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Question- Angular Simple Harmonic motion

  1. Apr 26, 2013 #1
    Question--- Angular Simple Harmonic motion..

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

    A spherical ball of mass "m" and radius "r" rolls without slipping on a rough concave surface of large radius "R". It makes small oscillations about the lowest point. Find the time period of such oscillations.

    2. Relevant equations

    http://en.wikipedia.org/wiki/Simple_harmonic_motion

    3. The attempt at a solution

    http://postimg.org/image/whgt6dvvb/ [Broken]

    In the last line of the image above, the formula, Time period = 2π√(I/Mgl)

    Where "l" is the distance between the point of suspension and centre of mass of physical pendulum. Why does this formula fail in this case ?

    It yields :

    l= R-r
    I= 2mr2/5 + m(R-r)2

    On putting these values in formula, Time period = 2π√(I/Mgl), why I keep getting the wrong answer ?

    Please helps !!

    Thanks in advance... :smile:
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Apr 26, 2013 #2

    BruceW

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    That formula is the formula for a compound pendulum. (Which is a pendulum that can move freely, but has some non-simple moment of inertia). In the problem you are doing, can the pendulum move freely? And does it even make sense to call it a pendulum? And how does this affect the result?

    hint: The time period for this problem is not immediately obvious from any kind of short-cut formula (as far as I know). So you might need to go through the full calculation to get the answer.
     
  4. Apr 26, 2013 #3
    I don't know but I think that it can be treated as a compound pendulum (physical pendulum), as the solid sphere is a rigid body and that its undergoing angular simple harmonic oscillations about the imaginary point of suspension, which is the centre of the concave surface. Right ?

    The question asks about ASHM about the lowest point. I have to calculate its moment of inertia about the lowest point and I calculated it about the centre of concavity !! :redface:

    Even with this correction, I did not get the correct answer. With this correction, moment of inertia = I= 2mr2/5 + mr2, which is not yielding the correct answer. Why ?
     
  5. Apr 27, 2013 #4

    ehild

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    No. All points of a compound pendulum (the right one) rotate with the same angular velocity so the linear velocity of a point is proportional to the distance from the suspension. The ball here rotates about the CM with a certain angular velocity ω while the CM moves along a circle of radius R-r with a different angular velocity Ω. The linear velocity of the point nearest to the centre moves with VCM+rω, the furthest one moves with VCM-rω. In case of pure rolling, that contact point is in rest.

    ehild
     

    Attached Files:

  6. Apr 27, 2013 #5

    BruceW

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    No, unfortunately not. If the pendulum was free to oscillate (i.e. if the concave surface was not there), then I would agree with you. But because of the concave surface, it is not a free pendulum, so it is not a compound pendulum. Another way to think of it, is that the motion of the pendulum will be affected by the concave surface, right? But the usual equation for a compound pendulum only takes the moment of inertia into account. So this is a reason to be immediately suspicious of that equation.
     
  7. Apr 27, 2013 #6
    This is a very interesting question from "H.C.Verma Concepts of physics".
    Hint-It is best to not compare this situation with that of a pendulum. Think conservation of energy. can you extract the equation you need out of that?
     
  8. May 2, 2013 #7
    Hello everyone again.

    Can I use free body diagrams of the sphere also, to evaluate for the acceleration on it due to the restoring force denoted by -kx ? Will it be OK ? consciousness, I was also thinking of conservation of energy equation and then differentiating it with respect to time and putting this derivative to zero. Can I also use free body diagram to solve for that ?

    Thanks.
     
  9. May 2, 2013 #8

    Yes you can! In fact the answer actually takes less time when done by making the FBD! So start making that FBD with the center of the sphere at a general angle with the vertical. But we have to use something else also...see the question. (we can make an approximation).

    (Your approach to conservation of energy method is correct.)
     
  10. May 12, 2013 #9
    Consciousness, I cannot fathom your point. I know that for small angular oscillations, sinθ≈θ, is the assumption.

    Let me frame equations from FBD :

    N-mg=mv2/(R-r)

    N is normal reaction.

    Again,

    (Torque about point of suspension)/I = angular acceleration

    {-mgsin(θ)*(R-r)}/{2mr2/5 + m(R-r)2}= α

    Or

    {-gθ(R-r)}/{2r2/5 + (R-r)2}= α

    ω(Angular Frequency) = {g(R-r)}/{2r2/5 + (R-r)2}

    T=2π/ω, then gives me incorrect answer from FBD. Where did I go wrong ?
     
  11. May 13, 2013 #10
    For small oscillations you can approximate the the path of the sphere to be a straight line. So it becomes a linear SHM. Then by making FBD prove F=-kx (where x is dipacement from lowest point) and thus find the time period.
     
  12. May 13, 2013 #11
    If it were simple harmonic motion, then there will not be role of moment of inertia at all. So,

    -mgsinθ = -kx

    For small oscillations,

    -mgx/(R-r) = -kx

    So, k = mg/(R-r)

    So, T= 2∏√{(R-r)/g}, which is not the correct answer. Please help !! Anyone !! Suggest me what to do.
     
  13. May 13, 2013 #12

    haruspex

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    No, the MI will have a role. You seem to be assuming ##F = m\ddot x##, but you're overlooking the force of friction from the surface. You need an equation relating acceleration of x to angular acceleration of the sphere.
     
  14. May 13, 2013 #13

    ehild

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    See picture. The ball rolls inside the sphere. The resultant of the forces (gravity, normal force and static friction) determines the acceleration of the CM, while the torque of the static friction accelerates its rotation. Apply the rolling condition to relate acceleration of the CM to the angular acceleration of the rotation.

    ehild
     

    Attached Files:

  15. May 14, 2013 #14
    ehild, have I to make another equation for static friction ?

    Alright so net torque will be : fs*R + mgθ*(R-r), correct ?

    Then I do not know coefficient of static friction. How will I evaluate for net torque ?

    Thanks once again...
     
  16. May 14, 2013 #15
    As we have approximated it into a linear SHM gravitational force has zero torque. You don't need to know the value of fs as it is just substituted into your main force equation.
     
  17. May 14, 2013 #16

    ehild

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    You need the torque on a moving object - it is the ball. The CM of the ball travels along a circle of radius R-r, and it also rotates around its centre. The torque with respect to the centre of the ball is fsr. That torque accelerates the rotation of the ball: fsr=Idω/dt if you consider the rotation positive when it corresponds to the velocity of the ball. You do not need the actual value of the torque, enough to know that the static friction always keeps the ball rolling.
    For the (tangential) acceleration of the CM of the ball along its orbit, you get ma=mgsinθ-fs. That acceleration is proportional to the second derivative of the angle θ :a=-(R-r)(d2θ/dt2).
    The rolling condition means that the speed of the CM is the same as the linear velocity of the perimeter of the ball, and the same holds for the accelerations: a=r(dω/dt).

    You can make a single equation in terms of the angle theta, by cancelling fs.
     
  18. May 15, 2013 #17
    consciousness, your logic does not appeal to me. If I consider it SHM only, then I will surely get wrong answer. Infact there is pure rolling in the sphere. Making it simple harmonic instead of angular harmonic, is an implication of incomplete concept. As ehild and haruspex point out, I have to consider angular harmonic and torques on the sphere. May be you understand this. Making an FBD does not necessarily imply that motion is plain SHM, we can make FBD if motion is ASHM as well.

    Ok, so I get the equation :

    fs = Idω/rdt

    So, ma=mgsinθ-Idω/rdt

    Again, dω/dt = a/(R-r)

    So,
    ma=mgsinθ-Ia/(R-r)
    a= mgsinθ/(m+I/(R-r))

    For small oscillations, a= mgθ/(m+I/(R-r))

    Is this equation correct ?
     
  19. May 15, 2013 #18

    ehild

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    ω is the angular speed of rotation about the centre of the ball. The rolling condition means that the CM of the ball moves with the same speed (and acceleration) as the perimeter of the ball. a=r(dω/dt).

    Relate a to theta, and write up the equation in terms of theta. The angle theta should change with time according to an SHM. Find the time period.

    ehild
     
  20. May 15, 2013 #19
    I have already got the answer by approximating the motion of the sphere to be a straight line. In fact this is same as the sinθ≈θ approximation. The easiest way to solve this problem is to prove that Net force on sphere=-kx where x is displacement from the lowest point(mean position). Note that to do it this way you have to consider the torque of static friction and apply condition of pure rolling etc. Why do you assume that for a linear SHM you wont have to do so?

    All angular SHMS with the small oscillations condition can be adequately represented by a linear SHM with the small displacements(amplitude) condition.
     
  21. May 15, 2013 #20
    consciousness, there was one question in H.C. Verma, not remembering question number. There was pendulum rectangular in shape having two point of suspensions etc.. Its time period was 2π√(L/g), independent of any other dimensions. It was the time period of SHM. If Mr. Verma would have asked time period for ASHM, then it would not be the same. In fact, you have to calculate moment of inertia of that rectangular lamina about point of suspension. So what I mean to say is that, we are calculating its time period for ASHM. In this question also, time period of SHM and time period of ASHM are not the same. Think. In fact what you did was to calculate time period of ASHM and thus you got the correct answer.

    @ehild,

    I already did this question by law of conservation of mechanical energy. I need to do it by free body diagram concept.
     
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