Uniform Circular Motion — Question 2

Question 2

Derive centripetal acceleration from a small turn

A particle traverses a circle of radius R at constant nonzero speed v. During a positive angular advance $\Delta\theta<\pi$, its velocity direction turns by the same angle. Use radians.

1. Find $|\Delta\vec v|$ from the isosceles triangle formed by the two velocity vectors.

2. Find the magnitude of average acceleration during that interval and take the limit as $\Delta\theta\to0$.

3. Compare the finite-interval direction with the instantaneous inward direction. Why is the average magnitude slightly smaller than $v^2/R$?

Question 2 - Solution

Derive centripetal acceleration from a small turn

(a) Use the chord in velocity space.

$$|\Delta\vec v|^2=v^2+v^2-2v^2\cos\Delta\theta=4v^2\sin^2(\Delta\theta/2)$$ The two velocity vectors both have length v. Their head-to-head chord has length $$|\Delta\vec v|=2v\sin(\Delta\theta/2).$$ The velocity difference is not the difference of their equal magnitudes.

(b) Divide by the arc travel time.

$$\Delta t=\frac{R\Delta\theta}{v},\quad\lim_{\Delta\theta\to0}|\vec a_{\mathrm{av}}|=\frac{v^2}{R}\lim_{u\to0}\frac{\sin u}{u}=\boxed{\frac{v^2}{R}}$$ $\Delta t=R\Delta\theta/v$, so $$|\vec a_{\mathrm{av}}|=\frac{v^2}{R}\frac{2\sin(\Delta\theta/2)}{\Delta\theta}.$$ Since $\sin u/u\to1$, the instantaneous magnitude is $a=v^2/R$. The limiting vector points toward the center.

(c) Compare a chord with an arc.

$$0<\frac{\sin u}{u}<1\quad(0<u<\pi/2),\quad |\vec a_{\mathrm{av}}|<\frac{v^2}{R}$$ For a symmetric circular interval, the average acceleration points inward at the midpoint angle. At finite positive $\Delta\theta$, $\sin(\Delta\theta/2)<\Delta\theta/2$, so its magnitude is smaller than the instantaneous value. Different acceleration directions partly cancel when averaged over the turn.