Motorbike momentum
A think about the physics of a motorbike riding upside down
Age
16 to 18
Challenge level
Problem
A motorbike stuntman goes from town to town riding his bike in a large sphere 6m in diameter made from steel wire. The wire is so thick the ball remains rigid while he rides inside it. See a video here.
The motorbike and rider are about 2m tall together. We might model this as a particle in the centre, so when the bike & rider are in the sphere, this particle is 2m from the centre of the sphere.
The wheels of the bike are 40cm in diameter. What is the minimum angular speed the wheels must rotate if the bike is to be able to go upside down over the top?
Can you draw a graph of the reaction force that the sphere exerts on the bike over time if the bike maintains this speed?
Student Solutions
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 $\sqrt{\frac{g}{r}}\;.$