Projectile and Uniform Circular Motion

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Senior 4 · Physics

Projectile and Uniform Circular Motion

5 lessons

Lesson 4 of 5

Centripetal Acceleration and the Forces in Circular Motion

25 min

Let go, and it flies straight

A hammer thrower spins round three or four times, leaning back against the pull of the heavy ball, then lets go. The hammer does not fly outwards, away from the centre of the circle, as many people expect. It flies off along the line it was moving along at the instant of release: the tangent. Before release the hammer was not trying to escape outwards; it was being pulled inwards, all the time, by the wire and the thrower's hands. In S3 you met this inward force and the formula for it. In this lesson you derive the acceleration of circular motion from the change in velocity, find out which real force supplies it in each situation, and extend the idea to bodies that speed up or slow down as they go round.

Why circular motion needs a force

A body moving at constant speed in a circle is accelerating, because its velocity keeps changing direction. By Newton's first law, a body with no resultant force moves in a straight line; so something must keep pulling it towards the centre. To find how large the acceleration is, compare the velocities at two nearby points A and B (Figure 4.1). The speed $v$ is the same at both, but the direction has turned through the same angle $\Delta\theta$ as the radius. Drawn from one point, the two velocity vectors and the change $\Delta\vec{v}$ form an isosceles triangle with apex angle $\Delta\theta$ . For a small angle the base is an arc of radius $v$ , so

|\Delta\vec{v}| \approx v\,\Delta\theta \qquad\Rightarrow\qquad a = \frac{|\Delta\vec{v}|}{\Delta t} = v\,\frac{\Delta\theta}{\Delta t} = v\omega

Using $v = r\omega$ , this gives three equivalent forms:

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

As $\Delta\theta \to 0$ , $\Delta\vec{v}$ becomes perpendicular to the velocity and points straight at the centre. This is the centripetal acceleration, from the Latin for "centre-seeking".

Figure 4.1: The velocity at A and at B, and the change in velocity between them. The change, and so the acceleration, points towards the centre.

Definition

A body moving with speed $v$ in a circle of radius $r$ has a centripetal acceleration $a_c = \frac{v^{2}}{r} = \omega^{2} r$ , directed towards the centre. The resultant force needed to give it this acceleration is the centripetal force:

F_c = m a_c = \frac{m v^{2}}{r} = m\omega^{2} r

It is not a new kind of force: it is the name for the role played by a real force (tension, friction, gravity, a normal reaction, an electric force) or by the resultant of several. Example: a 0.50 kg ball on a string moving at $4.0\,\mathrm{m\,s^{-1}}$ in a horizontal circle of radius 2.0 m needs $F_c = \frac{0.50 \times 4.0^{2}}{2.0} = 4.0\,\mathrm{N}$ towards the centre.

Which force does the job?

The first step in every circular-motion problem is to name the real force, or forces, that point towards the centre. Table 4.1 lists common cases.

Situation	Force that provides the centripetal force
a stone whirled on a string in a horizontal circle	tension in the string
a car or motorcycle on a flat bend	sideways friction of the road on the tyres
the Moon or a satellite in orbit	the gravitational pull of the Earth (S5)
an electron round a nucleus, in Bohr's model	the electric attraction of the nucleus
clothes in a spinning washing-machine drum	the inward push (normal reaction) of the drum wall
a charged particle in a magnetic field	the magnetic force on the moving charge (S5)

Table 4.1: The real forces that act as centripetal forces.

On a flat bend the friction cannot exceed $\mu mg$ , where $μ \mu$ is the coefficient of friction between tyres and road. Setting $\frac{mv^{2}}{r} \le \mu mg$ gives the greatest safe speed,

v_{\max} = \sqrt{\mu g r}

It does not depend on the mass, but it falls sharply when the road is wet, oily or gravelly and $μ \mu$ is small.

Worked example 1: a stone on a string

A stone of mass 0.20 kg is whirled on a smooth horizontal table in a circle of radius 0.80 m, making 3.0 revolutions per second. Find its speed, its acceleration and the tension in the string.

\omega = 2\pi f = 2\pi \times 3.0 = 18.8\,\mathrm{rad\,s^{-1}}, \qquad v = \omega r = 18.8 \times 0.80 = 15.1\,\mathrm{m\,s^{-1}}

a = \omega^{2} r = 18.8^{2} \times 0.80 = 284\,\mathrm{m\,s^{-2}}, \qquad T = ma = 0.20 \times 284 = 57\,\mathrm{N}

The acceleration is about 29 times $g$ : a light stone needs a strong pull when it is spun fast, because $F$ grows as $\omega^{2}$ .

Worked example 2: a bend on a dry and a wet road

A flat bend has radius 50 m. The coefficient of friction between tyres and road is 0.60 when dry and 0.20 when wet. Find the greatest safe speed in each case.

v_{\text{dry}} = \sqrt{0.60 \times 9.8 \times 50} = 17.1\,\mathrm{m\,s^{-1}} \;(62\,\mathrm{km\,h^{-1}}), \qquad v_{\text{wet}} = \sqrt{0.20 \times 9.8 \times 50} = 9.9\,\mathrm{m\,s^{-1}} \;(36\,\mathrm{km\,h^{-1}})

A third of the friction allows only $\sqrt{\frac{1}{3}} = 0.58$ of the speed. A driver who takes the bend at the dry-road speed in the rain needs three times more friction than the road can give, and the car slides off along a line close to the tangent.

Speeding up on a circle: tangential, centripetal and resultant forces

If the speed changes as the body goes round, as for a car accelerating round a curved track or a mill starting up, the acceleration has two perpendicular parts (Figure 4.2):

a tangential acceleration $a_t = r\alpha = \frac{\mathrm{d}v}{\mathrm{d}t}$ , along the tangent, which changes the speed;
a centripetal acceleration $a_c = \frac{v^{2}}{r}$ , towards the centre, which changes the direction.

They are produced by a tangential force $F_{t} = m a_{t}$ and a centripetal force $F_{c} = m a_{c}$ . The resultant acceleration and force are

a = \sqrt{a_t^{2} + a_c^{2}}, \qquad F = ma, \qquad \tan\varphi = \frac{a_t}{a_c}

where $\varphi$ is the angle between the resultant and the radius. In uniform circular motion $a_{t} = 0$ and the resultant force points exactly at the centre.

Figure 4.2: Non-uniform circular motion: the tangential and centripetal accelerations add as vectors.

Worked example 3: a car accelerating round a curve

A car of mass 1200 kg drives round a circular track of radius 100 m. At the moment its speed is $20\,\mathrm{m\,s^{-1}}$ it is speeding up at $1.5\,\mathrm{m\,s^{-2}}$ . Find the resultant horizontal force on the car and its direction.

a_c = \frac{v^{2}}{r} = \frac{20^{2}}{100} = 4.0\,\mathrm{m\,s^{-2}}, \qquad a_t = 1.5\,\mathrm{m\,s^{-2}}

a = \sqrt{4.0^{2} + 1.5^{2}} = 4.27\,\mathrm{m\,s^{-2}}, \qquad F = 1200 \times 4.27 = 5.1 \times 10^{3}\,\mathrm{N}

\tan\varphi = \frac{1.5}{4.0} \Rightarrow \varphi = 20.6^\circ

The resultant force, 5.1 kN, points $20.6^\circ$ forward of the line to the centre. All of it comes from the friction of the road on the tyres, so a driver who accelerates hard in a bend uses up grip that is needed for turning.

Is there a "centrifugal force"?

Passengers in a minibus taking a sharp bend feel pushed towards the outside of the bend. Seen from the road, no outward force acts on them. Their bodies tend to keep moving in a straight line (Newton's first law) while the bus turns beneath them, until the side of the seat or the door pushes them inwards and supplies their centripetal force. The outward "centrifugal force" is felt only by someone who describes the motion from inside the turning bus; in this course, always work from the ground, where the only forces are real pushes and pulls and the resultant points towards the centre.

Warning

Common mistakes. (1) Drawing "centripetal force" as an extra force on a free-body diagram alongside tension, weight and friction: it is the resultant of the real forces, not one more of them. (2) Drawing an outward centrifugal force when working from the ground. (3) Using the diameter instead of the radius in $\frac{v^{2}}{r}$ . (4) Forgetting to convert revolutions per second or per minute into $\mathrm{rad\,s^{-1}}$ before using $\omega^{2} r$ . (5) Saying a body in uniform circular motion has no acceleration because its speed is constant.

Tip

Three steps for any circle problem: (1) draw the real forces; (2) find their resultant component towards the centre; (3) set it equal to $\frac{mv^{2}}{r}$ or $m\omega^{2}r$ . If the speed is changing, add a fourth: the tangential component equals $m a_{t}$ .

I never understood the derivation of centripetal acceleration... until now! (FloatHeadPhysics). Watch how the change in velocity between two nearby points on the circle is drawn as a vector, giving a = v²/r towards the centre, as in Figure 4.1.

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Practice

A car of mass 900 kg goes round a flat bend of radius 40 m at $15\,\mathrm{m\,s^{-1}}$ . Find its centripetal acceleration and the centripetal force, and name the force that provides it.
A ball of mass 0.50 kg on a string is whirled on a smooth table in a horizontal circle of radius 1.2 m at 2.0 revolutions per second. Find the tension in the string.
A car takes a flat bend of radius 30 m at $12\,\mathrm{m\,s^{-1}}$ . What is the smallest coefficient of friction that lets it do so without skidding?
The Moon moves round the Earth in a nearly circular orbit of radius $3.84 \times 10^{8}\,\mathrm{m}$ once every 27.3 days. Find its angular velocity and centripetal acceleration.
A point on the rim of a wheel of radius 0.50 m is moving at $3.0\,\mathrm{m\,s^{-1}}$ and speeding up at $4.0\,\mathrm{m\,s^{-2}}$ along the circle. Find its centripetal acceleration, its resultant acceleration, and the angle between the resultant and the radius.
Using a sketch of the velocity vectors at two nearby points, derive $a = \frac{v^{2}}{r}$ and show that the acceleration points towards the centre.
Explain, without using the idea of a centrifugal force, why a passenger in a minibus slides towards the door on a sharp bend.
A road engineer suggests that drivers should halve their speed on a bend when rain lowers the coefficient of friction from 0.60 to 0.20. Evaluate this advice.

Show answersHide answers

$a = \frac{15^{2}}{40} = 5.6\,\mathrm{m\,s^{-2}}$ ; $F = 900 \times 5.63 = 5.1 \times 10^{3}\,\mathrm{N}$ , provided by sideways friction of the road on the tyres.
$\omega = 2\pi \times 2.0 = 12.6\,\mathrm{rad\,s^{-1}}$ ; $T = m\omega^{2}r = 0.50 \times 12.6^{2} \times 1.2 = 95\,\mathrm{N}$ .
$\mu \ge \frac{v^{2}}{gr} = \frac{12^{2}}{9.8 \times 30} = 0.49$ .
$\omega = \frac{2\pi}{27.3 \times 86\,400} = 2.66 \times 10^{-6}\,\mathrm{rad\,s^{-1}}$ ; $a = \omega^{2}r = (2.66 \times 10^{-6})^{2} \times 3.84 \times 10^{8} = 2.7 \times 10^{-3}\,\mathrm{m\,s^{-2}}$ , towards the Earth.
$a_c = \frac{3.0^{2}}{0.50} = 18\,\mathrm{m\,s^{-2}}$ ; $a = \sqrt{18^{2} + 4.0^{2}} = 18.4\,\mathrm{m\,s^{-2}}$ ; $\varphi = \tan^{-1}\frac{4.0}{18} = 12.5^\circ$ forward of the radius.
As in Figure 4.1: $|\Delta\vec{v}| \approx v\Delta\theta$ , so $a = v\frac{\Delta\theta}{\Delta t} = v\omega = \frac{v^{2}}{r}$ ; as $\Delta\theta \to 0$ , $\Delta\vec{v}$ becomes perpendicular to $\vec{v}$ , along the radius towards the centre.
The friction of a smooth seat is too small to give the passenger the centripetal force needed, so the passenger keeps moving nearly in a straight line while the bus turns; relative to the bus this looks like sliding outwards, until the door pushes the passenger inwards and supplies the centripetal force.
$v_{\max} = \sqrt{\mu g r} \propto \sqrt{\mu}$ : the safe speed falls to $\sqrt{\frac{0.20}{0.60}} = 0.58$ of the dry value. Halving the speed (to 0.50) keeps the driver just inside the limit, so the advice is safe but with little margin; worn tyres, mud or a sharper part of the bend would leave none. A larger reduction is wiser.

Quick check

A car goes round a flat, level bend at a steady speed. What provides its centripetal force?

Key points

Circular motion at constant speed is accelerated motion: $a_c = \frac{v^{2}}{r} = \omega^{2}r$ , towards the centre.
The centripetal force $\frac{mv^{2}}{r}$ is the resultant of real forces (tension, friction, gravity, normal reaction), never an extra force.
On a flat bend $v_{\max} = \sqrt{\mu g r}$ .
If the speed changes, $a_t = r\alpha$ is added at right angles: $a = \sqrt{a_t^{2} + a_c^{2}}$ , $F = m a$ . Released bodies leave along the tangent.

Sources and credits

Photo: Hammer Throw Austin Celtic Festival Texas 2023 by Larry D. Moore, Wikimedia Commons, licensed under CC BY 4.0.
Video: I never understood the derivation of centripetal acceleration... until now!, FloatHeadPhysics, embedded from YouTube.

Written by Ganzaa: the lesson text, derivations, diagrams, table, worked examples and practice exercises.

1Let go, and it flies straight

2Why circular motion needs a force

3Which force does the job?

4Speeding up on a circle: tangential, centripetal and resultant forces

5Is there a "centrifugal force"?

Let go, and it flies straight

Why circular motion needs a force

Which force does the job?

Speeding up on a circle: tangential, centripetal and resultant forces

Is there a "centrifugal force"?