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Period of ball on track

  1. May 30, 2015 #1
    Hi guys,

    My logic is obviously flawed here I'm just not sure what I'm missing. I'd love a hint but don't tell me too much as I want to try to figure this out once I'm heading in the right direction :) I'm really upset with myself that I couldn't solve this, so I'd love more SHM problems to practice. Does anyone know of a good resource(s)?

    Thanks for reading!

    1. The problem statement, all variables and given/known data


    A small solid ball of mass M and radius R rolls without slipping on the track whose height y as a function of horizontal position x is given by y=ax^2 where a is a constant with the units of inverse length. The mass is given an initial displacement from the bottom of the track and then released. Find an expression for the period of the resulting motion.

    2. Relevant equations

    U=mgh (gravitational)
    U=(1/2)kx^2 (SHM)
    T=2pi*sqrt(m/k)

    3. The attempt at a solution

    U = mgh = mgy = mgax^2 = (1/2)kx^2 => k = 2mga => T = 2pi/sqrt(2ga)
     
  2. jcsd
  3. May 30, 2015 #2

    haruspex

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    What makes you think your answer is wrong?
     
  4. May 31, 2015 #3
    Because when I input the answer into Mastering Physics it says incorrect, that I'm off by a numerical factor or something. Does it look logical to you?
     
  5. May 31, 2015 #4

    TSny

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    Did you include the effect of rolling without slipping as opposed to just sliding?
     
  6. May 31, 2015 #5
    I was wondering about that and the moment of inertia of the sphere, I'm just not totally sure how to include it. I tried playing around with KE and stuff but didn't really know where I was going with it so stopped. Could you give me a hint? Thanks!
     
  7. May 31, 2015 #6

    TSny

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    Can you write the total KE (translation + rotation)? Try to express the total KE as (1/2)Av2 where A is some constant related to the mass M. The constant A will take the place of M in the period formula.
     
  8. May 31, 2015 #7
    I ended up with something like (7/10)Mv^2 when I tried that, thing is I don't quite know what to do with the "v".
     
  9. May 31, 2015 #8

    ehild

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    How does v depend on x and dx/dt?
    The ball performs horizontal SHM for small displacements from equilibrium, that is, from x=0. Write the conservation of energy in terms of x only. Find the equation of motion ##m\ddot x = F(x)## by differentiating the equation for the energy. For small displacements, it must have the form ##m \ddot x = -kx##.
     
    Last edited: May 31, 2015
  10. May 31, 2015 #9
    Darn I'm sorry I'm still just a little perplexed. Could you help clarify for me the relationship mgax^2 has with (1/2)kx^2 in the problem? I thought mgax^2 acted as the restoring force thus was equivalent to (1/2)kx^2, and that as the potential energy it was proportional to both the rotational and translational KE which is why I neglected worrying about KE, if that makes any sense. I'm still confused as to where I went wrong with my logic there, so I'm embarrassed to admit I'm having a little trouble writing the conservation of energy for the system. Also, when you say write it in terms of x, do you mean KE=bm(dx/dt)^2? Thanks so much!
     
  11. May 31, 2015 #10

    ehild

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    Yes, this is a strange problem. You have a potential function proportional to x2, but it is not the real potential for the motion.
    If a point mass moves along the x axis, and the force acting on it has the potential 0.5kx2, then it corresponds to the restoring force -kx, which acts along the x axis.
    In this problem, the ball moves along a parabolic track so its displacement can be written as the length of arc of the parabola. Moreover, it rolls, so you have angular displacement, too. As the displacement is not along x, mgax2 is not potential for this motion.
    It is not only gravity applied on the ball, but there is the normal force from the track, too. The net force has both horizontal and vertical components.
    Still, the ball has potential energy mgy and kinetic energy which is equal to the translational energy of the CM plus the rotational energy about the CM. As the ball rolls, the angular velocity ω and the translational velocity v are related as v = -Rω. With these, you can write the kinetic energy of the ball.
    Initially, the ball was displaced from the bottom of the track and had zero speed. And the energy is conserved, it is equal to the initial potential energy.
    Once you you have the equation for conservation of energy you can find the speed of the ball in terms of x, and using the equation of the track, you get dx/dt. Differentiating, you get d2x/dt2 which is equal to F(x)/m.
    You can assume that the displacement is very small so you can neglect higher order powers of x.
     
  12. May 31, 2015 #11
    Thank you so much for your reply! I feel I'm starting to understand how to approach this, however I'm a very slow learner and definitely not the smartest cookie.
    When you say
    I'm just a little confused. I thought potential is independent of direction? Should I be imagining say a projection of the ball on the horizontal surface the track rests on, and think in terms of that?
    Why the minus sign?
    I'm the worst! I'm still not quite grasping how this should look unfortunately. I was thinking E_tot = KE + U, where KE=(7/10)mv^2, but U is still perplexing me. Is it 0.5kx^2 then? I'm also unsure about E_tot...I would've thought it would be mgax_int^2, but now I'm second guessing that idea (and no info on x_int is given).

    EDIT: I know this is ridiculous but I'm also confused about why I'm differentiating with respect to time not x, as although d/dt would turn the velocity into acceleration, wouldn't it mess up the x terms I'm writing the equation in terms of? Also I thought force was related to the derivative of kinetic energy with respect to x?
     
  13. May 31, 2015 #12

    ehild

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    Well, that is the trickiest thing in this problem. The potential energy of the ball is mgy, but the ball moves along the track, in a plane. The restoring force has to be along the tangent of the track, has both horizontal and vertical components. And there is also rolling, with angular velocity changing along the track because of the torque of forces applied.
    You can think in terms of the horizontal projection of the ball, but you have to count also the vertical component of the velocity when determining the kinetic energy.
    You are right, Etot=KE+U. KE=(7/10)mv2, and U = mgy=mgax2, but it is not 0.5kx2.
    And you are right, the total energy is mgaxi2 (xi is the initial horizontal displacement).

    Remember the ball moves in two dimensions, its velocity has both x and y components. And force is related to the derivatives of the potential energy (it is the negative gradient of the potential energy).
    From conservation of energy, you get an expression for ##v^2={\dot x}^2+{\dot y}^2={\dot x}^2\left(1+(\frac{dy}{dx})^2\right)##.
     
  14. May 31, 2015 #13
    That makes sense, thank you!
    I think I see how you got that. I'm being ridiculous and having trouble fitting this all in to the big picture though. I didn't want to ask this but I'm obviously having trouble thinking this through, what's your equation for the conservation of energy? Hopefully that will help me visualize things a little more. Then I take the derivative of that with respect to time (if I understand correctly), and I end up with a=F/m=(k/m)x...is that right? Sorry for all the confusion. I don't think I've ever been so lost by a homework question is my life.
     
  15. May 31, 2015 #14

    ehild

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    You know everything already to write up conservation of energy. KE+U=Ui, which means ##0.7mv^2+mgax^2=mgax_i^2##.
    ##v^2={\dot x}^2+{\dot y}^2##. Apply chain rule to write ##\dot y## in terms of x and ##\dot x##.
    Isolate ##{\dot x }^2##, and differentiate the equation. You will see that you can eliminate ##\dot x##. Then assume that ax<axi<<1. You arrive at an equation of form ##\ddot x +kx=0##
     
    Last edited: May 31, 2015
  16. May 31, 2015 #15
    Solved it :D never thought I'd see the day. Thank you SO MUCH for all your help!!
     
  17. May 31, 2015 #16

    ehild

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    Congratulation! Now you learned the method. :oldsmile:
     
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