Consider a pendulum made by attaching a point mass \(m\) to a string with negligible mass and length \(L\) and fixing one end to the highest point of a cylinder of radius \(R\) and letting the mass hang to the side. Find the equation of motion of the mass and the frequency for small oscillations.
Consider the variables defined in the problem as follows
We write the \( x \) and \( y \) components of the position vector in terms of the variables
defined above.
\( x = c + d = Rsin\theta + (L-R\theta)cos\theta \)
\( y = -a - b = -(L-R\theta)sin\theta -R(1-cos\theta) \)
Notice that \( x \) and \( y \) are functions of \(\theta\), so that is the only coordinate.
There will be only one equation of motion. The velocities are:
\( \dot x = \dot \theta [Rcos\theta + (L-R\theta)sin\theta -Rcos\theta] = \dot \theta(L-R\theta)sin\theta \)
\( \dot y = \dot \theta [-Rsin\theta - (L-R\theta)cos\theta + Rsin\theta]= -\dot \theta(L-R\theta)cos\theta \)
Now we can write the kinetic energy in terms of these variables
\( K.E. = {1\over 2} m(\dot x^2 +\dot y^2) = {1\over 2} m \dot \theta^2 (L-R\theta)^2 \)
To find the Lagrangian, we also need the potential energy
\( P.E. = mgy = mg[-R(1-cos\theta)-(L-R\theta)sin\theta] \)
So, the Lagrangian is
\( \mathscr L = K.E. - P.E = {1\over 2} m \dot \theta^2 (L-R\theta)^2 + mg[R(1-cos\theta)+(L-R\theta)sin\theta] \)
To find the equation of motion we take derivatives. Partial derivative of \(\mathscr L\) with respect to \(\dot\theta\)
\( \dfrac{\partial \mathscr L}{\partial \dot\theta} = m\dot\theta(L-R\theta)^2 \)
Then derivative of this expression with respect to time
\(\dfrac{d}{dt} \left(\dfrac{\partial \mathscr L}{\partial \dot\theta}\right) = m\ddot\theta(L-R\theta)^2
- 2mR\dot\theta^2(L-R\theta) \)
We also need the partial derivative of \(\mathscr L\) with respect to \(\theta\)
\( \dfrac{\partial \mathscr L}{\partial \theta} = -mR\dot\theta^2(L-R\theta) + mg(L-R\theta)cos\theta \)
Then, the equation of motion is
\( m\ddot\theta(L-R\theta)^2 - 2mR\dot\theta^2(L-R\theta) =
-mR\dot\theta^2(L-R\theta) + mg(L-R\theta)cos\theta \)
After some simplification we get
\( \ddot\theta(L-R\theta) - R\dot\theta^2 - g cos\theta = 0 \)
We can change variables to simplify the equation
\( \theta = \dfrac {\pi}{2} + \phi \rightarrow cos\theta = -sin\phi \)
\( C = L - R\pi/2 \rightarrow L-R\theta = C - R\phi \)
With the changes, we get
\( (C-R\phi)\ddot\phi - R\dot\phi^2 + g sin\phi = 0 \)
If the oscillations are small, we can ignore second order terms and approximate \( sin\phi \approx \phi \), so
\( C\ddot\phi + g \phi = 0 \)
The frequency of oscillation is
\( \omega = \sqrt{\dfrac{g}{C}} = \sqrt{\dfrac{g}{L-R\pi /2}}\)
We notice that this is the same equation that we get for a simple pendulum with length
\(L-R\pi /2\)