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# Motorbike Momentum

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 $\sqrt{\frac{g}{r}}\;.$

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Age 16 to 18

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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 $\sqrt{\frac{g}{r}}\;.$