Question

Particles of mud are thrown from the rim of a rolling wheel. If the forward speed of the wheel is v_0, and the radius of the wheel is b, show that the greatest height above ground the mud can go is:
\[b+\frac{v_0^2}{2g}+\frac{gb^2}{2v_0^2}\]\]

At what point on the rolling wheel does this mud leave?
(Note: it is necessary to assume that
\[v_0^2\ge bg\]

(my workings in next post)

Answer 1

So we have two co-ordinate systems: one polar co-ordinates relative to the centre of the wheel, and another 1D Cartesian on the ground.
Relative to centre of wheel:
\[r=b\cos(\theta)i-b\sin(\theta)j\]
\[\theta=\omega t=\frac{v_0}{b}t\]
\[r^.=-v_0\sin(\theta)i-v_0\cos(\theta)j\]
Relative to ground:
\[r^.=(v_0-v_0\sin(\theta))i-v_0\cos(\theta)j\]

So where do I go from here?