Uniform Circular Motion — Question 6

Question 6

Average vectors over a quarter turn are chords

A particle moves counterclockwise on a circle of radius 4 m at speed 2 m/s. It travels from $(4,0)$ m to $(0,4)$ m through a quarter turn.

1. Find the elapsed time and average velocity.

2. Find the average acceleration from the endpoint velocities.

3. Compare the two average magnitudes with the constant speed and instantaneous acceleration magnitude.

Question 6 - Solution

Average vectors over a quarter turn are chords

(a) Divide the position chord by arc travel time.

$$\Delta t=\frac{R\pi/2}{v}=\pi\,\mathrm s,\quad\boxed{\vec v_{\mathrm{av}}=\frac{(-4,4)}\pi\,\mathrm{m/s}}$$ The arc length is $R\pi/2=2\pi$ m, so $\Delta t=\pi$ s. Displacement is $(-4,4)$ m, giving $\vec v_{\mathrm{av}}=(-4/\pi,4/\pi)$ m/s and magnitude $4\sqrt2/\pi\approx1.801$ m/s.

(b) Use the velocity chord separately.

$$\vec a_{\mathrm{av}}=\frac{(-2,0)-(0,2)}\pi=(-2/\pi,-2/\pi)\,\mathrm{m/s^2}$$ The endpoint velocities are $(0,2)$ and $(-2,0)$ m/s. Thus $\vec a_{\mathrm{av}}=(-2/\pi,-2/\pi)$ m/s$^2$, magnitude $2\sqrt2/\pi\approx0.9003$ m/s$^2$. It points inward at the midpoint of the quarter arc.

(c) Explain both reductions.

$$|\vec v_{\mathrm{av}}|=4\sqrt2/\pi<2\,\mathrm{m/s},\quad|\vec a_{\mathrm{av}}|=2\sqrt2/\pi<1\,\mathrm{m/s^2}$$ The instantaneous speed is 2 m/s and acceleration magnitude is $v^2/R=1$ m/s$^2$. Both average magnitudes are smaller, since the vectors change direction during the interval and partly cancel. Averaging vector magnitudes would not produce the magnitudes of these vector averages.