IMAT Physics · Kinematics

Projectile Motion and Uniform Circular Motion

A projectile moves steadily sideways while falling freely; a body going round a circle at constant speed is still accelerating, towards the centre.

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Why this matters

The 2021 paper asked for the speed and acceleration of a ball whirled in a horizontal circle at a given angular velocity. Projectiles have not been asked yet, but they are on the syllabus and follow directly from free fall.

Concept 1 of 3: Projectile motion: horizontal and vertical parts are independent

Throw a ball sideways and drop another from the same height at the same moment: they land together. Gravity acts only downwards, so it changes the vertical motion and leaves the horizontal motion alone. Treat a projectile as two separate motions sharing one clock: steady speed across, free fall down.

Definition

Ignoring air resistance:

Horizontal: no force, so the horizontal velocity $u_x$ is constant: $x = u_x t$ .
Vertical: acceleration $g$ downwards, exactly as in free fall.
The time of flight is set by the vertical motion alone.
Launched at speed $u$ and angle $\theta$ : $u_x = u\cos\theta$ , $u_y = u\sin\theta$ .
At the highest point the vertical velocity is zero, but the body still moves sideways at $u_x$ .
On level ground the range is greatest at a launch angle of $45^\circ$ .

Launch from level ground

t_{\text{flight}} = \frac{2u_y}{g} \qquad R = u_x\, t_{\text{flight}}

$u_x, u_y$ horizontal and vertical components of the launch velocity
$R$ horizontal range on level ground

Worked example

A marble rolls off a table 0.80 m high at 3.0 m/s horizontally. Taking

g = 10\ \text{m/s}^2

, how long is it in the air and how far from the table does it land?

Practice this conceptself-check · 3 quick reps

At the top of its path a projectile is still moving

Only the vertical velocity is zero at the highest point. The horizontal velocity is unchanged, so the speed there equals

u\cos\theta

, not zero.

Concept 2 of 3: Period, frequency and angular velocity in circular motion

A point on a spinning wheel repeats the same journey again and again. The period is the time for one turn and the frequency is how many turns happen each second. Angular velocity measures the angle swept per second; points further from the centre sweep the same angle but cover more distance, so they move faster.

Definition

For uniform circular motion (constant speed round a circle of radius $r$ ):

Period $T$ : time for one revolution, in s. Frequency $f = 1/T$ , in hertz (revolutions per second).
Angular velocity $\omega$ : angle swept per second, in rad/s. One revolution is $2\pi$ rad, so $\omega = 2\pi/T = 2\pi f$ .
Speed along the circle: $v = 2\pi r/T = \omega r$ .
Revolutions per minute (rpm): divide by 60 to get hertz.

Circular motion quantities

f = \frac{1}{T} \qquad \omega = \frac{2\pi}{T} = 2\pi f \qquad v = \omega r

$T$ period, in s
$f$ frequency, in Hz
$\omega$ angular velocity, in rad/s
$r$ radius of the circle, in m

Worked example

A bicycle wheel of radius 0.30 m turns at 5.0 revolutions per second. Find its period, its angular velocity and the speed of a point on the rim.

Practice this conceptself-check · 3 quick reps

Change rpm into hertz before using 2π

Angular velocity in rad/s needs revolutions per SECOND. A value in revolutions per minute must be divided by 60 first, or the answer is 60 times too big.

Concept 3 of 3: Centripetal acceleration: always towards the centre

On a circle the speed can stay constant while the direction keeps turning. A change of direction is a change of velocity, so the body is accelerating. The change always points towards the centre, which is why this acceleration is called centripetal (centre-seeking).

Definition

A body in uniform circular motion has constant speed but a changing velocity, so it accelerates.
The centripetal acceleration points towards the centre of the circle, at right angles to the velocity.
Its size is $a = v^2/r = \omega^2 r$ .
The velocity at any instant is along the tangent to the circle.

Centripetal acceleration

a = \frac{v^2}{r} = \omega^2 r

$v$ speed along the circle, in m/s
$r$ radius, in m
$\omega$ angular velocity, in rad/s

Worked example

A car takes a flat circular bend of radius 80 m at a steady 20 m/s. Find the size and direction of its acceleration.

Practice this conceptself-check · 3 quick reps

Constant speed in a circle still means acceleration

Uniform circular motion has a constant speed but not a constant velocity. The acceleration

v^2/r

points towards the centre. An option giving zero acceleration for a body moving steadily round a circle is wrong.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Projectile motion: horizontal and vertical parts are independent
Launch from level ground
$t_{\text{flight}} = \frac{2u_y}{g} \qquad R = u_x\, t_{\text{flight}}$
Period, frequency and angular velocity in circular motion
Circular motion quantities
$f = \frac{1}{T} \qquad \omega = \frac{2\pi}{T} = 2\pi f \qquad v = \omega r$
Centripetal acceleration: always towards the centre
Centripetal acceleration
$a = \frac{v^2}{r} = \omega^2 r$

Watch out for (3)

At the top of its path a projectile is still moving
→ Projectile motion: horizontal and vertical parts are independent
Change rpm into hertz before using 2π
→ Period, frequency and angular velocity in circular motion
Constant speed in a circle still means acceleration
→ Centripetal acceleration: always towards the centre

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