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STEP III 1997 question 5 solution

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TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP III 1997 question 5 solution


Wlog, the wheel has radius 1. Let the angle turned through be t. If the wheel wasn't rolling, the position of a point on the rim is (-sin t, 1-cos t). As the wheel is rolling, we have to add t to the x coord to get (t-sin t, 1 - cos t).

Distance travelled by point on rim =

\displaystyle \int_0^{2\pi}\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2} dt = \int_0^{2\pi}\sqrt{(1-\cos t)^2+\sin^2 t} dt \\
= \int_0^{2\pi}\sqrt{2 - 2 \cos t} dt = 2 \int_0^{2\pi}\sqrt{\frac{1}{2}(1-\cos t)} dt \\
= 2 \int_0^{2\pi}\sqrt{\sin^2 \frac{t}{2}}\, dt = 2 \int_0^{2\pi} \left|\sin \frac{t}{2}\right| dt

(note that we must take the positive root).

This = \displaystyle 2\int_0^{2\pi} \sin \frac{t}{2} dt = -4\left[ \cos \frac{t}{2}\right]_0^{2\pi} = 8.

Distance traveled by center = 2 pi. So ratio is  \pi / 4.

Solution by DFranklin

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