First the problem I almost have solved:

    While visiting friends at Cal State Chico, you pay a visit to the Crazy Horse Saloon. This fine establishment features a 200 kg mechanical bucking bull, that has a mechanism that makes it move vertically in simple harmonic motion. Whether the "bull" has a rider or not, it moves with the same amplitude (0.250 m) and frequency (1.50 Hz). After watching other saloon patrons hold on to the "bull" while riding, you (mass 75.0 kg) decide to ride it the macho way by not holding on. No one is terribly surprised when you come out of the saddle.

    What is your speed relative to the saddle at the instant you return?

    Now, what I got was that you're .11m above the equilibrium point at the time you leave the seat with an upward velocity of 2.11 m/s. I used what is given to find the phase angle.

    [tex] \phi = arctan(-2.11/(1.5* 2*\pi*.11)) [/tex] which gives me:

    [tex] \phi = -1.114 radians [/tex] *edit* was in degrees

    I got omega = 1.5 * 2* pi

    I then use x = A*Cos(omega*t+ phi) I was given that the time in the air was .538 seconds.

    I think I'm close, but I can't seem to get the right answer. For the person, I used [tex] V = V_0 + a*t [/tex] which gave me -3.1624 m/s But when I try to find the velocity of the bull by using v = -omega * A *sin(omega *t + phi).

    Anyone see what I'm doing wrong? Am I even close?

    The next problem is a bit simpler but it seems to be going over my head:

    A holiday ornament in the shape of a hollow sphere with mass 1.10×10^(-2)and radius 4.50×10^(-2) is hung from a tree limb by a small loop of wire attached to the surface of the sphere. If the ornament is displaced a small distance and released, it swings back and forth as a physical pendulum.

    Calculate its period. (You can ignore friction at the pivot. The moment of inertia of the sphere about the pivot at the tree limb is 5/3 M*R^2.)

    I'm trying this approach cause none of the other ones would seem to work, too many unknowns.

    [tex] \omega^2 = m*g*d/I [/tex]

    it seems simple since everything is given except d. I just can't seem to eliminate d or solve for it in any way. Any ideas?


    Last edited: Jul 7, 2004
  2. jcsd
  3. Jul 8, 2004 #2

    Doc Al

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    Staff: Mentor

    Comment on the first problem:
    How did you get the 0.11m?

    Comment on the second problem:
    I assume they want you to use the small angle approximation for the physical pendulum, in which case the period is independent of the displacement. Rederive your equation for [itex]\omega^2[/itex]: it's almost right. (You can derive it just like you would for an ordinary pendulum.)
  4. Jul 8, 2004 #3


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    Perhaps he meant pendulum length instead of displacement with [tex]d[/tex] ?
  5. Jul 8, 2004 #4
    That is correct, I used d cause it was what my instructor used in class. d is the length of the pendulum.

    I'm at work right now so don't have my paper work in front of me, but I did get .11m above the equilibrium point as a correct answer (answers submitted via a website and confirms correctness/incorrectness as you submit answers) I'll try to get the formula I used for you later if still needed.
  6. Jul 8, 2004 #5

    Doc Al

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    For the physical pendulum:
    In that case, your formula is correct (properly interpreted). (Thanks, NateTG!) For a physical pendulum, d will be the length to the center of mass. Express d in terms of the radius R.

    For the bucking bull:
    My mistake: 0.11m is fine. You method looks good. What value did you get for the speed of the bull?
  7. Jul 10, 2004 #6
    sorry, first chance I've had in the past few days to post.

    I figured out what I was doing wrong with the bull one, its always something simple. I was subtracting the two speeds and not adding them :smile:

    Just to make sure for the other problem I should use

    [tex] \omega^2 = m*g*d/I [/tex]

    but in that formula, d actually equals d+R?

    so do this:

    [tex] \omega^2 = m*g*(d+r)/(5/3*M*r^2) [/tex]

    Gonna give it a try.

    Will let you know
  8. Jul 11, 2004 #7

    Doc Al

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    In that formula, d is the length from the pivot point to the center of mass. A better way to write that formula is:
    [tex] \omega^2 = m*g*L_{cm}/I_{pivot} [/tex]
    Since the object is just a sphere, its cm is in the middle so [itex]L_{cm} = R[/itex]. Make sense?

    You should compare this formula to that for a simple pendulum.
  9. Jul 11, 2004 #8
    K, that was way easier then I thought it was. I kept thinking the Length to the center of mass was the Radius + some unknown distance which was the length of the string.

    Thanks for the clarification.
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