STEP Maths I, II, III 1997 Solutions

1997 III, Q5 (finally):

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 =

$\int_0^{2\pi}\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2} dt = \int_0^{2\pi}\sqrt{(1-\cos t)^2+\sin^2 t} dt$

$= \int_0^{2\pi}\sqrt{2 - 2 \cos t} dt = 2 \int_0^{2\pi}\sqrt{\frac{1}{2}(1-\cos t)} dt$

$= 2 \int_0^{2\pi}\sqrt{\sin^2 \frac{t}{2}}\, dt = 2 \int_0^{2\pi} |\sin \frac{t}{2}| dt$

(note that we must take the positive root).

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

(N.B. Corrections due to ukgea's post below...)

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

[OK, what have I missed? Far, far too easy for STEP III]

Shouldn't it be

$\displaystyle = \int_0^{2\pi} 2\sqrt{\sin^2 \frac{t}{2}}dt = \cdots$

which gives the final answer $\pi/16$ instead? Other than that, it looks correct. I guess the hard part is realising that the wheel is moving with a speed $\omega r$ or something like that. And the fact that the integral will look more intimidating if you don't assume $\omega = r = 1$ as you have done. But yes, it seems rather easy.

Edit: Oh now I realise t doesn't mean time, but angle. My integrals have t = time and consequently a lot of $\omega$s floating around. Nifty. Never mind the part about $\omega$ then.