If the speed is constant, why is there an acceleration?

Acceleration is rate of change of velocity, and velocity includes direction. Moving in a circle constantly changes direction, so there is an acceleration towards the centre.

What provides the centripetal force?

Whatever real force points to the centre in that situation — tension, friction, gravity, a normal reaction, or a combination. Centripetal force is a description of the resultant, not a separate force.

Why do you feel pushed outwards on a roundabout?

Your body tends to continue in a straight line due to inertia. The roundabout pushes you inwards to change your direction; there is no actual outward force on you.

Why does centripetal force do no work?

It acts perpendicular to the velocity, so there is no component of force along the direction of motion and the speed does not change.

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103–107MechanicsStretch

Circular Motion

An object moving in a circle at constant speed is still $\textbf{accelerating}$ , because its direction — and therefore its velocity — changes continuously. That acceleration points towards the centre, and it requires a resultant force towards the centre: the $\textbf{centripetal force}$ .

40 min Topic 6 · Further Mechanics

What you'll be able to do

Measure angular displacement in radians and use $\omega = v/r$
Explain why uniform circular motion involves acceleration
Use $a = v^{2}/r = r\omega^{2}$ and $F = mv^{2}/r$
Identify the source of centripetal force in horizontal and vertical circles

Angular quantities

Angles in circular motion are measured in $\textbf{radians}$ : an angle in radians is arc length divided by radius, so a full circle is $2\pi$ rad.

$\textbf{Angular velocity}$ $\omega$ is the angle swept per second, in $\mathrm{rad\,s^{-1}}$ . For one revolution in time $T$ , $\omega = 2\pi/T = 2\pi f$ .

Linear speed and angular velocity are linked by $v = r\omega$ : points further from the centre move faster for the same rotation rate.

\omega = \dfrac{2\pi}{T} = 2\pi f

Angular velocity.

v = r\omega

Linear speed.

Example 1A wheel of radius 0.35 m turns at 120 revolutions per minute. Find

\omega

and the speed of a point on the rim.

f = 120/60 = 2.0

Hz, so

\omega = 2\pi \times 2.0 = 12.6\ \mathrm{rad\,s^{-1}}

v = r\omega = 0.35 \times 12.57

Answer

\omega = 12.6\ \mathrm{rad\,s^{-1}}

;

v = 4.4\ \mathrm{m\,s^{-1}}

Tip — Convert revolutions per minute to hertz by dividing by 60 before multiplying by $2\pi$ .

Why there is an acceleration

Velocity is a vector. Even if the speed is constant, a change of direction is a change of velocity — and a change of velocity over time is an acceleration.

For an object moving in a circle, the velocity at each instant is tangential, and the change in velocity over a short interval points towards the centre. So the $\textbf{centripetal acceleration}$ is directed towards the centre.

Its magnitude is $a = v^{2}/r = r\omega^{2}$ . Faster motion or a tighter circle both demand greater acceleration.

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

Centripetal acceleration, towards the centre.

Centripetal force

By Newton’s second law, a centripetal acceleration needs a resultant force towards the centre: $F = mv^{2}/r = mr\omega^{2}$ .

This is $\textbf{not}$ an additional force to draw on a free-body diagram. It is the resultant of the real forces. For a ball on a string it is the tension; for a car on a flat bend it is friction; for a satellite it is gravity.

If the centripetal force is removed — the string snaps — the object does not fly outwards. It continues in a straight line along the tangent, as Newton’s first law requires.

F = \dfrac{mv^{2}}{r} = mr\omega^{2}

Resultant force needed, towards the centre.

Example 1A 1200 kg car rounds a flat bend of radius 50 m at

15\ \mathrm{m\,s^{-1}}

. Find the friction needed, and the maximum speed if friction can provide at most 6000 N.

F = \dfrac{1200 \times 15^{2}}{50} = 5400

2Maximum:

6000 = \dfrac{1200v^{2}}{50}

, so

v^{2} = 250

Answer5400 N of friction; maximum speed

15.8\ \mathrm{m\,s^{-1}}

Tip — Never draw "centripetal force" as an extra arrow. Draw the real forces, then say which of them (or which combination) provides the centripetal force.

Vertical circles

In a vertical circle, weight acts downwards throughout while the direction to the centre changes, so the forces must be analysed at specific points.

At the $\textbf{top}$ of a loop, both weight and the normal reaction (or tension) point towards the centre: $N + mg = mv^{2}/r$ . The minimum speed to stay on the track is when $N = 0$ , giving $v_{\min} = \sqrt{gr}$ .

At the $\textbf{bottom}$ , the reaction points up towards the centre and weight points away: $N - mg = mv^{2}/r$ . The reaction exceeds the weight, which is why riders feel heavier at the bottom of a dip.

N + mg = \dfrac{mv^{2}}{r}

At the top.

N - mg = \dfrac{mv^{2}}{r}

At the bottom.

v_{\min} = \sqrt{gr}

Minimum speed at the top.

Example 1A rollercoaster loop has radius 8.0 m. Find the minimum speed at the top, and the reaction on a 60 kg rider at the bottom travelling at

18\ \mathrm{m\,s^{-1}}

. Take

g = 9.81

v_{\min} = \sqrt{9.81 \times 8.0} = 8.86\ \mathrm{m\,s^{-1}}

2Bottom:

N = mg + \dfrac{mv^{2}}{r} = 588.6 + \dfrac{60 \times 324}{8.0}

N = 588.6 + 2430 = 3019

Answer

8.9\ \mathrm{m\,s^{-1}}

; about 3000 N, over five times the rider’s weight.

Equation recap

\omega = 2\pi f = v/r

Angular velocity.

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

Centripetal acceleration.

F = mv^{2}/r

Centripetal force.

v_{\min} = \sqrt{gr}

Top of a vertical loop.

Common mistakes to avoid

Saying an object in uniform circular motion is not accelerating.

Its velocity direction changes, so it accelerates towards the centre.

Drawing an outward "centrifugal force".

There is no outward force; the resultant is inward.

Claiming a released object flies outwards radially.

It continues along the tangent.

Using the same force equation at the top and bottom of a loop.

Weight points towards the centre at the top and away at the bottom.

Key takeaways

Angular velocity $\omega = 2\pi f = v/r$ , measured in rad s⁻¹.
Uniform circular motion is accelerated motion, directed to the centre.
$a = v^{2}/r$ and $F = mv^{2}/r$ , provided by real forces.
Released objects move along the tangent; there is no outward force.
In vertical circles, weight helps at the top and opposes at the bottom; $v_{\min} = \sqrt{gr}$ .

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