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2 problems on simple harmonic motion

  1. Aug 28, 2005 #1

    Wen

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    these are 2 qn on simple harmonic motion which i am not sure if my solution is right or not.
    1) A thin rod, length L hinged at one end, connected to a spring at the other end. Take the spring constant to be K. What is the angular frequency?

    My solution:

    let the displacement from equil. to be X
    F=-kX
    Torque= K.X.L=I a
    a=K.X.L/I =L @ @ is the angul. aceleeration
    @= KX/( 1/3 ML^2) = W^2X
    W= (3k/ML^2)^0.5
    2) Find the angular frequency of a small cylinder mass m, radius r, rolling in a cylindrical shell of radius R.

    My sol.

    aceleration=gsin@ let @ be the small change in angle
    gsin@=R( Aceleration ,A)
    A= gsin@/R
    A= R angular acele. Since @is small, Sin@=@

    g@/R=R angular a
    angular a = w^2 @
    w=(g/R^2)^0.5
     
  2. jcsd
  3. Aug 28, 2005 #2

    lightgrav

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    1) angular frequency omega (w) should have units of [radian/sec].
    Torque causes rotational inertia (I) to have angular acceleration alpha (@).
    So you wanted to write kx L = I @ ... but this is rotation around the pivot.
    The connection at the spring has acceleration a = L @ ,
    and it is this acceleration that = - w^2 x ... [from 2nd derivitive of x(t)].

    2) You never used the cylinder mass or radius, or its Inertia term "1/2". hmm
    It might as well be sliding ... a = g sin(theta) is only for sliding or hanging.
    at theta from the bottom, torque around contact point mg r sin(theta) = I alpha.
    This alpha is the cylinder's rotational *spin* acceleration; "cylinder rolls" if
    arclength s = R*theta = r*(phi+theta), where phi is the cylinder spin angle.
     
  4. Aug 28, 2005 #3

    Wen

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    For qn 2, if mgsin (theta)=I @
    i can obtain the rotational angular acele
    using mgsin phi= ma
    i can obtain the translational; acele
    but how should i equate to w^2x because are in different direction?
     
  5. Aug 28, 2005 #4

    mukundpa

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    First question 1

    What about the mass and weight of the rod and its torque?
    If the mass of the rod is zero then what will be its moment of inertia?

    You have to consider rotetional SHM, similar equation with
    torque proportional to - angular displacement
    and with that you will get
    freq. n = (1/2Pi) (K/I)^(1/2) Where K is torque constant.
     
  6. Aug 29, 2005 #5

    Wen

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    for qns 1 , i tried using 3 method,
    1) KX= ma 3) KX=Ia
    2) KX.L=I a 4) KX.L= ma

    which one is wrong and why.
    For qn 1, the spring is above the rod. Hence the rod is hanging from the spring.

    Mukundpa, what is SHM?

    Could someone enlighten me on qn 2?
     
  7. Aug 29, 2005 #6

    mukundpa

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    Now you are saying "the spring is above the rod. Hence the rod is hanging from the spring."
    So where is the hinge? Can you draw a diagram to show the position of spring and the rod?
     
  8. Aug 29, 2005 #7

    Wen

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    for qn 1, its a horizontal beam , pivoted at one end on the wall. Its other end is supported by a spring from the top so that it is horizontal.
     
  9. Aug 29, 2005 #8

    mukundpa

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    whether the other end is displaced slightly horizontally or vertically?
     
  10. Aug 29, 2005 #9

    mukundpa

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    If the end is displaced slightly such that the rod is rotated by small angle@ then the extension in the spring will be L@ and hence the tension in the spring is KL@.
    This provides a restoring torque KL^2@ on the rod about the hinge.
    So we can write the equation torque = - KL^2@ which is similar to F = - Kx

    We can not take force Kx because there is a reaction force at the hinge which is not accounted.
     
  11. Aug 29, 2005 #10

    mukundpa

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    As for SHM angular frequency is given by squar rt (K/m) similarly for rotetional SHM it is given by squar rt (K/I) where K is torque constant and I is moment of inertia.

    In your problem Torque constant is KL^2 and I = mL^2/3
     
  12. Aug 30, 2005 #11

    Wen

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    thanks. what about qn 2?

    if i use gsin(thetre)= R angular a

    since the cylinder also rotates abt the axis through its centre,

    mgsin( thetre).r=torque=I. angular acele abt its axis.
    As the two acele. are in different directions, how should i combine them together?
     
  13. Aug 30, 2005 #12

    mukundpa

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    Rolling without slipping is not possible without friction. It will be better to find acceleration of centre of mass of the sphere. Do you have any idea for that?


    Consider the axis of rotation passing through the point of contect and find angular acceleration and then linear acceleration.
     
  14. Aug 30, 2005 #13

    Wen

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    okay, if i find the angular aceleration and translational a, should i sum them up to equal to w^2 thetre. But the direction of the acele is different.
     
  15. Aug 30, 2005 #14

    mukundpa

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    Let the line joining the axis of the small cylinder and the axis of the surface makes an angle @ with vertical. the torque on the small cylinder about the point of contact is
    mgr sin@
    and angular acceleration is
    mgrsin@/(3mr^2/2) = 2gsin@/3r
    hence linear acceleration of center of mass will be.......( radius R-r)
    2gsin@/3 = 2g@/3 = 2g[x/(R-r)]/3
    where x is displacement of C.M. from equilibrium position and as @ is small enough
    Hence the restoring force F = - m[2g/3(R-r)] x which is the equation of SHM
    so angular frequency is sq.rt.of 2g/3(R-r).

    EEEnjoy !!!
     
  16. Aug 31, 2005 #15

    Wen

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    isn't the I of cylinder 1/2mr^2.

    However, if i use the same way to work out one qn ( which is very popular in all textbooks) A solid sphere radius R rolls without slipping in a cylinderical trough, radius 5R. show that the period is T=2pi (28R/5g)^0.5.

    Since Rmgsin@=I a.
    a= Rmgsin@/(2/5 mR^2)=5gsin@/2R=5g@/2R
    @= X/(5R-R)
    a= 5g/2R(X/4R)=w^2X

    But cannot prove to the eqn stated.
     
  17. Aug 31, 2005 #16

    mukundpa

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    Torque is taken about the point of contect hence the I is to be taken about the same point i.e. (according to parallel axes theoren) 1/2mr^2 + mr^2 = 3/2 mr^2
     
  18. Aug 31, 2005 #17

    mukundpa

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    The simelar problem
    The moment of inertia of the sphere about a tangent is

    [tex]\frac{\ 2mR^2}{5} + mR^2 = \frac{\ 7mR^2}{5}[/tex]
     
    Last edited: Aug 31, 2005
  19. Aug 31, 2005 #18

    Wen

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    thank you . thank you
     
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