Circular and projectile motion

Velocity is a vector, so it changes whenever either its magnitude or its direction changes; acceleration is the rate of change of that vector. In uniform circular motion the speed is constant, yet the direction turns continuously, so $\Delta \vec v = \vec v_2 - \vec v_1$ points toward the centre and the centripetal acceleration has magnitude $a_c = v^2/r = \omega^2 r$. In projectile motion near the Earth the only acceleration is the free-fall acceleration $\vec g$, directed downward, while the horizontal velocity component stays constant because no horizontal force acts. The horizontal and vertical motions are independent, and each is described by the uniformly accelerated equations you already know: $v = v_0 + at$ and $s = v_0 t + \frac{1}{2}at^2$. A useful reminder: $a = 0$ is impossible both on a circular path (the direction changes there) and on a projectile path (the vertical component changes there).

A ball on a string moves in a circle of radius $r = 0.50$ m with constant speed $v = 2.0$ m/s. Find the centripetal acceleration and state its direction.

A ball is launched from the ground with speed $v_0 = 20$ m/s at an angle $30^\circ$ above the horizontal. Taking $g = 9.8$ m/s$^2$ and neglecting air resistance, find the time of flight, the maximum height and the range.

A stone is thrown horizontally at $15$ m/s from a cliff of height $45$ m. Find the time of fall, the horizontal distance from the foot of the cliff to the landing point, and the magnitude and direction of the stone's velocity at the moment it hits the ground. Take $g = 9.8$ m/s$^2$.

Self-check with a simulation. Open the PhET: Projectile Motion simulation (or Ladybug Revolution for circular motion). Work in the “no air resistance” mode.

1. Fix $v_0 = 20$ m/s and vary the angle from $15^\circ$ to $75^\circ$ in steps of $15^\circ$. Record the range in a table. Confirm that the maximum occurs near $45^\circ$ and explain it with the formula $R = v_0^2 \sin 2\theta / g$.
2. Turn on the velocity vectors. Check that the horizontal arrow never changes, while the vertical component changes linearly: it decreases to zero at the apex of the trajectory and then grows in the opposite (downward) direction, that is, it reverses sign at the top.
3. In the circular-motion mode set a constant angular velocity and verify that doubling the radius doubles the centripetal acceleration $a_c = \omega^2 r$, even though the number of revolutions per second is unchanged.
4. Write a one-paragraph conclusion: which vector changes on a circular path and which one changes on a projectile path.

Safety. All measurements are made in the simulation only. Do not throw objects across roads, from balconies, or toward people; if you do film a ball rolling off a table with a phone, do it on the floor and away from walkways.

To make the vector nature of acceleration visible: to show that even at constant speed the velocity vector still changes, and that on a projectile path only one of its components changes.

A two-part diagram.

Left: a circle of radius $r$ with points $A$, $B$, $C$. At each point an arrow $\vec v$ is drawn along the tangent, all of the same length. Beside it is a vector triangle: $\vec v_1$ and $\vec v_2$ are drawn from a common tail, and the vector $\Delta\vec v = \vec v_2 - \vec v_1$ points toward the centre of the circle, labelled $a_c = v^2/r$.

Right: a parabolic trajectory of an angled launch. At three points (launch, apex, landing) the velocity vector is resolved into a constant horizontal component $v_x$ (equal arrows) and a vertical component $v_y$ that decreases, becomes zero at the apex and then reverses. Below, the single acceleration vector $\vec g$ points vertically downward.

During a free kick a player strikes the ball at $20\ \mathrm{m/s}$ at $30^\circ$ above the ground from pitch level. The velocity splits into $v_x = 20\cos 30^\circ \approx 17\ \mathrm{m/s}$ along the ground and $v_y = 20\sin 30^\circ = 10\ \mathrm{m/s}$ upwards. With $g = 10\ \mathrm{m/s^2}$ the ball climbs for $1.0\ \mathrm{s}$, reaches a peak of $h = v_y^2/(2g) = 5\ \mathrm{m}$ and lands after $2.0\ \mathrm{s}$ about $17 \times 2.0 = 34\ \mathrm{m}$ away — almost at the edge of the penalty area.

This is projectile motion: the horizontal component of the velocity stays constant while the vertical component is uniformly accelerated with $\vec a = \vec g$, so $\vec r(t) = \vec r_0 + \vec v_0 t + \tfrac{1}{2}\vec g t^2.$ The two components are independent — uniform motion along the ground and uniformly accelerated motion vertically add up to a parabola — and the velocity vector $\vec v$ turns continuously along the trajectory.

On the spin cycle a washing machine drum of radius $r = 0.25\ \mathrm{m}$ turns at $1400\ \mathrm{rpm}$. In one second the drum makes $1400/60 \approx 23$ turns, so $\omega = 2\pi \times 23 \approx 147\ \mathrm{rad/s}$. A water droplet on the drum wall at $0.25\ \mathrm{m}$ from the axis moves at $v = \omega r \approx 37\ \mathrm{m/s}$ and has a centripetal acceleration $a = \omega^2 r \approx 5.4\times10^3\ \mathrm{m/s^2}$, about $550$ times $g$. The wall has to push the droplet towards the axis with that acceleration; where a hole stops pushing it, the droplet leaves along the tangent — which is why the water flies out through the holes.

This exercises the circular-motion part: at constant speed $v$ the velocity vector changes direction at every instant, so the acceleration $a_c = v^2/r = \omega^2 r$ points towards the centre and is perpendicular to $\vec v$. The magnitude of the velocity is unchanged while its direction rotates — which is exactly why the acceleration in uniform circular motion is not zero.

A car of mass $m = 1200\ \mathrm{kg}$ takes a bend of radius $R = 80\ \mathrm{m}$ at $20\ \mathrm{m/s}$ ($72\ \mathrm{km/h}$). Its centripetal acceleration is $a_c = v^2/R = 400/80 = 5\ \mathrm{m/s^2}$, so the road must supply a sideways force $F = m a_c = 6000\ \mathrm{N}$ from friction, which needs a friction coefficient of at least $a_c/g \approx 0.51$. On the same bend at $30\ \mathrm{m/s}$ ($108\ \mathrm{km/h}$) the required acceleration is $900/80 = 11.25\ \mathrm{m/s^2}$, larger than $g$ itself, and no dry road surface can hold the car.

The car's velocity is tangent to the circle and its magnitude does not change on a steady bend, yet its direction changes at every point, so the acceleration vector points to the centre with $a_c = v^2/R$. If friction is too small, the car stops following the circle and leaves along the tangent — the straight-line motion that belongs to the projectile half of this concept.