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Pendulum issues - find acceleration and restoring force

  1. Apr 7, 2005 #1
    I am given that the mass of a simple pendulum is 0.25kg, length 1m and displaced 15 degrees then released.

    a) max angular acceleration?
    b) max restoring force?

    for a... I vound that the max velocity is 0.817m/s

    [tex]\omega=\frac{v}{r}[/tex]
    [tex]\omega=\frac{0.817m/s}{1m}[/tex]
    [tex]\omega=0.817rad/s[/tex]

    [tex]\alpha=r \omega ^2[/tex]
    [tex]\alpha=1m * 0.817^2[/tex]
    [tex]\alpha=0.668rad/s^2[/tex]

    I know that's wrong but why?

    for b...
    I just multiply alpha by the radius to get acceleration then multiply that by the mass.
     
  2. jcsd
  3. Apr 8, 2005 #2

    HallsofIvy

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    It's hard to tell what you are doing wrong because it's not clear what you did.

    You say you got "max velocity is 0.817m/s". How? Are you using the exact equation for the motion or the linear approximation.

    It seems strange to me that you would find the linear velocity, v, when the problem asks for the angular velocity and it is more direct to find it directly.

    The exact equation of motion, F= ma, for a pendulum is [tex]ml\frac{d^2\theta}{dt^2}= -mg sin(\theta)[/tex]. The "m"s cancel so we have [tex]\frac{d^2\theta}{dt^2}= -\frac{g}{l}sin(\theta)[/tex]. Let [tex]\omega= \frac{d\theta}{dt}[/tex] (the angular velocity). By the chain rule [tex]\frac{d^2\theta}{dt^2}= \frac{d\omega}{dt}= \frac{d\omega}{d\theta}\frac{d\theta}{dt}= \omega\frac{d\omega}{d\theta}[/tex] so the equation becomes [tex]\omega\frac{d\omega}{d\theta}= -\frac{g}{l}sin \theta[/tex].

    That's a separable equation: [tex]\omega d\omega= -\frac{g}{l}sin \theta d\theta[/tex] which integrates to [tex]\frac{1}{2}\omega^2= \frac{g}{l}cos \theta[/tex].
    You should be able to find the maximum value of [tex]\omega[/tex] directly from that.

    The "linearized" version of that is [tex]l\frac{d^2\theta}{dt}= -g \theta [/tex] (since , for small [tex]\theta[/tex], [tex]sin \theta[/tex] is approximately [tex]\theta[/tex]) which is the "linear homogeneous d.e. with constant coefficients", [tex]\frac{d^2\theta}{dt^2}+ \frac{g}{l}\theta= 0[/tex]. The general solution for that is [tex]\theta= Ccos(\sqrt{\frac{g}{l}}t)+ Dsin(\sqrt{\frac{g}{l}}t)[/tex]. The angular velocity is [tex]\omega= \frac{d\theta}{dt}= (\sqrt{\frac{g}{l}})(-Csin(\sqrt{\frac{g}{l}}t)+ Dcos(\sqrt{\frac{g}{l}}t))[/tex]. Again, you can find the maximum angular velocity directly from that.
     
  4. Apr 8, 2005 #3

    dextercioby

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    I think he used conservation of energy.At zero PE (the lowest point of the circular trajectory),the KE is maximum and -------->max.linear velocity...

    My guess.

    Daniel.
     
  5. Apr 8, 2005 #4

    BobG

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    Maximum velocity doesn't have to occur at the same point as your maximum acceleration. HallsofIvy gives the mathematical explanation, but you should be able to see this visually, as well. Gravity is providing the force that accelerates the pendulum, but the pendulum is restrained by the rod holding it to the center. Your maximum acceleration will come when the pendulum's motion is straight down ; when the motion is aligned with the force of gravity (or when it's as close to straight down as it's ever going to get). For a pendulum, your acceleration will hit it's maximum velocity when it's at the bottom of its swing, a point where the force of gravity is perpendicular to the pendulum's motion, meaning acceleration will be zero.

    You should be able to see where your maximum acceleration has to occur and figure it out from there.
     
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