I tried to draw the situation in GeoGebra for visualization

Imagine a ball is travelling in a bumpy road like the graph $\sin x$ (I'm using $\sin x$ just as an example, it has no significance in the question I guess).

So, a ball is travelling in this way and a bob of pendulum is hanging from it and having a harmonic motion. Can anyone give me idea how we can find the bob's equation of motion here?

  • 1
    $\begingroup$ The shape of the road is definitely significant. $\endgroup$
    – R.W. Bird
    Aug 15 at 13:15

Assume the ball of is rolling on the graph of the function $y = f(x)$. Let $m_1$ be the mass of the ball with radius $R$ and denote $\vec{r}_1 = (x_1,y_1)$ the position of its center of mass.

Let $m_2$ be the mass of the bob and let $\vec{r}_2 = (x_2,y_2)$ be its position. Assume $\theta$ is the displacement angle of the bob from the vertical and let $L$ be the length of the string.

If the ball rotates with angular velocity $\omega$, the kinetic and potential energy of the system are $$T = \frac12I\omega^2 + \frac12m_1\dot{r}_1^2 + \frac12m_2\dot{r}_2^2, \qquad V = mgy_1 - m_2g L\cos\theta.$$

  1. In the first term we have $I = \frac25m_1R^2$ and if the ball is rolling without slipping we have $R\omega = v$ where $v$ is the velocity of the ball along the curve. The ball traces the arc length of the curve so we have $$\int_0^t v(t)\,dt = \int_{x_T(0)}^{x_T(t)}\sqrt{1+f'(x)^2}\,dx$$ where $(x_T, y_T)$ is the position of the contact point of the sphere and the curve. Applying $\frac{d}{dt}$ gives $$v(t) = \sqrt{1+f'(x_T(t))^2}\dot{x}_T(t)$$ so $\omega = \frac1R \sqrt{1+f'(x_T)^2}\dot{x}_T.$

  2. For the second term, the hardest part is to connect $(x_T, y_T)$ with $(x_1, y_1)$. If $\phi$ is the angle of the slope at that point, from a right triangle we get $$\frac{x_T-x_1}{R} = \sin\phi = \frac{\tan\phi}{\sqrt{1+\tan^2\phi}} = \frac{f'(x_T)}{\sqrt{1+f'(x_T)^2}}$$ since $\tan\phi = f'(x_T)$. The same right triangle gives $$\frac{y_1-y_T}{R} = \cos\phi = \frac{1}{\sqrt{1+\tan^2\phi}} = \frac{1}{\sqrt{1+f'(x_T)^2}}$$ so in total $$(x_1,y_1) = (x_T,y_T) + \frac{R}{\sqrt{1+\tan^2\phi}}(-1,f'(x_T)).$$

  3. For the third term, it is easy to see that $$(x_2,y_2) = (x_1+L\sin\theta, y_1-L\cos\theta).$$

The crucial step is to notice that $y_T = f(x_T)$ so you can express the Lagrangian in terms of two generalized coordinates: $x_T$ and $\theta$. No such simple relationship exists between $x_1$ and $y_1$.


Not the answer you're looking for? Browse other questions tagged or ask your own question.