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Spring pendulum, Kinetic Energy

  1. Apr 4, 2015 #1
    In the included picture, I don't get how they get to the kinetic energy part. I would say, that the travelled distance is equal to (l + x(t))*θ. Then I would take the time derivative, resulting in dx(t)/dt * θ + (l + x(t))* dθ/dt. Then I would square this result and multiply that with 1/2 m. But then I would get a totally different kinetic energy.
     

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  3. Apr 4, 2015 #2

    mfb

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    2016 Award

    Staff: Mentor

    What does that mean? The travelled distance depends on the trajectory. You'll need a position (as vector) to get a meaningful derivative.
     
  4. Apr 5, 2015 #3
    You must also consider the radial component. The velocity component in the radial direction is the time derivative of x(t) (since it is a spring, it can compress or expand). Hence the total kinetic energy is the sum of energies in both radial and angular directions.
     
  5. Apr 5, 2015 #4

    vanhees71

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    In polar coordinates, with the polar angle relative to the direction poiinting downwards, you have
    $$\vec{x}=r \begin{pmatrix}
    \sin \varphi \\ \cos \varphi
    \end{pmatrix}.$$
    Then, after straight-forward algebra, you have
    $$T=\frac{m}{2} \dot{\vec{x}}^2=\frac{m}{2} (\dot{r}^2+r^2 \dot{\varphi}^2).$$
    Now in the textbook, they set
    $$r=l+x,$$
    where ##l## is the length of the relaxed spring.

    For the potential energy you have the part from the gravitational field of the earth and the spring:
    $$V=-m g x_2+\frac{k}{2} x^2 = -m g (x+l) \cos \varphi+\frac{k}{2} x^2.$$
     
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