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Possible to derive pendulum velocity in a dynamic approach?

  1. Nov 23, 2015 #1
    Hi,

    Using conservation of energy $$m\cdot\ g\cdot\left(h_0-h\right)=\frac{1}{2}m\cdot\ v^2$$ it's easy to find the exact velocity of a pendulum $$v\left(h\right)=\sqrt{2g\cdot\left(h_0-h\right)}$$
    at height ##h## above the minimum when it was let go from the inital height ##h_0##. Is it possible to derive this result from a dynamic point of view, i.e. looking at the forces at every height and integrate?

    I know there is no elementary solution to the differential equation of the pendulum for angles beyond the small-angle approximation, but the simplicity of above expression for ##v\left(h\right)## might suggest that there's a way around this if one's only interested in the velocity.
     
  2. jcsd
  3. Nov 26, 2015 #2
    Gave it a shot..
    So we're looking to integrate the tangential acceleration over time in order to estimate the speed of the mass. Given the symmetry of the problem, we know that the magnitude of the velocity will be equal for angles ##\theta## and ##-\theta##, i.e. ##v(\theta)=v(-\theta)##, with ##\theta = 0## corresponding to the situation where the gravitational force and the string are in line with each other. Given this symmetry we can then assert that $$v(t) = \int_0^t dt g sin\theta(t) = \int_0^{\theta} d\theta' g \frac{sin\theta'}{\dot{\theta}'}.$$
    We also know that the velocity of the mass at any time is in fact ##v(t) = l\dot{\theta}##, where ##l## is the length of the string. This gives us the following relation:$$\int_0^{\theta} d\theta' g \frac{sin\theta'}{\dot{\theta}'} = l\dot{\theta}.$$ If we now differentiate both sides with respect to ##\theta##, we find that $$g\frac{sin\theta}{\dot{\theta}} = l\frac{d\dot{\theta}}{d\theta}.$$ Rearranging this expression in order to separate dynamical variables, we see that $$l^2\dot{\theta}d\dot{\theta} = gl sin\theta d\theta.$$
    Given that coordinates and velocities are independent dynamical variables, we can integrate both sides: $$l^2\int_0^{\dot{\theta}}\dot{\theta}'d\dot{\theta}' = gl\int_{\theta_0}^{\theta}d\theta' sin\theta' ,$$ where we've assumed an initial velocity of 0. Thus, we find that $$l^2\dot{\theta}^2 = 2gl(cos\theta_0 - cos\theta)$$ which is identical to the expression you obtain from energy conservation.
     
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