Lesson 4 of 5Dynamics

Unit 1 · Lesson 4 of 5

Uniform Circular Motion

Why an object moving in a circle at constant speed is accelerating toward the centre, the equations that connect speed, radius, period and frequency, and how real forces such as friction, tension, gravity and the normal force keep things on a curved path.

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Key things to remember

1Uniform circular motion has constant speed but changing velocity, so there is an acceleration toward the centre.
2 $a_{c} = \frac{v ^{2}}{r} = \frac{4 π ^{2} r}{T ^{2}} = 4 π^{2} r f^{2},$ with $v = \frac{2 π r}{T}$ and $f = \frac{1}{T}$ .
3 $F_{c} = \frac{m v ^{2}}{r}$ is supplied by real forces such as friction, tension, gravity or the normal force; it is never an extra force.
4Flat curve: $v_{ma x} = μ_{s} g r$ .
5Vertical circle: the minimum speed at the top is $v = g r$ ; tension is greatest at the bottom.

A car taking a highway ramp, a satellite circling Earth, clothes pressed against a spinning washer drum: all move along circles. Uniform circular motion means moving in a circle at constant speed. It sounds like the simplest possible motion, yet it always involves an acceleration and always needs a net force. Understanding why is the key to this whole topic.

Constant speed, changing velocity

Fig 1Velocity is tangent to the circle; acceleration and net force point to the centre. If the force stops, the object continues in a straight line along the tangent.

Velocity is a vector, so a change in direction is a change in velocity, even when the speed stays the same. Take two points a short distance $Δ s$ apart on the circle. The two velocity vectors have the same length $v$ but point in slightly different directions, and the change $Δ v$ between them points toward the centre. The triangle formed by the two velocities has the same shape as the triangle formed by the two radii, so $\frac{Δ v}{v} = \frac{Δ s}{r}$ . Dividing by the time interval, with $Δ s /Δ t = v$ , gives $a = \frac{v ^{2}}{r}$ , directed toward the centre. This is centripetal (centre-seeking) acceleration.

a_{c} = \frac{v ^{2}}{r} = \frac{4 π ^{2} r}{T ^{2}} = 4 π^{2} r f^{2} v = \frac{2 π r}{T} f = \frac{1}{T}

T is the period in seconds for one revolution and f is the frequency in hertz. Substitute v equals 2 pi r over T into v squared over r to get the middle form.

Centripetal force is a job, not a new force

Newton's second law gives

F_{c} = m a_{c} = \frac{m v ^{2}}{r}

, but you never draw an arrow labelled 'centripetal force' on a free-body diagram. Instead, ask which real force, or combination of forces, points toward the centre. That net inward force is doing the centripetal job.

Situation	Real force supplying the centripetal force
Car rounding a flat curve	Static friction between the tires and the road
Ball whirled on a string in a horizontal circle	Tension in the string
Satellite in a circular orbit	Gravity from Earth
Rider at the bottom of a roller-coaster dip	Normal force up minus gravity down
Clothes against a spinning washer drum	Normal force from the drum wall

Flat curves

On a level road, the only horizontal force on a turning car is static friction from the road, so it must supply $\frac{m v ^{2}}{r}$ . Friction has a maximum of $μ_{s} m g$ , which sets a maximum safe speed: $μ_{s} m g = \frac{m v _{ma x}^{2}}{r}$ , so $v_{ma x} = μ_{s} g r$ . The mass cancels, but the coefficient does not: on ice $μ_{s}$ is small, so the safe speed drops sharply. Highway ramps are often banked (tilted) so that part of the normal force points toward the centre and less friction is needed.

Vertical circles

Fig 2At the top, gravity and the string or track both point to the centre. At the bottom, they point in opposite directions.

ATop of the circle (centre is below)

Gravity and tension (or the normal force) both point down, toward the centre
$F_{T} + m g = \frac{m v ^{2}}{r}$ , so $F_{T} = \frac{m v ^{2}}{r} - m g$
Tension is smallest here
Riders in a loop feel lightest here

BBottom of the circle (centre is above)

Tension (or the normal force) points up, gravity points down
$F_{T} - m g = \frac{m v ^{2}}{r}$ , so $F_{T} = \frac{m v ^{2}}{r} + m g$
Tension is largest here, which is where a string is most likely to break
Riders feel heaviest here

Minimum speed at the top

The slowest an object can go at the top and still follow the circle is when the string goes slack (

F_{T} = 0

) and gravity alone supplies the centripetal force:

m g = \frac{m v ^{2}}{r}

, so

v_{min} = g r

. Any slower and gravity pulls it inward more than the circle needs, so it falls off the path.

Centrifugal force is an apparent force

In a turning car you feel thrown outward. From the ground, no outward force acts on you: your body tends to keep going straight along the tangent while the car turns inward, and the door pushes you toward the centre. The 'centrifugal force' exists only as an apparent force in the rotating frame of the car. A washer's spin cycle works the same way: the drum wall pushes the clothes around the circle, but water that reaches a hole has no wall there to push it inward, so it continues along the tangent and out through the hole.

Worked examples

Reveal one step at a time and try to beat it to the next line.

Example 1

0.25 kg

puck on a frictionless horizontal table is tied to a

0.80 m

string and completes one revolution every

0.50 s

. Find its speed, its centripetal acceleration and the tension in the string.

Example 2

A car rounds a flat curve of radius

50 m

. The coefficient of static friction between the tires and the dry road is

0.70

. What is the maximum speed at which the car can take the curve without skidding?

Example 3

0.40 kg

ball on a

0.90 m

string is swung in a vertical circle. When it passes the top and the bottom it is moving at

5.0 m/s

. Find the tension at each point, and the minimum speed at the top.

Example 4

70 kg

rider passes through the bottom of a roller-coaster dip with radius

15 m

12 m/s

. What normal force does the seat exert on the rider, and how does it compare with the rider's weight?

Where marks get lost

Drawing a separate 'centripetal force' arrow on a free-body diagram in addition to the real forces. The centripetal force is the net inward result of those real forces.
Saying there is no acceleration because the speed is constant. The direction of the velocity changes, so the object accelerates toward the centre.
Drawing a 'centrifugal force' pointing outward in the ground frame. No such force acts; the object's inertia makes it tend to go straight along the tangent.
Using the same equation at the top and bottom of a vertical circle. At the top gravity helps supply the centripetal force; at the bottom it works against it.
Plugging in the diameter instead of the radius, or a period in minutes instead of seconds.

Flashcards 8

Uniform circular motion

Centripetal acceleration

Period and frequency

Speed in a circle

+ 4 more

Practice

A cyclist rides around a circular track of radius

30 m

at a constant

12 m/s

. What is her centripetal acceleration, in

m/s^{2}

? Give three significant figures.

m/s²

1200 kg

car rounds a curve of radius

45 m

15 m/s

. What centripetal force is needed, in newtons?

A child sits

2.5 m

from the centre of a merry-go-round that turns once every

4.0 s

. What is the child's centripetal acceleration, in

m/s^{2}

? Give three significant figures.

m/s²

What is the minimum speed a roller-coaster car must have at the top of a vertical loop of radius

8.0 m

to stay on the track, in m/s? Give three significant figures.

m/s

70 kg

rider goes over the top of a roller-coaster hill (radius

15 m

, centre below the rider) at

12 m/s

. What normal force does the seat exert, in newtons? Give three significant figures.

What is the maximum speed, in m/s, at which a car can round a flat curve of radius

60 m

if the coefficient of static friction is

0.50

? Give three significant figures.

m/s

A ball is whirled in a horizontal circle on a string. At the instant the string breaks, in which direction does the ball move (seen from above)?

If a car doubles its speed around the same curve, how does the centripetal force required change?

A passenger in a car turning left feels pushed against the right-hand door. What is really happening, as seen from the road?

For a car rounding a flat, unbanked curve, which force supplies the centripetal force?

Match each circular motion to the real force that supplies its centripetal force.

The Moon orbiting Earth

A hammer thrower's hammer

A car on a flat exit ramp

Clothes in a spinning washer drum

Order these objects from largest centripetal acceleration to smallest.

12 m/s on a circle of radius 1 m
26 m/s on a circle of radius 4 m
34 m/s on a circle of radius 1 m
410 m/s on a circle of radius 2 m
510 m/s on a circle of radius 5 m

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