### Whirlyball

Whirl a conker around in a horizontal circle on a piece of string. What is the smallest angular speed with which it can whirl?

### Earth Orbit

Follow in the steps of Newton and find the path that the earth follows around the sun.

### Gravity Paths

Where will the spaceman go when he falls through these strange planetary systems?

# Motorbike Momentum

##### Stage: 5 Challenge Level:

For the minimum speed, the only force on the bike should be weight at the top. So we can equate weight to the centripetal force required to keep the bike moving in a circular arc:
$$mg = m\omega^2r \quad\therefore \omega = \sqrt{\frac{g}{r}}$$
where r is the position of the particle we have modeled, i.e. 2m from the centre.

So the speed at the bottom of the wheels is $3\omega$
And if the edges of the wheels are moving at $3\omega$ then the angluar speed of the wheels is $$\omega_{wheels} = v_{edge}/r_{wheel} = 3\omega/0.2 = 15\omega = 15\sqrt{g/2} = 33.20\textrm{ rad/s} =317.1\textrm{ rev/min}\;.$$

If you draw a free body diagram of the particle at a general position $\theta$ measured from the downward vertical, you can see that the reaction force is $R = mg\cos{\theta} + m\omega^2r$. The 2nd term is constant, so the graph is just $mg\cos{2\pi/\omega}$ shifted up by $mg$, i.e. sitting on the x-axis. The period is $2\pi/\omega$, where $\omega$ is $\sqrt{\frac{g}{r}}\;.$