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5th Year OL · Physics
Interaction · Kinematics

Uniform circular motion

The sliotar on the end of a hurler's swinging arm. The chairs on a funfair ride at a Galway fair. The drum of a washing machine. All of them move in a circle and, if they always do so at the same speed, they have a type of motion with its own name: uniform circular motion (UCM). Unlike motion in a straight line at constant velocity, here the velocity, as a vector, changes direction continuously, even though its magnitude (the speed) is constant. Later we will see how a car stays on a bend on the M50: it needs a force towards the centre.

What UCM is

Look at the second hand of a wall clock: it always takes exactly 60 seconds to complete a full turn without slowing down or speeding up, but at every instant it points in a different direction. How can we describe a motion that keeps its speed constant but never stops changing direction? (We answer this below.)

Animation of two points moving round a circle in the same direction and at constant speed, a quarter of a turn apart; the velocity vector of each one rotates with it and is always tangent to the curve, perpendicular to the radius. v r
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The horses on a merry-go-round always rotate at the same speed, describing circles. a) What type of motion do they describe? b) Explain in your own words what the period…

The horses on a merry-go-round always rotate at the same speed, describing circles.

  1. a) What type of motion do they describe?
  2. b) Explain in your own words what the period and frequency of this motion are.
Solve
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Observe the second hand of an analogue clock. a) What is its period, that is, how long does it take to make one complete turn? b) How many turns does it make in one hour?

Observe the second hand of an analogue clock.

  1. a) What is its period, that is, how long does it take to make one complete turn?
  2. b) How many turns does it make in one hour?
Solve
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Two children get on the same merry-go-round: one on an inner horse, $2\ \text{m}$ from the centre, and another on an outer horse, $5\ \text{m}$ from the centre. The…

Two children get on the same merry-go-round: one on an inner horse, $2\ \text{m}$ from the centre, and another on an outer horse, $5\ \text{m}$ from the centre. The diagram is shown in the figure.

2 m inner 5 m outer
  1. a) Do they take the same time to complete one turn? What physical quantity do they share?
  2. b) Which of the two moves faster? Explain using the distance covered by each in the same amount of time.
Solve

Period and frequency

Diagram of a circular path showing the period as the time needed to complete one full turn. one full turn = T seconds
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A wheel completes $50$ turns in $10\ \text{s}$. a) Calculate its period, in seconds. b) Calculate its frequency, in hertz ($\text{Hz}$).

A wheel completes $50$ turns in $10\ \text{s}$.

  1. a) Calculate its period, in seconds.
  2. b) Calculate its frequency, in hertz ($\text{Hz}$).
Solve
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A Ferris wheel takes $40\ \text{s}$ to complete one full turn. a) How many turns does it complete in $10\ \text{min}$? b) How long does it take to make $6$ turns?…

A Ferris wheel takes $40\ \text{s}$ to complete one full turn.

  1. a) How many turns does it complete in $10\ \text{min}$?
  2. b) How long does it take to make $6$ turns? Express the result in minutes.
Solve
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A turntable rotates at $45$ revolutions per minute ($\text{rpm}$). a) Express its frequency in hertz (revolutions per second). b) Calculate its period, rounded to one…

A turntable rotates at $45$ revolutions per minute ($\text{rpm}$).

  1. a) Express its frequency in hertz (revolutions per second).
  2. b) Calculate its period, rounded to one decimal place.
Solve

Linear speed

On a funfair ride in motion, two people take exactly the same number of seconds to complete a turn, but the one on the outer edge feels the wind lashing at them much more strongly than the one nearer the middle. How do we calculate the actual distance each of them covers per second along the curve?

Animation of a rotating platform: two points, one inner and one outer, turn together with the same period; their tangent velocity vectors rotate with them and the one for the outer point is 2.25 times longer, like its radius. inner point (small r): lower v outer point (large r): higher v
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A horse on a merry-go-round rotates at $4\ \text{m}$ from the centre, and the merry-go-round takes $10\ \text{s}$ to make one turn, as shown in the figure. Calculate its…

A horse on a merry-go-round rotates at $4\ \text{m}$ from the centre, and the merry-go-round takes $10\ \text{s}$ to make one turn, as shown in the figure.

r = 4 m v = ? T = 10 s

Calculate its linear speed using $v = \frac{2 \pi r}{T}$. Take $\pi = 3.14$ and round the result to one decimal place.

Solve
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A bicycle travels at $6\ \text{m/s}$ and its wheels have a radius of $0.3\ \text{m}$. a) Calculate the rotational period of the wheels, that is, the time they take to…

A bicycle travels at $6\ \text{m/s}$ and its wheels have a radius of $0.3\ \text{m}$.

  1. a) Calculate the rotational period of the wheels, that is, the time they take to complete one turn. Hint: in one turn, a point on the wheel travels $2 \pi r$.
  2. b) Calculate the rotational frequency, in $\text{Hz}$.

Take $\pi = 3.14$ and round the results to one decimal place.

Solve
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The Earth rotates on its axis with a period of $24\ \text{h}$ and its equatorial radius is $6370\ \text{km}$. The diagram is shown in the figure. a) Calculate the linear…

The Earth rotates on its axis with a period of $24\ \text{h}$ and its equatorial radius is $6370\ \text{km}$. The diagram is shown in the figure.

R = 6370 km v = ? T = 24 h
  1. a) Calculate the linear speed of a point on the equator, in $\text{km/h}$. Take $\pi = 3.14$ and round to the nearest whole number.
  2. b) If this speed is so high, why do we not feel it? Give reasons for your answer.
Solve

Angular velocity

Circle with the radius in two positions: the initial radius and the radius after a time interval, with the angle θ swept out and the angular velocity ω as that angle divided by the time. θ O (axis) position at t₀ position at t r r ω ω = θ / t One turn: θ = 2π rad in T seconds
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A child on a merry-go-round is $3\ \text{m}$ from the centre and moves at $1.5\ \text{m/s}$. a) Calculate the period of the merry-go-round, rounded to one decimal place.…

A child on a merry-go-round is $3\ \text{m}$ from the centre and moves at $1.5\ \text{m/s}$.

  1. a) Calculate the period of the merry-go-round, rounded to one decimal place. Take $\pi = 3.14$.
  2. b) Calculate its angular velocity using $\omega = \frac{v}{r}$ and state its unit ($\text{rad/s}$).
Solve
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A wheel rotates with an angular velocity of $\omega = 2\ \text{rad/s}$. Consider two points: one at $0.2\ \text{m}$ from the centre and another at $0.4\ \text{m}$. The…

A wheel rotates with an angular velocity of $\omega = 2\ \text{rad/s}$. Consider two points: one at $0.2\ \text{m}$ from the centre and another at $0.4\ \text{m}$. The diagram is shown in the figure.

0.2 m 0.4 m v₁ = ? v₂ = ? 2 rad/s
  1. a) Calculate the linear speed of each point using $v = \omega \times r$.
  2. b) Explain why both points have the same angular velocity but different linear speeds.
Solve
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Two friends get on a merry-go-round with a period of $8\ \text{s}$: one at $2\ \text{m}$ from the centre and the other at $5\ \text{m}$. a) Calculate the linear speed of…

Two friends get on a merry-go-round with a period of $8\ \text{s}$: one at $2\ \text{m}$ from the centre and the other at $5\ \text{m}$.

  1. a) Calculate the linear speed of each person. Take $\pi = 3.14$ and round to one decimal place.
  2. b) Calculate the ratio of the two speeds and compare it with the ratio of the radii.
  3. c) Calculate the angular velocity of the merry-go-round using $\omega = \frac{2 \pi}{T}$ and explain why it is the same for both friends even though their linear speeds are different.
Solve

Applications: comparing radii and limiting speeds

Graph of linear speed against radius showing a straight line through the origin whose gradient is equal to the angular velocity omega. rv v = ω·r (straight line through the origin)

Centripetal force

Since the velocity keeps changing direction, the object is accelerating, even at constant speed. The change in velocity over a short time points towards the centre of the circle, so the acceleration points towards the centre too. Its size is $a = \dfrac{v^2}{r}$, or equivalently $a = \omega^2 r$, which follows by substituting $v = \omega r$. So a force towards the centre is needed: the centripetal force, $F = \dfrac{mv^2}{r} = m\omega^2 r$. It is not a new kind of force: it is the resultant of real forces such as tension in a string, friction on a bend or gravity. The greater the speed, the greater the force needed: that is why a car on a bend of a motorway has a limiting speed, beyond which friction cannot supply enough force.

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A microwave turntable has a radius of $0.15\ \text{m}$ and takes $10\ \text{s}$ to complete one revolution. a) Calculate the linear speed of a point on the edge, in…

A microwave turntable has a radius of $0.15\ \text{m}$ and takes $10\ \text{s}$ to complete one revolution.

  1. a) Calculate the linear speed of a point on the edge, in $\text{m/s}$, and then express it in $\text{cm/s}$. Take $\pi = 3.14$.
  2. b) If a plate of food is heated for $2.5\ \text{min}$, how many revolutions does it make?
  3. c) Explain the purpose of having the turntable rotate while heating.
Solve
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A large Ferris wheel with a radius of $60\ \text{m}$ takes $30\ \text{min}$ to complete one full revolution. a) Calculate the linear speed of a capsule, in $\text{m/s}$.…

A large Ferris wheel with a radius of $60\ \text{m}$ takes $30\ \text{min}$ to complete one full revolution.

  1. a) Calculate the linear speed of a capsule, in $\text{m/s}$. Take $\pi = 3.14$ and round to two decimal places.
  2. b) A person walking travels at about $1.4\ \text{m/s}$. How many times faster is the person walking than the capsule?
  3. c) In light of the result, explain why the passengers on the Ferris wheel hardly notice the movement.
Solve
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A ceiling fan rotates at a frequency of $4\ \text{Hz}$. Consider a point A halfway along the blade ($0.3\ \text{m}$ from the axle) and a point B at the tip (…

A ceiling fan rotates at a frequency of $4\ \text{Hz}$. Consider a point A halfway along the blade ($0.3\ \text{m}$ from the axle) and a point B at the tip ($0.6\ \text{m}$).

  1. a) Calculate the linear speed of each point. Take $\pi = 3.14$ and round to one decimal place.
  2. b) For safety reasons, the tip of the blade must not exceed $20\ \text{m/s}$. Calculate the maximum frequency at which the fan can rotate, rounded to one decimal place.
  3. c) Justify, using the relationship between $v$, $\omega$ and $r$, why the critical point is the tip of the blade and not the centre.
Solve

Summary

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