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Small particle, circular bowl

  1. May 4, 2016 #1
    • Thread moved from the technical forums, so no HH Template is shown.
    Having trouble with the following problem:

    A small particle slides along the bottom of a 10 inch radius circular bowl. Neglect friction and assume small oscillations. If the particle has a speed of 15 in/s when it is at the bottom of the bowl, determine

    a. The differential equation governing the motion
    b. The period and amplitude of the resulting vibration
    c. The position of the particle as a function of time

    So far I've got:

    at = gcosΘ

    Since a = dv/dt and v= r##\dot Θ##
    d/dt (r##\dot Θ##) = gcosΘ
    r##\ddot Θ## -gcosΘ = 0
    ##\ddot Θ## -g/r cosΘ= 0

    To me this looks like a second order differential equation. I'm taking differential equations now and not doing too well in the class, so I'm struggling. Plus, I don't think we've had one like this in class yet.
     
  2. jcsd
  3. May 4, 2016 #2

    Ray Vickson

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    The solution of the DE ##d^2 f(t)/dt^2 = c \cos(f(t))## involves Elliptic functions; see, eg., http://www.mhtlab.uwaterloo.ca/courses/me755/web_chap3.pdf for material on elliptic functions and elliptic integrals and https://en.wikipedia.org/wiki/Pendulum_(mathematics) for the relation between a pendulum and elliptic functions.
     
  4. May 4, 2016 #3

    Simon Bridge

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    Please be careful to define your terms.
    If ##\theta## is the angular position of the particle, with positive angles read anticlockwise from the bowl's symmetry axis, taken about the bowl's center - and the particle is always moving in a plane the symmetry axis lies in.
    ... then ##a_t=g\cos\theta## means the acceleration is max when ##\theta = 0##
    ... i.e. at the bottom of the bowl. Is that right?

    Also - what does "assume small oscillations" mean in this context?
     
  5. May 4, 2016 #4

    mfb

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    That is a key point. You'll have to approximate the cosine or sine to get a closed form for the solution.
     
  6. May 4, 2016 #5
    I wish I knew what assume small oscillations meant. Our teacher has a tendency of giving us random problems from other books he has, so I'm not used to that terminology. From what I can see Θ is just an angle that increases as the particle slides down the curve. At Θ= 90° the particle is at the bottom of the bowl.

    I look at this ode as:

    x'' - kcosx =0.

    We had the same problem with a drag coefficient a couple months ago. For it the original equation was: at =gcosΘ - kv/m

    With some substitution it came out to:

    Θ'' + .11Θ' -6.54Θ = 0
     
  7. May 4, 2016 #6
    "Small oscillations means you need to linearize the trig function. For example, if you have sinx, for small x this is just sinx=x." <-- I have no idea what that means
     
  8. May 4, 2016 #7

    Simon Bridge

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    You have an equation that is hard to do because of the trig function.
    Small oscillations means that ##x## is always so small that ##\sin x \approx x## and ##\cos x \approx 1## ... you can get rid of the annoying trig function by substituting with one of these relations as if they were exact equalities. It would be better to say "substitute: ##x\to\sin x## perhaps.
    So if you have ##\ddot x = \cos x##, then "small oscillations" means you can write ##\ddot x = 1##...

    Beware: you cannot just reuse equations derived in class without understanding how they were derived - you are supposed to use the derivation, not the end result.

    Anyone from a physics backgrund can tell just by looking that the cosine in your equation is wrong for this problem unless you used a really unusual definition for the angle. Recheck your geometry.
     
    Last edited: May 4, 2016
  9. May 5, 2016 #8

    mfb

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    See post #5, the definition of angle is unusual.
     
  10. May 7, 2016 #9

    Simon Bridge

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    Oh I must have missed that, I'll take another look:
    ... yeah that's pretty unusual. May explain the confusion about what "small oscillations" would mean.

    Hopefully the angle is still taken about the center of the sphere the bowl is a part of.
    I think the observation about understanding how the equations used are derived is still most important.

    So taking ##\theta## about the center, clockwise from the horizontal - then:
    Then ##\ddot\theta - k\cos\theta = 0 : k=g/r## ... and "small oscillations" means oscillations about the 90deg mark.

    Close to 90deg ##\cos\theta## looks like ##90-\theta## ... which I think is safe to reveal at this stage.
    Though should really do angles in radians... so the small oscillation approximation of the DE goes something like:

    ##\ddot\theta -k(\frac{\pi}{2}-\theta) = 0:\theta(0)=\frac{\pi}{2}, \dot\theta(0)=15in/s##
    ... which should be easier to solve.


    In general - an equation of form ##\ddot x = f(x)## ... the function on the RHS can be approximated by a power series: ##f(x)=\sum_{n=0}^N a_nx^n## ... where N is the "order" of the approximation.
    This is desirable since polynomials are usually easier to handle.
    The downside is that the approximation may not be very good.

    The N=1 order approximation, is ##f(x) = a_0+a_1x## ... which is the equation of a straight line, so it is sometimes called "linearizing the function".
    To find the right coefficients, you have to pick a specific value for x where the approximation needs to be very close.

    Look up "Taylor series".
     
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