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The Not-so-simple Pendulum 2

Things are roughened up and friction is now added to the approximate simple pendulum

Falling Beads

Stage: 5 Challenge Level: Challenge Level:2 Challenge Level:2

First of all let's choose a convenient Cartesian coordinate system as shown in the picture and suppose that the bead which is sliding through the needle inclined at an angle $\alpha$ to x-axis is at distance $r(t)$ from the origin.

The gravity force is acting on the bead and it makes the bead accelerate along the needle.  Write II-Newton's law in the direction along the needle and in the direction perpendicular to the needle.

$$ma = mg\sin(\alpha)$$

$0 = N - mg\cos(\alpha)$ where $N$ is a reaction force.

Beads start from rest, so $r(t) = \frac{at^2}{2} = \frac{g\sin(\alpha) t^2}{2}$.

Change the polar coordinates to Cartesian coordinates by substituting $r = \sqrt{x^2 + y^2}$ and $\sin(\alpha) = \frac{y}{\sqrt{x^2 + y^2}}$.

$$\sqrt{x^2 + y^2} = \frac{y}{\sqrt{x^2 + y^2}}\frac{gt^2}{2}$$ Multiply both sides by $\sqrt{x^2 + y^2}$ and complete the square.

$$x^2 + \left(y -\frac{gt^2}{4}\right)^2 = \left(\frac{gt^2}{4}\right)^2$$

  • The shaped formed by beads after time $t$ is a circle of radius $\frac{gt^2}{4}$ with centre coordinates $\left(0, \frac{gt^2}{4}\right)$.
  • From the previous answer we deduce that there is a circle whose centre is falling with the acceleration $\frac{g}{2}$ and this circle is expanding with the same acceleration. A picture shows the shape of beads after time $0.5\textrm{ s}$, $0.75\textrm{ s}$, $1\textrm{ s}$, $1.25\textrm{ s}$, $1.5\textrm{ s}$, $1.75\textrm{ s}$, $2\textrm{ s}$.
    Extension: Now we have a friction force due to sliding. We modify our previous equations:

$$ma = mg\sin(\alpha) -\mu N$$

$0 = N - mg\cos(\alpha)$ where $N$ is a reaction force.

$$\therefore\qquad a = g(\sin(\alpha) - \mu\cos(\alpha))$$ It is important to notice that if $a < 0$ then the bead is not even starting to slide. Thus, the beads will slide for $\alpha$ from $\tan^{-1}(\mu)$ to $(\pi - \tan^{-1}(\mu))$. To simplify the situation, notice that the shape will be symmetrical in the y-axis. Thus, we need to analyse what happens for $\alpha$ from $\tan^{-1}(\mu)$ to $\frac{\pi}{2}$. $$r(t) = \frac{at^2}{2} = (\sin(\alpha) - \mu \cos(\alpha)) \frac{gt^2}{2}$$ Change to the Cartesian coordinate system $x^2 + y^2 = \frac{gt^2}{2} (y - \mu x)$ for $\frac{\pi}{2} > \tan^{-1} (\frac{y}{x}) > \tan^{-1} (\mu)$. Complete the squares to get

$$\left(x+\frac{\mu gt^2}{4}\right)^2+\left(y-\frac{gt^2}{4}\right)^2=\left(1+\mu^2\right)\left(\frac{gt^2}{4}\right)^2\;.$$ It is easier to plot a whole circle and then to take a suitable arc. Similarly, to reflect a circle in y-axis change $x$ to $-x$ then we have an equation $$\left(-x+\frac{\mu gt^2}{4}\right)^2 + \left(y - \frac{gt^2}{4}\right)^2 = \left(1+ \mu^2\right)\left(\frac{gt^2}{4}\right)^2$$ and we take a suitable arc again. Let's take $\mu = \frac{\sqrt3}{3}$ then $\tan^{-1}\left(\frac{\sqrt3}{3}\right) = \frac{\pi}{6}$. We plot circles which help to show how the shape of beads in red changes. These graphs represent the shape of beads after time $0.5\textrm{ s}$, $0.75\textrm{ s}$, $1\textrm{ s}$, $1.25\textrm{ s}$, $1.5\textrm{ s}$, $1.75\textrm{ s}$, $2\textrm{ s}$. Circles are traveling along green lines which equations can be found knowing that the coordinates of center is $\left(\frac{\mu gt^2}{4}, \frac{gt^2}{4}\right) = \frac{gt^2}{4}\left(\mu, 1\right)$ and  $\frac{gt^2}{4}\left(-\mu, 1\right)\;.$

To sum up:

  • The shape of beads after time $t$ is an union of two arcs.
  • Centre of circles are traveling with $\frac{g}{2} \sqrt{1+\mu^2}$ acceleration and expanding with the same acceleration.