Pendulum (derivations)

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Derivations from Pendulum (mathematics).

[edit] Analysis of a simple gravity pendulum

Figure 2. Force diagram of a simple gravity pendulum.
Figure 2. Force diagram of a simple gravity pendulum.

To begin, we shall make three assumptions about the simple pendulum

  • The rod/string/cable on which the bob is swinging is massless and always remains taut;
  • The bob is a point mass;
  • Motion occurs in a 2 dimensional plane, i.e. pendulum does not swing into and out of the page.

Consider Figure 2. The blue arrow is the gravitational force acting on the bob, violet arrows are that same force resolved into components parallel and perpendicular to the bob's instantaneous motion, the motion along the red axis, which is always perpendicular to the cable/rod. Newton's second law

F=ma\,

where F is the force acting on mass m, causing it to accelerate at a meters per second2. Because the bob is constrained to move on the green circular arc, there is no need to consider any force other than the one responsible for instantaneous acceleration parallel to the known path, the short violet arrow in our case

F = -mg\sin\theta = ma\,
a = -g \sin\theta\,

where

g is the acceleration due to gravity near the surface of the earth. It is negative because it is pointing downward.

This linear acceleration a along the red axis can be related to the change in angle θ by the arc length formulas:

s = \ell\theta\,
v = {ds\over dt} = \ell{d\theta\over dt}
a = {d^2s\over dt^2} = \ell{d^2\theta\over dt^2}

thus:

\ell{d^2\theta\over dt^2} = - g \sin\theta
{d^2\theta\over dt^2}+{g\over \ell} \sin\theta=0 \quad\quad\quad (1)
Figure 3. Trigonometry of a simple gravity pendulum.
Figure 3. Trigonometry of a simple gravity pendulum.

This is the differential equation which, when solved for θ(t), will yield the motion of the pendulum. It can also be obtained via the conservation of mechanical energy principle: any given object which fell a vertical distance h would have acquired kinetic energy equal to that which it lost to the fall. In other words, gravitational potential energy is converted into kinetic energy. Change in potential energy is given by

\Delta U = mgh\,

change in kinetic energy (body started from rest) is given by

\Delta K = {1\over2}mv^2

Since no energy is lost, those two must be equal

{1\over2}mv^2 = mgh
v = \sqrt{2gh}\,

Using the arc length formula above, this equation can be re-written in favor of {d\theta\over dt}

{d\theta\over dt} = {1\over \ell}\sqrt{2gh}

but what is h? It is the vertical distance the pendulum fell. Consider Figure 3. If the pendulum starts its swing from some initial angle θ0, then y0, the vertical distance from the screw, is given by

y_0 = \ell\cos\theta_0\,

similarly, for y1, we have

y_1 = \ell\cos\theta\,

then h is the difference of the two

h = \ell\left(\cos\theta-\cos\theta_0\right)

substituting this into the equation for {d\theta\over dt} gives

{d\theta\over dt} = \sqrt{{2g\over \ell}\left(\cos\theta-\cos\theta_0\right)}\quad\quad\quad (2)

This equation is known as the first integral of motion, it gives the velocity in terms of the location and includes an integration constant related to the initial displacement (θ0). We can differentiate, by applying the chain rule, with respect to time to get the acceleration

{d\over dt}{d\theta\over dt} = {d\over dt}\sqrt{{2g\over \ell}\left(\cos\theta-\cos\theta_0\right)}
{d^2\theta\over dt^2} = {1\over 2}{-(2g/\ell) \sin\theta\over\sqrt{(2g/\ell) \left(\cos\theta-\cos\theta_0\right)}}{d\theta\over dt} = {1\over 2}{-(2g/\ell) \sin\theta\over\sqrt{(2g/\ell) \left(\cos\theta-\cos\theta_0\right)}}\sqrt{{2g\over \ell} \left(\cos\theta-\cos\theta_0\right)} = -{g\over \ell}\sin\theta
{d^2\theta\over dt^2} = -{g\over \ell}\sin\theta

same as obtained through force analysis.

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