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Physics · Secondary School · Kinematics

Circular Motion

A fan, a Ferris wheel, the hand of a clock and the Earth going around the Sun have something in common, since everything that spins repeats the same path over and over. We will measure that spinning, see why whoever is on the edge moves faster and find out that turning a corner takes acceleration, even when the speedometer doesn't move, which may surprise you at first.

  1. 1Period and frequency
  2. 2The radian
  3. 3Angular × linear speed
  4. 4Tangential velocity
  5. 5Centripetal acceleration
  6. 6Centripetal force
  7. 7Pulleys and gears
  8. 8Sine and cosine
  9. 9Accelerated circular motion
  10. 10Globe of death
  11. 11Kepler's laws
  12. ✓Challenges
STEP 1

How long does one turn take?

We call the time for one complete turn the period (T), and the number of turns that happen in one second the frequency (f). Each is the inverse of the other, so if each turn takes 0.5 s, 2 turns fit into 1 s.

\(T = \dfrac{1}{f}\)\(f = \dfrac{1}{T}\)T in seconds (s) · f in hertz (Hz = turns per second) · rpm = revolutions per minute = f × 60

Let's discuss

  • Pause halfway, count how many turns the disc made in 10 s and divide by 10. Does the result match f?
  • If we double the frequency, what happens to the period?
  • How many turns per second does a fan at 1200 rpm make?
Period T
rpm
Time
Turns
Exercises for step 1 10 questions · 4 basic · 4 intermediate · 2 challenge
  1. basic

    A fan makes 20 complete turns in 4 s. What are the frequency and the period of the fan?

    Show solution
    \(f = \dfrac{\text{no. of turns}}{\text{time}} = \dfrac{20}{4} = 5\ \text{Hz}\)
    \(T = \dfrac{1}{f} = \dfrac{1}{5} = 0.2\ \text{s}\)
    \(f = 5\ \text{Hz}\) and \(T = 0.2\ \text{s}\)
  2. basic

    A Ferris wheel completes one turn every 40 s. What is its frequency, in Hz?

    Show solution
    \(f = \dfrac{1}{T} = \dfrac{1}{40}\)
    \(f = 0.025\ \text{Hz}\)
  3. basic

    The blade of a blender spins at 3000 rpm. What are the frequency in Hz and the period?

    Show solution
    \(f = \dfrac{3000\ \text{turns}}{60\ \text{s}} = 50\ \text{Hz}\)
    \(T = \dfrac{1}{f} = \dfrac{1}{50} = 0.02\ \text{s}\)
    \(f = 50\ \text{Hz}\) and \(T = 0.02\ \text{s}\)
  4. basic

    Conceptual: if the rotation frequency of a wheel doubles, what happens to its period?

    Show solution
    Since \(T = \dfrac{1}{f}\), period and frequency are inversely proportional.
    Making twice as many turns per second means spending half the time on each turn.
    The period is cut in half.
  5. intermediate

    A vinyl LP spins at 33⅓ rpm. Calculate its frequency in Hz and the time for one turn.

    Show solution
    \(f = \dfrac{33.33}{60} \approx 0.556\ \text{Hz}\)
    \(T = \dfrac{1}{f} = \dfrac{60}{33.33} = 1.8\ \text{s}\)
    \(f \approx 0.56\ \text{Hz}\) and \(T = 1.8\ \text{s}\)
  6. intermediate

    On the spin cycle, the drum of a washing machine turns at 1200 rpm. How many turns does it make in 5 minutes? What is the period?

    Show solution
    \(\text{Turns} = 1200\ \text{turns/min} \cdot 5\ \text{min} = 6000\)
    \(f = \dfrac{1200}{60} = 20\ \text{Hz} \to T = \dfrac{1}{20} = 0.05\ \text{s}\)
    6000 turns; \(T = 0.05\ \text{s}\)
  7. intermediate

    The minute hand of a clock makes one turn per hour. What is its frequency in Hz? How many turns does it make in a day?

    Show solution
    \(T = 1\ \text{h} = 3600\ \text{s}\)
    \(f = \dfrac{1}{3600} \approx 2.78 \cdot 10^{-4}\ \text{Hz}\)
    In 24 h, the hand makes \(24 \cdot 1\ \text{turn} = 24\ \text{turns}\)
    \(f \approx 2.78 \cdot 10^{-4}\ \text{Hz}\); 24 turns per day
  8. intermediate

    The platter of a hard drive spins at 7200 rpm. What is the period of one turn, in milliseconds? How many turns does it make in 0.5 s?

    Show solution
    \(f = \dfrac{7200}{60} = 120\ \text{Hz}\)
    \(T = \dfrac{1}{120} \approx 0.00833\ \text{s} \approx 8.33\ \text{ms}\)
    Turns in 0.5 s \(= f \cdot \Delta t = 120 \cdot 0.5 = 60\)
    \(T \approx 8.33\ \text{ms}\); 60 turns
  9. challenge

    Fan A spins at a frequency of 15 Hz; fan B has a period of 0.05 s. In 1 minute, how many more turns does B make than A?

    Show solution
    A: \(\text{turns} = 15 \cdot 60 = 900\)
    B: \(f = \dfrac{1}{0.05} = 20\ \text{Hz} \to \text{turns} = 20 \cdot 60 = 1200\)
    \(\text{Difference} = 1200 - 900 = 300\)
    B makes 300 more turns
  10. challenge

    A strobe light flashes 10 times per second, lighting up a fan with one marked blade. The marked blade seems to stand still. What is the lowest possible frequency of the fan, in Hz and in rpm?

    Show solution
    Between two flashes, \(\dfrac{1}{10} = 0.1\ \text{s}\) go by.
    The blade seems to stand still if, in that interval, the fan makes a whole number of turns, and the smallest such case is 1 turn.
    \(T = 0.1\ \text{s} \to f = \dfrac{1}{0.1} = 10\ \text{Hz}\)
    \(\text{rpm} = f \cdot 60 = 600\)
    \(f = 10\ \text{Hz} = 600\ \text{rpm}\)
STEP 2

The radian: measuring angles with the radius itself

Take a piece of string as long as the radius and lay it along the edge of the circle; the angle it ‘hugs’ is what we call 1 radian (≈ 57.3°). If we keep laying radii around the whole circle, a little more than 6 of them fit, exactly 2π ≈ 6.28.

\(\theta = \dfrac{s}{R}\)\(1\ \text{turn} = 2\pi\ \text{rad} = 360^\circ\)s = arc length · R = radius · the angle in radians tells you ‘how many radii’ of arc were covered

Let's discuss

  • Drag the slider until the arc is as long as the orange radius and read how many degrees it shows.
  • How many radians make half a turn, and how many make a quarter turn?
  • Why might physicists prefer radians? (Hint: s = θ·R needs no conversion.)
In radians
In degrees
Arc (in radii)
Exercises for step 2 10 questions · 4 basic · 4 intermediate · 2 challenge
  1. basic

    Convert \(90^\circ\) to radians. Use \(\pi \approx 3.14\).

    Show solution
    \(360^\circ = 2\pi\ \text{rad} \to 1^\circ\) \(= \dfrac{\pi}{180}\ \text{rad}\)
    \(90^\circ = 90 \cdot \dfrac{\pi}{180} = \dfrac{\pi}{2}\ \text{rad}\)
    \(\dfrac{\pi}{2} \approx \dfrac{3.14}{2} = 1.57\ \text{rad}\)
    \(90^\circ = \dfrac{\pi}{2}\ \text{rad} \approx 1.57\ \text{rad}\)
  2. basic

    Convert \(\pi/3\) rad to degrees.

    Show solution
    \(\pi\ \text{rad} = 180^\circ\)
    \(\dfrac{\pi}{3}\ \text{rad} = \dfrac{180^\circ}{3}\)
    \(\dfrac{\pi}{3}\ \text{rad} = 60^\circ\)
  3. basic

    A point on the edge of a wheel of radius 15 cm covers an arc of 30 cm. What angle did the wheel turn through, in radians and in degrees? Use \(\pi \approx 3.14\).

    Show solution
    \(\theta = \dfrac{s}{R} = \dfrac{30\ \text{cm}}{15\ \text{cm}} = 2\ \text{rad}\)
    Converting to degrees, \(2 \cdot \dfrac{180}{3.14} \approx 114.6^\circ\)
    \(\theta = 2\ \text{rad} \approx 115^\circ\)
  4. basic

    Conceptual: why is one complete turn worth \(2\pi\) rad?

    Show solution
    \(\theta = \dfrac{s}{R}\), and in one complete turn the arc covered is the whole circumference, \(s = 2\pi R\).
    \(\theta = \dfrac{2\pi R}{R} = 2\pi\), whatever the radius.
    Because the length of the circumference is \(2\pi\) times the radius.
  5. intermediate

    The minute hand of a wall clock is 12 cm long. How far does its tip travel in 15 minutes? Use \(\pi \approx 3.14\).

    Show solution
    \(15\ \text{min} = \dfrac{1}{4}\) of a turn \(\to \theta = \dfrac{2\pi}{4}\) \(= \dfrac{\pi}{2} \approx 1.57\ \text{rad}\)
    \(s = \theta \cdot R = 1.57 \cdot 0.12\ \text{m}\)
    \(s \approx 0.188\ \text{m}\)
    \(s \approx 18.8\ \text{cm}\)
  6. intermediate

    A bicycle wheel has a radius of 0.35 m and makes 3 complete turns. What angle does it turn through in radians, and how far does a point on the tyre travel? Use \(\pi \approx 3.14\).

    Show solution
    \(\theta = 3 \cdot 2\pi\) \(= 6\pi \approx 6 \cdot 3.14\) \(= 18.84\ \text{rad}\)
    \(s = \theta \cdot R = 18.84 \cdot 0.35\)
    \(s \approx 6.59\ \text{m}\)
    \(\theta \approx 18.8\ \text{rad}\); \(s \approx 6.6\ \text{m}\)
  7. intermediate

    A pizza of radius 20 cm is cut into 8 equal slices. What is the angle of each slice in radians, and the length of the crust of each slice? Use \(\pi \approx 3.14\).

    Show solution
    \(\theta = \dfrac{2\pi}{8} = \dfrac{\pi}{4} \approx 0.785\ \text{rad}\)
    \(s = \theta \cdot R = 0.785 \cdot 0.20\ \text{m}\)
    \(s \approx 0.157\ \text{m}\)
    \(\theta \approx 0.79\ \text{rad}\); crust \(\approx 15.7\ \text{cm}\)
  8. intermediate

    The Earth turns once every 24 h and has a radius of 6400 km. In 6 hours, what angle does it turn through, and what arc does a city on the equator travel? Use \(\pi \approx 3.14\).

    Show solution
    \(6\ \text{h} = \dfrac{6}{24} = \dfrac{1}{4}\) of a turn \(\to \theta = \dfrac{2\pi}{4} \approx 1.57\ \text{rad}\)
    \(s = \theta \cdot R = 1.57 \cdot 6400\ \text{km}\)
    \(s \approx 10\,048\ \text{km}\)
    \(\theta = \dfrac{\pi}{2} \approx 1.57\ \text{rad}\); \(s \approx 1.0 \cdot 10^4\ \text{km}\)
  9. challenge

    The arm of a windscreen wiper is 50 cm long and sweeps an angle of \(120^\circ\). Convert the angle to radians and calculate the arc traced by the tip. Use \(\pi \approx 3.14\).

    Show solution
    \(\theta = 120 \cdot \dfrac{\pi}{180}\) \(= \dfrac{2\pi}{3} \approx 2.09\ \text{rad}\)
    \(s = \theta \cdot R = 2.0933 \cdot 0.50\)
    \(s \approx 1.05\ \text{m}\)
    \(\theta \approx 2.09\ \text{rad}\); \(s \approx 1.05\ \text{m}\)
  10. challenge

    A car drives 157 m around a circular track of radius 50 m. What angle did it sweep, in radians and in degrees? What fraction of a lap did it complete? Use \(\pi \approx 3.14\).

    Show solution
    \(\theta = \dfrac{s}{R} = \dfrac{157}{50} = 3.14\ \text{rad}\)
    In degrees, \(3.14 \cdot \dfrac{180}{3.14} = 180^\circ\)
    As a fraction of a lap, \(\dfrac{180^\circ}{360^\circ} = \dfrac{1}{2}\)
    \(\theta \approx 3.14\ \text{rad} = 180^\circ\): half a lap
STEP 3

On the merry-go-round, everyone turns together, but not at the same speed

Ana and Estela sit on the same radius of the merry-go-round and sweep the same angle in the same time, so we say they have the same angular speed ω. Estela, farther from the centre, covers a longer path on each turn, and her linear speed v is therefore greater.

\(\omega = \dfrac{\Delta\theta}{\Delta t} = \dfrac{2\pi}{T} = 2\pi f\)\(v = \omega\,R\)ω in rad/s · v in m/s · the trails show the distance covered in the last second

Ana   Estela   velocity v

Let's discuss

  • Put Estela at twice Ana's radius and watch what happens to v and to ω.
  • Where on the merry-go-round do you get the most ‘butterflies in your stomach’, and what could explain it?
  • On Earth, who moves faster, someone at the Equator or someone at the North Pole?
Period T
Ana's v
Estela's v
v Estela / v Ana
Exercises for step 3 10 questions · 4 basic · 4 intermediate · 2 challenge
  1. basic

    A fan has a period of 0.2 s. What is its angular speed? Use \(\pi \approx 3.14\).

    Show solution
    \(\omega = \dfrac{2\pi}{T} = \dfrac{2 \cdot 3.14}{0.2}\)
    \(\omega = 31.4\ \text{rad/s}\)
    \(\omega \approx 31.4\ \text{rad/s}\)
  2. basic

    A merry-go-round turns with \(\omega = 0.5\ \text{rad/s}\). What is the speed of a child sitting 4 m from the centre?

    Show solution
    \(v = \omega \cdot R = 0.5 \cdot 4\)
    \(v = 2\ \text{m/s}\)
  3. basic

    A Ferris wheel of radius 20 m makes one turn every 60 s. What is the speed of a seat? Use \(\pi \approx 3.14\).

    Show solution
    \(v = \dfrac{2\pi R}{T} = \dfrac{2 \cdot 3.14 \cdot 20}{60}\)
    \(v = \dfrac{125.6}{60} \approx 2.09\ \text{m/s}\)
    \(v \approx 2.1\ \text{m/s}\)
  4. basic

    Conceptual: on a merry-go-round, Ana is at the edge and Estela is near the centre. Who has the greater angular speed? And the greater linear speed?

    Show solution
    Both make one turn in the same time, so they have the same \(\omega\).
    Since \(v = \omega \cdot R\), whoever is farther from the centre goes around a bigger circle in the same time.
    Same \(\omega\); Ana (at the edge) has the greater \(v\).
  5. intermediate

    The Earth turns once every 24 h and has a radius of 6400 km. Calculate the Earth's angular speed and the speed of a point on the equator.

    Show solution
    \(T = 24 \cdot 3600 = 86\,400\ \text{s}\)
    \(\omega = \dfrac{2\pi}{T} \approx 7.27 \cdot 10^{-5}\ \text{rad/s}\)
    \(v = \omega \cdot R = 7.27 \cdot 10^{-5} \cdot 6.4 \cdot 10^{6}\ \text{m}\)
    \(v \approx 465\ \text{m/s}\)
    \(\omega \approx 7.27 \cdot 10^{-5}\ \text{rad/s}\); \(v \approx 470\ \text{m/s}\) (\(\approx 1680\ \text{km/h}\))
  6. intermediate

    The second hand of a clock is 10 cm long. What is its angular speed and the speed of its tip? Use \(\pi \approx 3.14\).

    Show solution
    \(T = 60\ \text{s} \to \omega\) \(= \dfrac{2\pi}{60}\) \(= \dfrac{6.28}{60} \approx 0.105\ \text{rad/s}\)
    \(v = \omega \cdot R = 0.1047 \cdot 0.10\)
    \(v \approx 0.0105\ \text{m/s}\)
    \(\omega \approx 0.105\ \text{rad/s}\); \(v \approx 1.05\ \text{cm/s}\)
  7. intermediate

    An LP spins at 33⅓ rpm. Compare the speed of a point 15 cm from the centre with that of a point 5 cm from it. Use \(\pi \approx 3.14\).

    Show solution
    \(f = \dfrac{33.33}{60} \approx 0.556\ \text{Hz}\)
    \(\omega = 2\pi f \approx 3.49\ \text{rad/s}\) (the same for both points)
    \(v_{15} = 3.49 \cdot 0.15 \approx 0.52\ \text{m/s}\)
    \(v_{5} = 3.49 \cdot 0.05 \approx 0.174\ \text{m/s}\)
    \(v_{15} \approx 0.52\ \text{m/s}\) and \(v_{5} \approx 0.17\ \text{m/s}\): the outer point is 3 times faster
  8. intermediate

    A cyclist rides at 7 m/s. The wheel has a radius of 0.35 m. What is the angular speed of the wheel, and how many turns per second does it make? Use \(\pi \approx 3.14\).

    Show solution
    \(v = \omega \cdot R \to \omega\) \(= \dfrac{v}{R}\) \(= \dfrac{7}{0.35}\) \(= 20\ \text{rad/s}\)
    \(f = \dfrac{\omega}{2\pi} = \dfrac{20}{6.28} \approx 3.18\ \text{Hz}\)
    \(\omega = 20\ \text{rad/s}\); \(f \approx 3.2\) turns per second
  9. challenge

    The platter of a hard drive spins at 7200 rpm. What is the speed of a point 4.5 cm from the axis, in m/s and in km/h? Use \(\pi \approx 3.14\).

    Show solution
    \(f = \dfrac{7200}{60} = 120\ \text{Hz}\)
    \(\omega = 2\pi f = 2 \cdot 3.14 \cdot 120 = 753.6\ \text{rad/s}\)
    \(v = \omega \cdot R\) \(= 753.6 \cdot 0.045 \approx 33.9\ \text{m/s}\)
    In km/h: \(33.9 \cdot 3.6 \approx 122\ \text{km/h}\)
    \(v \approx 34\ \text{m/s} \approx 122\ \text{km/h}\)
  10. challenge

    On a spinning disc, point B is 0.30 m farther from the centre than point A, and B's speed is 1.2 m/s greater than A's. What are the angular speed of the disc and its frequency? Use \(\pi \approx 3.14\).

    Show solution
    Both points have the same \(\omega\), so \(v_B - v_A = \omega\,(R_B - R_A)\)
    \(\omega = \dfrac{1.2}{0.30} = 4\ \text{rad/s}\)
    \(f = \dfrac{\omega}{2\pi} = \dfrac{4}{6.28} \approx 0.637\ \text{Hz}\)
    \(\omega = 4\ \text{rad/s}\); \(f \approx 0.64\ \text{Hz}\)
STEP 4

Cut the string: where does the stone go?

A stone spins tied to a string, and before cutting it you should make a guess about where it will go. Then watch.

In circular motion, the velocity is always tangent to the circle (perpendicular to the radius), and the string is what keeps ‘pulling’ the stone towards the centre and bending its path. Without it, the stone goes straight on, by inertia.

Let's discuss

  • Why does water fly off ‘along the tangent’ from a spinning umbrella?
  • In which direction would you expect the sparks from a grinding wheel to fly?
  • In the hammer throw, at what moment does the athlete need to let go?
Pick a guess and then cut the string.
Exercises for step 4 10 questions · 4 basic · 4 intermediate · 2 challenge
  1. basic

    Conceptual: seen from above, you spin a stone tied to a string. At the instant the stone is at the northernmost point of the circle, moving east, the string snaps. Where does it go?

    Show solution
    Without the string, there is no longer any force pulling the stone towards the centre.
    By inertia, it carries on in a straight line with the velocity it had, which is tangent to the circle.
    It goes straight east (along the tangent), not north and not towards the centre.
  2. basic

    Conceptual: when you sharpen a knife on a grinding wheel, in which direction do the sparks fly relative to the wheel?

    Show solution
    The sparks are hot particles that break off from the edge of the wheel.
    At the instant they break off, nothing holds them on the circle any more, so they carry on with the velocity they had.
    They fly off along the tangent to the wheel, perpendicular to the radius at that point.
  3. basic

    Conceptual: you spin a wet umbrella. Do the drops get ‘flung’ outwards along the radius, or along the tangent? Why?

    Show solution
    While stuck to the fabric, the drop moves in a circle and its velocity is tangent to the edge.
    When adhesion can no longer pull it towards the centre, it carries on in a straight line with that velocity.
    Along the tangent: the drop simply stops following the curve.
  4. basic

    A stone spins in a circle of radius 0.8 m and completes one turn every 0.5 s. With what speed does it fly off if the string is released? Use \(\pi \approx 3.14\).

    Show solution
    When released, it leaves with the same speed it had on the circle.
    \(v = \dfrac{2\pi R}{T} = \dfrac{2 \cdot 3.14 \cdot 0.8}{0.5}\)
    \(v \approx 10.05\ \text{m/s}\)
    \(v \approx 10\ \text{m/s}\), tangent to the circle
  5. intermediate

    In the hammer throw, the athlete swings the ball in a circle of radius 2 m, making 2 turns per second. With what speed does the hammer leave? At what moment should the athlete let go? Use \(\pi \approx 3.14\).

    Show solution
    \(v = 2\pi R f = 2 \cdot 3.14 \cdot 2 \cdot 2\)
    \(v \approx 25.1\ \text{m/s}\)
    The hammer leaves along the tangent, so the athlete should let go when the velocity (perpendicular to the wire) points towards the field.
    \(v \approx 25\ \text{m/s}\); let go when the wire is perpendicular to the direction of the field
  6. intermediate

    Conceptual: on the spin cycle, the drum of a washing machine turns fast. Why does the water leave the clothes while the clothes stay pressed against the wall?

    Show solution
    The wall of the drum pushes the clothes towards the centre, keeping them in circular motion.
    At the holes there is no wall, and nothing pushes the water towards the centre there, so it carries on in a straight line along the tangent and out through the holes.
    The water escapes along the tangent through the holes; the clothes stay, because the wall holds them.
  7. intermediate

    A car at 72 km/h is going round a bend and drives over an oil patch, losing friction for 2 s. Where does it go and how far does it travel in that time?

    Show solution
    Without friction, there is no force towards the centre of the bend, and the car goes straight on along the tangent.
    \(v = \dfrac{72}{3.6} = 20\ \text{m/s}\)
    \(d = v \cdot t = 20 \cdot 2 = 40\ \text{m}\)
    It goes in a straight line along the tangent and travels 40 m
  8. intermediate

    On a frictionless air table, a puck tied to a thread moves in a circle of radius 0.5 m, making 2 turns per second. The thread snaps. How far does the puck travel in 0.3 s? Use \(\pi \approx 3.14\).

    Show solution
    \(v = 2\pi R f = 2 \cdot 3.14 \cdot 0.5 \cdot 2 = 6.28\ \text{m/s}\)
    Without the thread, it moves in a straight line at that speed.
    \(d = v \cdot t = 6.28 \cdot 0.3 \approx 1.88\ \text{m}\)
    \(d \approx 1.9\ \text{m}\), in a straight line along the tangent
  9. challenge

    On an air table, a puck circles point O in a circle of radius 1 m, at 2 m/s. The thread snaps when the puck is 1 m to the right of O, moving ‘up’ (on the screen). How far from O will the puck be 1 s later?

    Show solution
    It leaves along the tangent, going straight ‘up’, perpendicular to the radius.
    In 1 s it travels \(d = 2 \cdot 1 = 2\ \text{m}\).
    This forms a right triangle with legs 1 m (the radius) and 2 m.
    \(\text{Distance} = \sqrt{1^2 + 2^2}\) \(= \sqrt{5} \approx 2.24\ \text{m}\)
    \(\approx 2.2\ \text{m}\) from O
  10. challenge

    A bicycle is held on a stand, with its rear wheel (radius 0.35 m) spinning in the air at 20 rad/s. A drop of mud comes off the top of the wheel, 0.90 m above the ground. How far away horizontally does it land? Use \(g = 10\ \text{m/s}^2\).

    Show solution
    At the top, the velocity is horizontal (tangent), with \(v = \omega \cdot R = 20 \cdot 0.35 = 7\ \text{m/s}\)
    The fall of 0.90 m takes \(t = \sqrt{\dfrac{2h}{g}}\) \(= \sqrt{\dfrac{2 \cdot 0.90}{10}} \approx 0.424\ \text{s}\)
    \(x = v \cdot t = 7 \cdot 0.424 \approx 2.97\ \text{m}\)
    \(x \approx 3.0\ \text{m}\)
STEP 5

The speedometer stays still, and yet there is acceleration

Velocity is a vector, with a size and a direction. In uniform circular motion (UCM) the size stays the same while the direction changes all the time, and changing the velocity, in any way, is accelerating.

We call this acceleration, which always points towards the centre, the centripetal acceleration. In a car, the friction between the tyres and the road is what supplies it.

When you turn on Step by step, time is chopped into pieces Δt. During each piece the car moves straight, and at the end of it the acceleration gives a little ‘nudge’ Δv = a·Δt pointing towards the centre, which, added to the old velocity, turns the velocity without changing its size.

It is worth looking carefully at the drawing. In the sum vafter = vbefore + Δv, the Δv arrow sits at the tip of vbefore (the rule for adding vectors), which is why it may seem ‘out of place’. The same Δv, drawn starting from the car, points exactly towards the centre, and that is the direction of the acceleration.

\(\vec a = \dfrac{\Delta \vec v}{\Delta t}\)\(\vec v_{\text{after}} = \vec v_{\text{before}} + \Delta \vec v\)\(a_c = \dfrac{v^2}{R} = \omega^2 R\)ac in m/s² · doubling v quadruples ac · doubling R cuts ac in half

velocity v (tangent)   centripetal acceleration (towards the centre)

Let's discuss

  • In step-by-step mode, use ‘Advance 1 step’ and check, at several positions, where each Δv points.
  • With 6 steps per turn the path becomes a hexagon. As we increase the steps, what happens to the path, and how does Δv/Δt compare with v²/R?
  • Double the speed and check whether ac got 2× or 4× bigger.
  • Why do you think bends on fast roads are so wide (large R)?
  • If the road is wet and friction can't keep up, what does the car do? (Go back to step 4!)
ac from 0 to 30 m/s²| = 1 g (9.8 m/s²)
v in km/h
ω
ac
ac in ‘g’
Exercises for step 5 10 questions · 4 basic · 4 intermediate · 2 challenge
  1. basic

    A car goes round a bend of radius 50 m at 10 m/s, with constant speed. What is its centripetal acceleration?

    Show solution
    \(a_c = \dfrac{v^2}{R} = \dfrac{10^2}{50} = \dfrac{100}{50}\)
    \(a_c = 2\ \text{m/s}^2\), pointing towards the centre of the bend
  2. basic

    A stone spins at \(\omega = 4\ \text{rad/s}\) in a circle of radius 0.5 m. What is the centripetal acceleration?

    Show solution
    \(a_c = \omega^2 \cdot R = 4^2 \cdot 0.5 = 16 \cdot 0.5\)
    \(a_c = 8\ \text{m/s}^2\)
  3. basic

    Conceptual: in UCM the speedometer always shows the same value. Is there acceleration anyway? Where does it point?

    Show solution
    Acceleration is \(a = \dfrac{\Delta v}{\Delta t}\), and \(v\) is a vector.
    In UCM the magnitude of \(v\) stays the same while its direction changes all the time.
    Yes, there is: it is the centripetal acceleration, which points towards the centre.
  4. basic

    On a bend, a car has a centripetal acceleration of \(2\ \text{m/s}^2\). If it takes the same bend at twice the speed, what will the centripetal acceleration be?

    Show solution
    Since \(a_c = \dfrac{v^2}{R}\), the acceleration depends on the square of \(v\).
    Doubling \(v \to (2)^2 = 4\) times bigger.
    \(a_c = 4 \cdot 2\)
    \(a_c = 8\ \text{m/s}^2\)
  5. intermediate

    The Earth turns once every 24 h and has a radius of 6400 km. What is the centripetal acceleration of a person on the equator? Compare it with \(g = 10\ \text{m/s}^2\).

    Show solution
    \(\omega = \dfrac{2\pi}{86\,400} \approx 7.27 \cdot 10^{-5}\ \text{rad/s}\)
    \(a_c = \omega^2 \cdot R\) \(= (7.27 \cdot 10^{-5})^2 \cdot 6.4 \cdot 10^{6}\)
    \(a_c \approx 0.0338\ \text{m/s}^2\)
    \(\dfrac{a_c}{g} \approx 0.00338 \approx 0.34\%\)
    \(a_c \approx 0.034\ \text{m/s}^2\), about 0.3% of \(g\)
  6. intermediate

    The drum of a washing machine has a radius of 0.25 m and spins at 1200 rpm. What is the centripetal acceleration of the clothes stuck to the wall? How many times bigger than g is that? Use \(\pi \approx 3.14\). Use \(g = 10\ \text{m/s}^2\).

    Show solution
    \(f = \dfrac{1200}{60}\) \(= 20\ \text{Hz} \to \omega\) \(= 2\pi f\) \(= 2 \cdot 3.14 \cdot 20\) \(= 125.6\ \text{rad/s}\)
    \(a_c = \omega^2 \cdot R\) \(= 125.6^2 \cdot 0.25 \approx 3944\ \text{m/s}^2\)
    \(\dfrac{a_c}{g} = \dfrac{3944}{10} \approx 394\)
    \(a_c \approx 3.9 \cdot 10^3\ \text{m/s}^2\), about 390 times \(g\)
  7. intermediate

    A Ferris wheel of radius 20 m makes one turn every 60 s. What is the centripetal acceleration of a passenger? Compare it with \(g\). Use \(\pi \approx 3.14\). Use \(g = 10\ \text{m/s}^2\).

    Show solution
    \(v = \dfrac{2\pi R}{T} = \dfrac{2 \cdot 3.14 \cdot 20}{60} \approx 2.093\ \text{m/s}\)
    \(a_c = \dfrac{v^2}{R}\) \(= \dfrac{2.093^2}{20} \approx 0.219\ \text{m/s}^2\)
    \(\dfrac{a_c}{g} \approx 0.0219 \approx 2.2\%\)
    \(a_c \approx 0.22\ \text{m/s}^2\), only about 2% of \(g\)
  8. intermediate

    A car at 20 m/s is heading north and, 4 s later, is heading east at 20 m/s, after a quarter turn. What is the magnitude of \(\Delta v\) and of the average acceleration? Where does \(\Delta v\) point?

    Show solution
    In \(\Delta v = v_{\text{final}} - v_{\text{initial}}\), the two velocities are perpendicular.
    \(|\Delta v| = \sqrt{20^2 + 20^2}\) \(= 20\sqrt{2} \approx 28.3\ \text{m/s}\)
    \(a_{\text{avg}} = \dfrac{|\Delta v|}{\Delta t}\) \(= \dfrac{28.3}{4} \approx 7.07\ \text{m/s}^2\)
    \(\Delta v\) points southeast, that is, towards the centre of the bend (at the middle of the stretch).
    \(|\Delta v| \approx 28\ \text{m/s}\); \(a_{\text{avg}} \approx 7.1\ \text{m/s}^2\), pointing towards the centre
  9. challenge

    An object in UCM has a speed of 10 m/s. In 0.05 s, its velocity turns through a small angle of 0.1 rad. Estimate \(|\Delta v|\), the acceleration and the radius of the circle. Where does \(\Delta v\) point?

    Show solution
    For a small angle, \(|\Delta v| \approx v \cdot \Delta\theta = 10 \cdot 0.1\) \(= 1\ \text{m/s}\)
    \(a = \dfrac{|\Delta v|}{\Delta t} = \dfrac{1}{0.05} = 20\ \text{m/s}^2\)
    \(\omega = \dfrac{\Delta\theta}{\Delta t}\) \(= \dfrac{0.1}{0.05}\) \(= 2\ \text{rad/s} \to R\) \(= \dfrac{v}{\omega}\) \(= \dfrac{10}{2}\) \(= 5\ \text{m}\)
    As a check, \(\dfrac{v^2}{R} = \dfrac{100}{5} = 20\ \text{m/s}^2\) ✓
    \(\Delta v\) is almost perpendicular to both velocities and points towards the centre.
    \(|\Delta v| \approx 1\ \text{m/s}\); \(a = 20\ \text{m/s}^2\); \(R = 5\ \text{m}\); \(\Delta v\) points towards the centre
  10. challenge

    For passengers to stay comfortable, a bus's centripetal acceleration should be at most \(0.5\,g\). What is the highest speed on a bend of radius 80 m? If the driver goes in at twice that speed, what will the acceleration be in units of \(g\)? Use \(g = 10\ \text{m/s}^2\).

    Show solution
    \(a_{\text{max}} = 0.5 \cdot 10 = 5\ \text{m/s}^2\)
    \(\dfrac{v^2}{R} = 5 \to v\) \(= \sqrt{5 \cdot 80}\) \(= \sqrt{400}\) \(= 20\ \text{m/s}\) \(= 72\ \text{km/h}\)
    At twice the speed (\(40\ \text{m/s}\)), \(a_c = \dfrac{40^2}{80} = 20\ \text{m/s}^2 = 2g\)
    \(v_{\text{max}} = 20\ \text{m/s}\) (\(72\ \text{km/h}\)); at twice the speed, \(a_c = 2g\) (four times bigger)
STEP 6

Who pulls towards the centre? The centripetal force

By Newton's 2nd law, every acceleration needs a force, F = m·a, so if circular motion has an acceleration towards the centre, some force must point towards the centre too.

Centripetal force is not a new force, and the name only describes the role that some real force plays, whether friction for the car, tension in the string or gravity for the satellite, in which case it is given by the law of gravitation, F = G·M·m/r². If that force can't keep up, the object leaves along the tangent, as in step 4.

\(F_c = m\,a_c = \dfrac{m\,v^2}{R}\)F in newtons (N) · m in kg · each situation has a limit: maximum friction (μ·m·g), the strength of the string, the strength of gravity

velocity v   force towards the centre

Let's discuss

  • For the car, increase v until it skids, and then check whether it skids sooner or later on a wet road.
  • For the stone, increase v until the string breaks. If the string could hold twice as much, would the maximum speed double? (Hint: vmax = √(Tmax·R/m).)
  • For the satellite, change the mass and see whether the force of gravity changes, and whether the orbit does. How would you explain what you see?
  • Launch the satellite at 7.7 km/s, then at 9, 11 and 6 km/s. What seems to be the lowest speed that lets it escape from the Earth?
  • Why doesn't the satellite fall, if gravity pulls on it all the time?
  • Is the ‘centrifugal force’ that throws you outwards on the bus a real force, or is it your body trying to go straight on?
  • Turn on ‘View from inside the car’, where the car is standing still for the passenger. What force do they need to invent to explain that, and what do they see when you make the car skid?
Supplied by
Exercises for step 6 10 questions · 4 basic · 4 intermediate · 2 challenge
  1. basic

    A 2 kg object moves in a circle of radius 1 m at a speed of 4 m/s. What is the centripetal force on it?

    Show solution
    \(F_c = \dfrac{m \cdot v^2}{R}\)
    \(F_c = \dfrac{2 \cdot 4^2}{1} = \dfrac{2 \cdot 16}{1}\)
    \(F_c = 32\ \text{N}\)
  2. basic

    When the bus goes round a bend, you feel ‘thrown’ outwards. Is there a ‘centrifugal force’ pushing you?

    Show solution
    No. Because of inertia, your body tends to keep going in a straight line.
    The bus turns inwards, and the seat or the handrail is what pulls you towards the centre.
    The feeling is inertia, not an outward force.
  3. basic

    A 1,000 kg car takes a flat bend of radius 50 m at 10 m/s. What centripetal force is needed, and which real force supplies it?

    Show solution
    \(F_c = \dfrac{m \cdot v^2}{R} = \dfrac{1000 \cdot 10^2}{50}\)
    \(F_c = \dfrac{100\,000}{50}\)
    \(F_c = 2000\ \text{N}\), supplied by friction between the tyres and the road
  4. basic

    A 0.5 kg stone tied to a string spins in a horizontal plane, in a circle of radius 0.8 m, at 4 m/s. What is the tension in the string (ignore the weight)?

    Show solution
    \(T = \dfrac{m \cdot v^2}{R}\)
    \(T = \dfrac{0.5 \cdot 4^2}{0.8} = \dfrac{8}{0.8}\)
    \(T = 10\ \text{N}\)
  5. intermediate

    What is the maximum speed at which a car can take a flat bend of radius 80 m if the coefficient of friction is \(\mu = 0.5\)? Use \(g = 10\ \text{m/s}^2\).

    Show solution
    At the limit, maximum friction equals the centripetal force, \(\mu \cdot m \cdot g = \dfrac{m \cdot v^2}{R}\)
    \(v = \sqrt{\mu \cdot g \cdot R}\) \(= \sqrt{0.5 \cdot 10 \cdot 80}\) \(= \sqrt{400}\)
    \(v = 20\ \text{m/s} = 72\ \text{km/h}\)
  6. intermediate

    A string can hold at most 50 N. It holds a 0.2 kg stone spinning horizontally with a radius of 1 m. What is the highest speed before the string breaks?

    Show solution
    \(T_{\text{max}} = \dfrac{m \cdot v^2}{R}\)
    \(50 = \dfrac{0.2 \cdot v^2}{1}\)
    \(v^2 = \dfrac{50 \cdot 1}{0.2} = 250\)
    \(v \approx 15.8\ \text{m/s}\) (above that, the string breaks)
  7. intermediate

    A cyclist and bicycle together have a mass of 80 kg and take a flat bend of radius 12 m at 6 m/s. With \(\mu = 0.4\) and \(g = 10\ \text{m/s}^2\), can friction hold the bend?

    Show solution
    The force needed is \(F_c = \dfrac{m \cdot v^2}{R} = \dfrac{80 \cdot 36}{12} = 240\ \text{N}\)
    The maximum friction is \(\mu \cdot m \cdot g = 0.4 \cdot 80 \cdot 10 = 320\ \text{N}\)
    \(240\ \text{N} \lt 320\ \text{N}\)
    Yes, the maximum friction (320 N) is more than the 240 N needed.
  8. intermediate

    Two satellites, one of 500 kg and the other of 1,000 kg, are in the same circular orbit. Which one needs the greater speed?

    Show solution
    \(\dfrac{m \cdot v^2}{r} = \dfrac{G \cdot M \cdot m}{r^2}\)
    The satellite's mass \(m\) appears on both sides and cancels, leaving \(v = \sqrt{\dfrac{G \cdot M}{r}}\).
    Both have the same speed: the satellite's mass does not change the orbit.
  9. challenge

    A satellite moves in a circular orbit of radius \(8.0 \cdot 10^6\ \text{m}\) around the Earth. Given that \(G \cdot M_{\text{Earth}} \approx 4.0 \cdot 10^{14}\ \text{m}^3/\text{s}^2\), calculate its speed and its period. Use \(\pi \approx 3.14\).

    Show solution
    \(\dfrac{m \cdot v^2}{r} = \dfrac{G \cdot M \cdot m}{r^2}\) \(\Rightarrow v = \sqrt{\dfrac{G \cdot M}{r}}\)
    \(v = \sqrt{\dfrac{4.0 \cdot 10^{14}}{8.0 \cdot 10^6}}\) \(= \sqrt{5.0 \cdot 10^7} \approx 7071\ \text{m/s}\)
    \(T = \dfrac{2\pi \cdot r}{v}\) \(= \dfrac{2 \cdot 3.14 \cdot 8.0 \cdot 10^6}{7071} \approx 7105\ \text{s}\)
    \(v \approx 7.1\ \text{km/s}\) and \(T \approx 7105\ \text{s}\) (about 2 h)
  10. challenge

    On a flat bend of radius 45 m, friction is \(\mu = 0.8\) on a dry road and \(\mu = 0.2\) on a wet road. Calculate the maximum speed in both cases (\(g = 10\ \text{m/s}^2\)) and compare.

    Show solution
    Dry: \(v = \sqrt{\mu \cdot g \cdot R}\) \(= \sqrt{0.8 \cdot 10 \cdot 45}\) \(= \sqrt{360} \approx 19\ \text{m/s}\)
    Wet: \(v = \sqrt{0.2 \cdot 10 \cdot 45}\) \(= \sqrt{90} \approx 9.5\ \text{m/s}\)
    The ratio between them is \(\sqrt{\dfrac{0.8}{0.2}} = \sqrt{4} = 2\)
    Dry \(\approx 19\ \text{m/s}\) (68 km/h), wet \(\approx 9.5\ \text{m/s}\) (34 km/h): the safe speed is cut in half.
STEP 7

Pulleys, belts and bicycle gears

When two pulleys are linked by a belt (or a chain, or touching teeth) and we assume the belt doesn't slip, the points on the rims of both pulleys have the same linear speed v, so the smaller pulley has to turn more times.

Pulleys on the same axle turn together, with the same ω and the same f, and in that case the bigger one has the faster rim.

Belt: \(f_A\,R_A = f_B\,R_B\)Same axle: \(f_A = f_B\)On a bicycle, the motion goes pedal → chainring (large) → chain → rear sprocket (small), and the sprocket is on the same axle as the rear wheel

Let's discuss

  • If the chainring has a radius 3× bigger than the sprocket, how many times does 1 pedal turn turn the wheel?
  • Why do we use the bigger sprocket going uphill (‘easy’ gear)?
  • On the same axle, which pulley has the faster rim?
f of A
f of B
v rim A
v rim B
Exercises for step 7 10 questions · 4 basic · 4 intermediate · 2 challenge
  1. basic

    A pulley A of radius 10 cm turns at 1200 rpm and is linked by a belt to a pulley B of radius 30 cm. What is the frequency of B?

    Show solution
    With a belt, both rims have the same speed \(\Rightarrow f_A \cdot R_A = f_B \cdot R_B\)
    \(1200 \cdot 10 = f_B \cdot 30\)
    \(f_B = \dfrac{12\,000}{30}\)
    \(f_B = 400\ \text{rpm}\)
  2. basic

    What is the difference between two pulleys fixed on the same axle and two pulleys linked by a belt?

    Show solution
    On the same axle, they turn together, with the same \(\omega\) and the same \(f\) (the bigger one has the faster rim).
    With a belt, the rims have the same linear speed \(v\) (the smaller one turns faster).
    Shared axle \(\to\) same \(\omega\); belt, chain or teeth \(\to\) same \(v\) at the rim.
  3. basic

    A gear with 12 teeth turns at 90 rpm and drives another with 36 teeth. What is the frequency of the bigger one?

    Show solution
    For meshed teeth, \(f_A \cdot n_A = f_B \cdot n_B\)
    \(90 \cdot 12 = f_B \cdot 36\)
    \(f_B = \dfrac{1080}{36}\)
    \(f_B = 30\ \text{rpm}\)
  4. basic

    On a bicycle, the chainring (at the pedals) has 48 teeth and the rear sprocket (at the wheel) has 16. How many turns does the wheel make for each complete pedal turn?

    Show solution
    The chain links the two, so \(f_{\text{chainring}} \cdot n_{\text{chainring}} = f_{\text{sprocket}} \cdot n_{\text{sprocket}}\)
    \(1 \cdot 48 = f_{\text{sprocket}} \cdot 16\) \(\Rightarrow f_{\text{sprocket}} = 3\)
    The wheel is on the same axle as the sprocket.
    The wheel makes 3 turns per pedal turn
  5. intermediate

    A cyclist pedals at 1 turn per second. The chainring has 44 teeth, the sprocket 22, and the wheel has a radius of 0.35 m. What is the speed of the bicycle? Use \(\pi \approx 3.14\).

    Show solution
    \(f_{\text{sprocket}} = \dfrac{f_{\text{chainring}} \cdot n_{\text{chainring}}}{n_{\text{sprocket}}}\) \(= \dfrac{1 \cdot 44}{22}\) \(= 2\ \text{Hz}\)
    The wheel is on the same axle, so \(f_{\text{wheel}} = 2\ \text{Hz}\)
    \(v = 2\pi \cdot R \cdot f = 2 \cdot 3.14 \cdot 0.35 \cdot 2 \approx 4.4\ \text{m/s}\)
    \(v \approx 4.4\ \text{m/s} \approx 16\ \text{km/h}\)
  6. intermediate

    To climb a steep hill, the cyclist shifts into the ‘easy’ gear: small chainring and large sprocket. Why does pedalling get easier?

    Show solution
    With a small chainring and a large sprocket, each pedal turn turns the wheel fewer times.
    The bicycle goes less far per pedal turn, but the force needed on the pedal is smaller.
    Easy gear: less speed per pedal turn, less force; hard gear: the opposite.
  7. intermediate

    In a blender, the motor pulley (radius 2 cm) turns at 6000 rpm and a belt drives the blade pulley (radius 3 cm). Find the frequency of the blades and the speed of the belt. Use \(\pi \approx 3.14\).

    Show solution
    \(f_B = \dfrac{f_A \cdot R_A}{R_B} = \dfrac{6000 \cdot 2}{3} = 4000\ \text{rpm}\)
    \(f_A = \dfrac{6000}{60} = 100\ \text{Hz}\)
    \(v = 2\pi \cdot R_A \cdot f_A\) \(= 2 \cdot 3.14 \cdot 0.02 \cdot 100\) \(= 12.56\ \text{m/s}\)
    \(f_B = 4000\ \text{rpm}\) and \(v \approx 12.6\ \text{m/s}\)
  8. intermediate

    A pulley A of radius 0.05 m turns with \(\omega = 40\ \text{rad/s}\) and is linked by a belt to pulley B of radius 0.2 m. Calculate the speed of the belt, \(\omega_B\) and \(f_B\). Use \(\pi \approx 3.14\).

    Show solution
    \(v = \omega_A \cdot R_A = 40 \cdot 0.05 = 2\ \text{m/s}\)
    \(\omega_B = \dfrac{v}{R_B} = \dfrac{2}{0.2} = 10\ \text{rad/s}\)
    \(f_B = \dfrac{\omega_B}{2\pi}\) \(= \dfrac{10}{6.28} \approx 1.59\ \text{Hz}\)
    \(v = 2\ \text{m/s}\), \(\omega_B = 10\ \text{rad/s}\), \(f_B \approx 1.59\ \text{Hz}\)
  9. challenge

    In a clock, gear A (10 teeth) turns at 120 rpm and drives B (40 teeth). On the same axle as B is C (10 teeth), which drives D (30 teeth). What is the frequency of D?

    Show solution
    A \(\to\) B (teeth): \(f_B = \dfrac{120 \cdot 10}{40} = 30\ \text{rpm}\)
    B and C on the same axle: \(f_C = 30\ \text{rpm}\)
    C \(\to\) D (teeth): \(f_D = \dfrac{30 \cdot 10}{30} = 10\ \text{rpm}\)
    \(f_D = 10\ \text{rpm}\) (a total reduction of 12 times)
  10. challenge

    A cyclist wants to ride at 18 km/h. The wheel has a radius of 0.3 m, the chainring 42 teeth and the sprocket 14. How many pedal turns per minute must she make? Use \(\pi \approx 3.14\).

    Show solution
    \(v = \dfrac{18}{3.6} = 5\ \text{m/s}\)
    \(f_{\text{wheel}} = \dfrac{v}{2\pi \cdot R}\) \(= \dfrac{5}{6.28 \cdot 0.3} \approx 2.654\ \text{Hz}\)
    \(f_{\text{sprocket}} = f_{\text{wheel}}\); \(f_{\text{chainring}} = f_{\text{sprocket}} \cdot \dfrac{14}{42} \approx 0.885\ \text{Hz}\)
    In rpm, \(0.885 \cdot 60 \approx 53\)
    About 53 pedal turns per minute
STEP 8

Sine and cosine are the shadows of a spin

If we shine a torch sideways at the spinning point, its shadow on the wall only goes up and down, and that vertical shadow is y = R·sin θ. The shadow on the floor, in turn, goes back and forth horizontally, following x = R·cos θ.

Since the angle grows with time (θ = ω·t), each shadow draws a wave, and the two waves have the same shape, with the cosine a quarter turn (π/2) ahead of the sine.

\(x = R\cos(\omega t)\)\(y = R\,\sin(\omega t)\)\(\sin^2\theta + \cos^2\theta = 1\)The last one is Pythagoras' theorem, with the two shadows as the legs and the radius as the hypotenuse of the right triangle in the figure.

Let's discuss

  • Leave only the sine switched on and find where the point is when the wave reaches its maximum, and where it is when the wave crosses zero.
  • Drag θ to 90°, 180° and 270°. What are sin θ and cos θ in each case?
  • With both waves switched on, which one reaches its maximum first?
  • If we double ω, what happens to the length of each wave in time? (Think about the period T.)
  • What else in nature goes up and down like this? Think of springs, pendulums, tides and sound.

sine: vertical shadow   cosine: horizontal shadow

θ
sin θ
cos θ
sin² + cos²
Period T
Exercises for step 8 10 questions · 4 basic · 4 intermediate · 2 challenge
  1. basic

    A point moves in a circle of radius 2 m. When \(\theta = 30^\circ\), what are its \(x\) and \(y\) coordinates? Use \(\sqrt{3} \approx 1.73\).

    Show solution
    \(x = R \cdot \cos 30^\circ\) \(= 2 \cdot \left(\dfrac{\sqrt{3}}{2}\right)\) \(= 2 \cdot 0.865\)
    \(y = R \cdot \sin 30^\circ = 2 \cdot 0.5\)
    \(x \approx 1.73\ \text{m}\) and \(y = 1\ \text{m}\)
  2. basic

    A point moves on a circle of radius 1. Give \(\sin\theta\) and \(\cos\theta\) for \(\theta = 0^\circ\), \(90^\circ\), \(180^\circ\) and \(270^\circ\) and say where the point is.

    Show solution
    \(0^\circ\): \(\cos = 1\), \(\sin = 0 \to\) point on the right \((1, 0)\)
    \(90^\circ\): \(\cos = 0\), \(\sin = 1 \to\) at the top \((0, 1)\)
    \(180^\circ\): \(\cos = -1\), \(\sin = 0 \to\) on the left \((-1, 0)\)
    \(270^\circ\): \(\cos = 0\), \(\sin = -1 \to\) at the bottom \((0, -1)\)
    \(\cos\) is the \(x\) coordinate and \(\sin\) is the \(y\) coordinate of the point.
  3. basic

    A point in UCM with radius 5 m has \(\omega = \pi/2\ \text{rad/s}\) and starts from \(\theta = 0\). Where is it at \(t = 1\ \text{s}\)?

    Show solution
    \(\theta = \omega \cdot t\) \(= \left(\dfrac{\pi}{2}\right) \cdot 1\) \(= \dfrac{\pi}{2}\ \text{rad}\) \(= 90^\circ\)
    \(x = 5 \cdot \cos 90^\circ = 5 \cdot 0 = 0\)
    \(y = 5 \cdot \sin 90^\circ = 5 \cdot 1 = 5\)
    \((x, y) = (0,\ 5\ \text{m})\): at the top of the circle
  4. basic

    Why does \(\sin^2\theta + \cos^2\theta = 1\) hold for any angle?

    Show solution
    On the circle of radius 1, the point forms a right triangle with legs \(\cos\theta\) (horizontal) and \(\sin\theta\) (vertical).
    The hypotenuse is the radius itself, equal to 1.
    By Pythagoras' theorem, \(\cos^2\theta + \sin^2\theta = 1^2\).
    It is Pythagoras' theorem with the radius as the hypotenuse.
  5. intermediate

    A Ferris wheel of radius 10 m makes one turn in 40 s. A seat starts from position \(\theta = 0\) (level with the centre, on the right). At \(t = 5\ \text{s}\), how far is it horizontally and vertically from the centre? Use \(\sqrt{2} \approx 1.41\).

    Show solution
    \(\theta = \left(\dfrac{360^\circ}{40}\right) \cdot 5\) \(= 45^\circ\)
    \(x = 10 \cdot \cos 45^\circ\) \(= 10 \cdot \left(\dfrac{\sqrt{2}}{2}\right) \approx 10 \cdot 0.705\)
    \(y = 10 \cdot \sin 45^\circ \approx 10 \cdot 0.705\)
    \(x \approx 7.05\ \text{m}\) and \(y \approx 7.05\ \text{m}\) (the shadow on the ground is 7.05 m from the centre)
  6. intermediate

    A point in the first quadrant has \(\sin\theta = 0.6\). What is \(\cos\theta\)?

    Show solution
    \(\sin^2\theta + \cos^2\theta = 1\)
    \(\cos^2\theta = 1 - 0.6^2 = 1 - 0.36 = 0.64\)
    \(\cos\theta = \sqrt{0.64}\) (positive in the 1st quadrant)
    \(\cos\theta = 0.8\)
  7. intermediate

    A fan blade makes 5 turns per second. If we plot the height \(y(t)\) of the blade tip, we get a sine wave. What are the period and the frequency of that wave?

    Show solution
    The wave repeats with every complete turn.
    \(T = \dfrac{1}{f} = \dfrac{1}{5}\)
    \(T = 0.2\ \text{s}\) and \(f = 5\ \text{Hz}\) (the same as the spin)
  8. intermediate

    A point spins with period \(T = 8\ \text{s}\). The vertical shadow (sine) reaches its maximum at \(t = 3\ \text{s}\). Given that the cosine is \(\pi/2\) ahead of the sine, when did the horizontal shadow (cosine) reach its maximum?

    Show solution
    \(\dfrac{\pi}{2}\ \text{rad}\) is \(\dfrac{1}{4}\) of a turn \(\to\) equivalent to \(\dfrac{T}{4} = \dfrac{8}{4} = 2\ \text{s}\)
    Being ahead means it happens earlier.
    \(t = 3 - 2\)
    \(t = 1\ \text{s}\)
  9. challenge

    The shadow of a point in UCM (radius 0.2 m, period 2 s) projected on a wall performs simple harmonic motion. What are the amplitude, the period and the maximum speed of the shadow? Where is it fastest? Use \(\pi \approx 3.14\).

    Show solution
    Amplitude \(= R = 0.2\ \text{m}\); period \(= T = 2\ \text{s}\)
    The maximum speed of the shadow equals the speed of the point, \(v = \dfrac{2\pi \cdot R}{T} = \dfrac{2 \cdot 3.14 \cdot 0.2}{2} = 0.628\ \text{m/s}\)
    At the centre, the point's velocity is parallel to the wall; at the ends it is perpendicular (the shadow stops).
    \(A = 0.2\ \text{m}\), \(T = 2\ \text{s}\), \(v_{\text{max}} \approx 0.63\ \text{m/s}\), at the centre of the oscillation (and zero at the ends)
  10. challenge

    A point moves on a radius of 4 m with \(\omega = \pi/6\ \text{rad/s}\), starting from \(\theta = 0\). What are its coordinates at \(t = 4\ \text{s}\)? Use \(\sqrt{3} \approx 1.73\).

    Show solution
    \(\theta = \left(\dfrac{\pi}{6}\right) \cdot 4\) \(= \dfrac{2\pi}{3}\ \text{rad}\) \(= 120^\circ\)
    \(\cos 120^\circ = -\cos 60^\circ = -0.5\); \(\sin 120^\circ = \sin 60^\circ\) \(= \dfrac{\sqrt{3}}{2}\)
    \(x = 4 \cdot (-0.5) = -2\ \text{m}\)
    \(y = 4 \cdot 0.865 \approx 3.46\ \text{m}\)
    \((x, y) \approx (-2\ \text{m};\ 3.46\ \text{m})\), in the 2nd quadrant
STEP 9

Switching on and off: uniformly accelerated circular motion

When a fan is switched on, it doesn't reach top speed all at once, and ω keeps growing; when it is switched off, ω keeps dropping until it stops. If ω always changes at the same rate, we call the motion uniformly accelerated circular motion (UACM), and that rate is the angular acceleration α = Δω/Δt.

The acceleration now has two parts. The tangential at = α·R, along the velocity, changes the size of v, while the centripetal ac = ω²·R keeps changing its direction, and when we add the two, the total acceleration no longer points towards the centre.

\(\omega = \omega_0 + \alpha\,t\)\(\theta = \omega_0\,t + \dfrac{\alpha\,t^2}{2}\)\(\omega^2 = \omega_0^2 + 2\,\alpha\,\Delta\theta\)\(a_t = \alpha\,R\)These are the formulas of uniformly accelerated motion in different clothes, with position s → angle θ, velocity v → ω and acceleration a → α.

v   at tangential   ac centripetal   total a

Let's discuss

  • Right after ‘Speed up’, which part of the acceleration is bigger, and which one is bigger when ω is already high?
  • On the ω × t graph, what does the slope of each stretch represent, and what does the ‘Hold’ stretch look like?
  • Click ‘Brake’ and see which way at points now. How many turns does the wheel make before it stops? You can check the count with ω² = ω₀² + 2·α·Δθ.
  • If we double |α|, does the wheel stop in half the time, and in half the turns?
Time
α now
ω
v at the rim
at
ac
Turns since the last button
Exercises for step 9 10 questions · 4 basic · 4 intermediate · 2 challenge
  1. basic

    A fan starts from rest and, in 6 s, reaches an angular speed of 30 rad/s. What is its average angular acceleration?

    Show solution
    \(\alpha = \dfrac{\Delta\omega}{\Delta t}\)
    \(\alpha = \dfrac{30 - 0}{6}\)
    \(\alpha = 5\ \text{rad/s}^2\)
  2. basic

    A bicycle wheel spins at 20 rad/s when the cyclist brakes, with an angular acceleration of \(-4\ \text{rad/s}^2\). How long does the wheel take to stop?

    Show solution
    \(\omega = \omega_0 + \alpha \cdot t\)
    \(0 = 20 + (-4) \cdot t\)
    \(t = \dfrac{20}{4}\)
    \(t = 5\ \text{s}\)
  3. basic

    A blender is switched on from rest with a constant angular acceleration of \(50\ \text{rad/s}^2\). What is the angular speed of the blades after 4 s?

    Show solution
    \(\omega = \omega_0 + \alpha \cdot t\)
    \(\omega = 0 + 50 \cdot 4\)
    \(\omega = 200\ \text{rad/s}\)
  4. basic

    A centrifuge is speeding up its spin. Which component of the acceleration changes the magnitude of the velocity, and which changes its direction? Does the total acceleration point towards the centre?

    Show solution
    The tangential acceleration (\(a_t = \alpha \cdot R\)) is tangent to the path and changes the magnitude of \(v\).
    The centripetal acceleration (\(a_c = \omega^2 \cdot R\)) points towards the centre and changes the direction of \(v\).
    The total is the vector sum of the two, tilted forward.
    \(a_t\) changes the magnitude, \(a_c\) changes the direction, and the total does NOT point towards the centre.
  5. intermediate

    A centrifuge starts from rest with \(\alpha = 10\ \text{rad/s}^2\). What angle does it sweep in 8 s, and how many turns is that? Use \(\pi \approx 3.14\).

    Show solution
    \(\theta = \omega_0 \cdot t + \dfrac{\alpha \cdot t^2}{2}\)
    \(\theta = 0 + \dfrac{10 \cdot 8^2}{2} = 320\ \text{rad}\)
    \(\text{turns} = \dfrac{\theta}{2\pi}\) \(= \dfrac{320}{6.28} \approx 50.96\)
    \(\theta = 320\ \text{rad} \approx 51\) turns
  6. intermediate

    A fan spinning at 60 rad/s is switched off and stops after making 30 turns. What is its angular acceleration (assumed constant)? Use \(\pi \approx 3.14\).

    Show solution
    \(\Delta\theta = 30 \cdot 2\pi\) \(= 30 \cdot 6.28\) \(= 188.4\ \text{rad}\)
    \(\omega^2 = \omega_0^2 + 2 \cdot \alpha \cdot \Delta\theta\)
    \(0 = 60^2 + 2 \cdot \alpha \cdot 188.4\)
    \(\alpha = -\dfrac{3600}{376.8}\)
    \(\alpha \approx -9.55\ \text{rad/s}^2\) (negative sign: braking)
  7. intermediate

    A disc of radius 0.15 m speeds up with \(\alpha = 4\ \text{rad/s}^2\). At the instant when \(\omega = 2\ \text{rad/s}\), calculate \(a_t\), \(a_c\) and the total acceleration of a point on the rim.

    Show solution
    \(a_t = \alpha \cdot R = 4 \cdot 0.15 = 0.6\ \text{m/s}^2\)
    \(a_c = \omega^2 \cdot R = 2^2 \cdot 0.15 = 0.6\ \text{m/s}^2\)
    \(a = \sqrt{a_t^2 + a_c^2}\) \(= \sqrt{0.6^2 + 0.6^2}\) \(= \sqrt{0.72}\)
    \(a_t = 0.6\ \text{m/s}^2\), \(a_c = 0.6\ \text{m/s}^2\), \(a \approx 0.85\ \text{m/s}^2\)
  8. intermediate

    While braking, a wheel goes from 12 rad/s to 4 rad/s in 2 s. Using the analogy with uniformly accelerated motion (\(\theta \leftrightarrow s\), \(\omega \leftrightarrow v\), \(\alpha \leftrightarrow a\)), calculate \(\alpha\) and the angle turned in that interval.

    Show solution
    \(\alpha = \dfrac{\Delta\omega}{\Delta t}\) \(= \dfrac{4 - 12}{2}\) \(= -4\ \text{rad/s}^2\)
    \(\theta = \omega_0 \cdot t + \dfrac{\alpha \cdot t^2}{2}\) \(= 12 \cdot 2 + \dfrac{(-4) \cdot 2^2}{2}\)
    \(\theta = 24 - 8\)
    \(\alpha = -4\ \text{rad/s}^2\) and \(\theta = 16\ \text{rad}\)
  9. challenge

    The tip of a fan blade is 0.5 m from the axis. The fan starts from rest with \(\alpha = 2\ \text{rad/s}^2\). At \(t = 3\ \text{s}\), calculate \(\omega\), the speed of the tip and its total acceleration.

    Show solution
    \(\omega = \alpha \cdot t = 2 \cdot 3 = 6\ \text{rad/s}\)
    \(v = \omega \cdot R = 6 \cdot 0.5 = 3\ \text{m/s}\)
    \(a_t = \alpha \cdot R = 1\ \text{m/s}^2\)
    \(a_c = \omega^2 \cdot R = 36 \cdot 0.5 = 18\ \text{m/s}^2\)
    \(a = \sqrt{1^2 + 18^2} \approx 18.03\ \text{m/s}^2\)
    \(\omega = 6\ \text{rad/s}\), \(v = 3\ \text{m/s}\), \(a \approx 18\ \text{m/s}^2\) (almost all centripetal)
  10. challenge

    A laboratory centrifuge starts from rest and reaches 100 rad/s after 25 turns, with constant \(\alpha\). Calculate \(\alpha\) and the time taken. Use \(\pi \approx 3.14\).

    Show solution
    \(\Delta\theta = 25 \cdot 2\pi = 157\ \text{rad}\)
    \(\omega^2 = \omega_0^2 + 2 \cdot \alpha \cdot \Delta\theta\) \(\Rightarrow 100^2 = 2 \cdot \alpha \cdot 157\)
    \(\alpha = \dfrac{10\,000}{314} \approx 31.85\ \text{rad/s}^2\)
    \(t = \dfrac{\omega}{\alpha}\) \(= \dfrac{100}{31.85} \approx 3.14\ \text{s}\)
    \(\alpha \approx 31.85\ \text{rad/s}^2\) and \(t \approx 3.14\ \text{s}\)
STEP 10

Globe of death: why doesn't the motorcycle fall at the top?

In the globe of death (a motorcycle sphere), the bike rides around the inside of a sphere, in a vertical circle. If we set aside friction and air resistance, the bike is under two forces, the weight P, always downwards, and the normal force N from the wall, always pointing towards the centre.

At the bottom, the weight points out of the circle, so the normal force has to be large and the rider feels ‘heavier’. At the top, weight and normal force both point towards the centre, and if the bike goes slowly, the weight alone is already more than the curve asks for, so the bike comes away from the wall and falls.

The geometry at the top. Without the wall, the bike would become a projectile and fall along a parabola, and near the top that free fall curves like a circle of radius v²/g. If v²/g > R, the fall would be wider than the globe, so the bike would try to go through the wall and the wall pushes back. If v²/g < R, the fall curves more than the globe, and the bike pulls away from the wall and falls. The magnifier below the simulation lets us compare the two curves.

Top: \(N + P = \dfrac{m\,v^2}{R}\)Bottom: \(N - P = \dfrac{m\,v^2}{R}\)\(v_{\text{min}} = \sqrt{g\,R}\)free-fall radius at the top: \(\dfrac{v^2}{g}\)The speed at the top is minimum when N = 0, that is, when the weight alone provides the centripetal force, and it does not depend on the mass of the bike.

weight P   normal N   net force towards the centre   v

Let's discuss

  • Look at the dashed line ‘path without the wall’ and find when it goes into the wall (green) and when it comes loose into the globe (red).
  • In the magnifier, make the fall curve sit exactly on top of the globe. What speed is that, and how does it compare with vmin?
  • Lower the speed until the bike falls, note the point where it comes away and compare with vmin.
  • When you change the mass, does the minimum speed change? What might explain the result?
  • At what point in the globe does the rider feel heaviest, and how many times their own weight?
  • Why doesn't the water fall out of a bucket swung very fast in a vertical circle?
vmin at the top
N at the top
N at the bottom
Felt weight now
Exercises for step 10 10 questions · 4 basic · 4 intermediate · 2 challenge
  1. basic

    A globe of death has a radius of 3.6 m. What is the minimum speed of the bike at the top so it doesn't fall? Use \(g = 10\ \text{m/s}^2\).

    Show solution
    At the top, with \(N = 0\), we have \(P = \dfrac{m \cdot v^2}{R}\) \(\Rightarrow v_{\text{min}} = \sqrt{g \cdot R}\)
    \(v_{\text{min}} = \sqrt{10 \cdot 3.6} = \sqrt{36}\)
    \(v_{\text{min}} = 6\ \text{m/s}\)
  2. basic

    A light rider and a heavy rider enter the same globe of death. Does the heavier one need more speed at the top?

    Show solution
    In \(v_{\text{min}} = \sqrt{g \cdot R}\) the mass cancels, because weight and centripetal force are both proportional to \(m\).
    No. The minimum speed depends only on \(g\) and \(R\), not on the mass.
  3. basic

    You swing a bucket of water in a vertical circle of radius 0.9 m. What is the lowest speed at the top so the water doesn't fall out? Use \(g = 10\ \text{m/s}^2\).

    Show solution
    At the limit, the weight alone provides the centripetal force, so \(v_{\text{min}} = \sqrt{g \cdot R}\)
    \(v_{\text{min}} = \sqrt{10 \cdot 0.9} = \sqrt{9}\)
    \(v_{\text{min}} = 3\ \text{m/s}\)
  4. basic

    A 100 kg roller-coaster car passes the top of a loop of radius 5 m at 10 m/s. What is the normal force from the rails on it? Use \(g = 10\ \text{m/s}^2\).

    Show solution
    At the top: \(N + P = \dfrac{m \cdot v^2}{R}\)
    \(\dfrac{m \cdot v^2}{R} = \dfrac{100 \cdot 100}{5} = 2000\ \text{N}\); \(P = 1000\ \text{N}\)
    \(N = 2000 - 1000\)
    \(N = 1000\ \text{N}\)
  5. intermediate

    A 60 kg person passes the lowest point of a loop of radius 10 m at 15 m/s. Calculate the normal force from the seat and the felt weight \(N/P\). Use \(g = 10\ \text{m/s}^2\).

    Show solution
    At the bottom: \(N - P = \dfrac{m \cdot v^2}{R}\)
    \(\dfrac{m \cdot v^2}{R} = \dfrac{60 \cdot 225}{10} = 1350\ \text{N}\); \(P = 600\ \text{N}\)
    \(N = 1350 + 600 = 1950\ \text{N}\)
    \(\dfrac{N}{P} = \dfrac{1950}{600} = 3.25\)
    \(N = 1950\ \text{N}\); she feels 3.25 times heavier
  6. intermediate

    Bike and rider together have a mass of 200 kg and pass the top of a globe of radius 4 m at 8 m/s. What is the normal force from the globe on the bike? Use \(g = 10\ \text{m/s}^2\).

    Show solution
    At the top: \(N + P = \dfrac{m \cdot v^2}{R}\)
    \(\dfrac{m \cdot v^2}{R} = \dfrac{200 \cdot 64}{4} = 3200\ \text{N}\); \(P = 2000\ \text{N}\)
    \(N = 3200 - 2000\)
    \(N = 1200\ \text{N}\)
  7. intermediate

    A plane flies a vertical loop of radius 500 m. At what speed must it pass the top for the pilot to be ‘weightless’ (\(N = 0\)) in the seat? Use \(g = 10\ \text{m/s}^2\).

    Show solution
    \(N = 0\) \(\Rightarrow P = \dfrac{m \cdot v^2}{R}\) \(\Rightarrow v = \sqrt{g \cdot R}\)
    \(v = \sqrt{10 \cdot 500} = \sqrt{5000}\)
    \(v \approx 70.7\ \text{m/s} \approx 255\ \text{km/h}\)
  8. intermediate

    A roller-coaster loop has a radius of 8 m. What is the minimum speed at the top, in m/s and in km/h? Use \(g = 10\ \text{m/s}^2\).

    Show solution
    \(v_{\text{min}} = \sqrt{g \cdot R} = \sqrt{10 \cdot 8} = \sqrt{80}\)
    \(v_{\text{min}} \approx 8.94\ \text{m/s}\)
    In km/h: \(8.94 \cdot 3.6 \approx 32.2\ \text{km/h}\)
    \(v_{\text{min}} \approx 8.9\ \text{m/s} \approx 32\ \text{km/h}\)
  9. challenge

    A plane passes the lowest point of a loop of radius 400 m at 100 m/s. How many times heavier does the pilot feel (\(N/P\))? Use \(g = 10\ \text{m/s}^2\).

    Show solution
    At the bottom: \(N - m \cdot g = \dfrac{m \cdot v^2}{R}\) \(\Rightarrow N = m \cdot \left(g + \dfrac{v^2}{R}\right)\)
    \(\dfrac{N}{P} = 1 + \dfrac{v^2}{g \cdot R} = 1 + \dfrac{10\,000}{10 \cdot 400}\)
    \(\dfrac{N}{P} = 1 + 2.5\)
    \(\dfrac{N}{P} = 3.5\) (the pilot feels 3.5 times their own weight)
  10. challenge

    At the top of a loop of radius 6 m, a passenger feels the seat push on them with twice their weight (\(N = 2P\)). What is the speed of the car? Use \(g = 10\ \text{m/s}^2\).

    Show solution
    At the top: \(N + P = \dfrac{m \cdot v^2}{R}\)
    \(2 \cdot m \cdot g + m \cdot g = \dfrac{m \cdot v^2}{R}\) \(\Rightarrow v^2 = 3 \cdot g \cdot R\)
    \(v^2 = 3 \cdot 10 \cdot 6 = 180\)
    \(v \approx 13.4\ \text{m/s}\)
STEP 11

Kepler's laws: the real orbits

Up to now, we have treated orbits as circles. Around 1609, by analysing Tycho Brahe's measurements, Johannes Kepler discovered how the planets really move.

1st law, of orbits: each planet traces an ellipse, with the Sun at one of the foci. We call the point closest to the Sun the perihelion and the farthest one the aphelion, and a circle is simply an ellipse with zero eccentricity.

2nd law, of areas: the Sun–planet line sweeps out equal areas in equal times, which is why the planet moves faster near the Sun and more slowly far from it.

3rd law, of periods: the square of the period divided by the cube of the semi-major axis is the same for all planets, T²/a³ = constant, so planets farther out take much longer to go around.

\(\dfrac{T^2}{a^3} = \text{constant}\)T in years, a in AU: \(\dfrac{T^2}{a^3} = 1\)AU = astronomical unit = average Earth–Sun distance ≈ 150 million km. Newton later showed that all three laws follow from gravitation F = G·M·m/r².
3rd law in the Solar System
Planeta (AU)T (years)T²/a³
Mercury0.3870.2411.00
Venus0.7230.6151.00
Earth1.0001.0001.00
Mars1.5241.8811.00
Jupiter5.20311.861.00
Saturn9.53729.461.00

Let's discuss

  • With the areas switched on, every slice took the same time, so why do the slices near the Sun look short and wide?
  • Is the speed greater at perihelion or at aphelion? Look at the value of v.
  • If we double the semi-major axis a, does the period double too? (Hint: T = a1.5.)
  • The Earth is closest to the Sun in January, so why isn't January summer in the northern hemisphere?
Distance to the Sun r
Speed v
Period T
T² / a³
Exercises for step 11 10 questions · 4 basic · 4 intermediate · 2 challenge
  1. basic

    A classmate says: ‘The Earth goes around in a perfect circle with the Sun right at the centre.’ According to Kepler's 1st law, what is wrong with that sentence? Also explain what perihelion and aphelion are.

    Show solution
    The 1st law says that each planet's orbit is an ellipse, and the Sun is at one of the foci, not at the centre.
    So the planet–Sun distance changes along the orbit.
    Perihelion: the point of the orbit closest to the Sun.
    Aphelion: the point of the orbit farthest from the Sun.
    The Earth's orbit is an ellipse with very small eccentricity (almost a circle), but the Sun is still off-centre.
    The orbit is elliptical with the Sun at one focus; perihelion = closest point, aphelion = farthest point.
  2. basic

    A planet takes the same time to go from A to B (near perihelion) as from C to D (near aphelion). Using Kepler's 2nd law, compare the areas swept out and say on which stretch the planet moves faster.

    Show solution
    By the 2nd law, the Sun–planet segment sweeps out equal areas in equal times.
    Since the time intervals are equal \(\to\) the two areas are equal.
    Near perihelion, the segment is short; to sweep the same area, the planet has to cover a longer arc.
    Near aphelion, the segment is long; a shorter arc is enough.
    Longer arc in the same time \(\to\) greater speed.
    The areas are equal, and the planet is faster on stretch AB (near perihelion).
  3. basic

    The semi-major axis of Mars's orbit is \(a \approx 1.52\ \text{AU}\). Using Kepler's 3rd law (\(T^2/a^3 = 1\), with \(T\) in years and \(a\) in AU), calculate the period of Mars.

    Show solution
    \(\dfrac{T^2}{a^3} = 1\) \(\Rightarrow T^2 = a^3\)
    \(a^3 = 1.52^3 \approx 3.51\)
    \(T = \sqrt{3.51} \approx 1.87\)
    \(T \approx 1.87\) years (about 1 year and 10 months).
  4. basic

    Jupiter has a semi-major axis \(a \approx 5.2\ \text{AU}\) and a period \(T \approx 11.9\) years. Calculate \(T^2/a^3\) and say whether the result confirms Kepler's 3rd law.

    Show solution
    \(T^2 = 11.9^2 \approx 141.6\)
    \(a^3 = 5.2^3 \approx 140.6\)
    \(\dfrac{T^2}{a^3} = \dfrac{141.6}{140.6} \approx 1.01\)
    The value is practically 1, the same constant as for the Earth (\(T = 1\) year, \(a = 1\ \text{AU}\)); the small difference comes from rounding the data.
    \(\dfrac{T^2}{a^3} \approx 1.01 \approx 1\): it confirms the 3rd law.
  5. intermediate

    Halley's comet has a very elongated orbit, with semi-major axis \(a \approx 17.8\ \text{AU}\). Use Kepler's 3rd law to estimate its period and compare it with the observed value of about 75 years.

    Show solution
    Halley orbits the Sun, so \(T^2/a^3 = 1\) applies (\(T\) in years, \(a\) in AU).
    \(T^2 = a^3 = 17.8^3 \approx 5\,640\)
    \(T = \sqrt{5\,640} \approx 75.1\)
    The result matches the observed period (\(\approx 75\) years).
    \(T \approx 75\) years.
  6. intermediate

    At perihelion, Halley's comet is about 0.59 AU from the Sun; at aphelion, about 35 AU. Using \(v_p \cdot r_p = v_a \cdot r_a\), how many times faster is it at perihelion than at aphelion?

    Show solution
    \(v_p \cdot r_p = v_a \cdot r_a\) \(\Rightarrow \dfrac{v_p}{v_a} = \dfrac{r_a}{r_p}\)
    \(\dfrac{v_p}{v_a} = \dfrac{35}{0.59} \approx 59.3\)
    As a check, \(a = \dfrac{r_p + r_a}{2} = \dfrac{0.59 + 35}{2} \approx 17.8\ \text{AU}\), in agreement with the known semi-major axis.
    At perihelion, Halley is about 59 times faster than at aphelion.
  7. intermediate

    The Moon orbits the Earth with an average radius of 384,000 km and a period of 27.3 days. A geostationary satellite has a period of 1 day. Using Kepler's 3rd law for bodies orbiting the Earth, calculate the radius of that satellite's orbit.

    Show solution
    For bodies orbiting the Earth, \(T^2/a^3\) is also constant, but the constant is not the Solar System one. So we compare two bodies that orbit the Earth:
    \(\dfrac{T_s^2}{a_s^3} = \dfrac{T_L^2}{a_L^3}\) \(\Rightarrow a_s = a_L \cdot \left(\dfrac{T_s}{T_L}\right)^{2/3}\)
    \(\dfrac{T_s}{T_L} = \dfrac{1}{27.3}\)
    \(\left(\dfrac{1}{27.3}\right)^{2/3} \approx 0.1103\)
    \(a_s \approx 384\,000 \cdot 0.1103 \approx 42\,350\ \text{km}\)
    \(a \approx 42\,000\ \text{km}\) (measured from the centre of the Earth).
  8. intermediate

    The Earth passes perihelion in early January (≈ 147 million km from the Sun) and aphelion in early July (≈ 152 million km). If distance caused the seasons, what would January be like in both hemispheres? What really causes the seasons?

    Show solution
    If distance were the cause, January would be summer on the whole planet, including the northern hemisphere.
    But in January it is winter in the northern hemisphere and summer in the southern hemisphere, so the hemispheres have opposite seasons at the same time.
    Besides, the change in distance is small: \(\dfrac{152 - 147}{147} \approx 3.4\%\).
    The real cause is the tilt of the Earth's axis (\(\approx 23.5^\circ\)): over the year, each hemisphere gets more direct sunlight and longer days at one time, and more slanted sunlight and shorter days at another.
    The seasons are caused by the tilt of the Earth's axis, not by the distance to the Sun.
  9. challenge

    A hypothetical planet X orbits the Sun with a period of 8 years. Its perihelion is 2 AU from the Sun. Find the semi-major axis, the aphelion distance and how many times faster it is at perihelion than at aphelion.

    Show solution
    3rd law: \(a^3 = T^2 = 8^2 = 64\) \(\Rightarrow a = \sqrt[3]{64} = 4\ \text{AU}\)
    1st law (ellipse): \(r_p + r_a = 2a\) \(\Rightarrow r_a = 2 \cdot 4 - 2 = 6\ \text{AU}\)
    2nd law: \(v_p \cdot r_p = v_a \cdot r_a\) \(\Rightarrow \dfrac{v_p}{v_a} = \dfrac{6}{2} = 3\)
    \(a = 4\ \text{AU}\); aphelion at 6 AU; at perihelion it is 3 times faster.
  10. challenge

    A satellite in low orbit has an orbital radius of 6,800 km (measured from the centre of the Earth) and a period of about 92 minutes. Estimate the period of another satellite with an orbital radius of 42,200 km, in hours. Why can't we use \(T^2/a^3 = 1\) here?

    Show solution
    \(T^2/a^3 = 1\) only holds (in years and AU) for bodies orbiting the Sun. Both satellites orbit the Earth, so they share a different constant, the same for both.
    \(\dfrac{T_2}{T_1} = \left(\dfrac{a_2}{a_1}\right)^{3/2}\)
    \(\dfrac{a_2}{a_1} = \dfrac{42\,200}{6\,800} \approx 6.21\)
    \((6.21)^{3/2} \approx 15.46\)
    \(T_2 \approx 92 \cdot 15.46 \approx 1\,422\ \text{min}\)
    \(\dfrac{1\,422}{60} \approx 23.7\ \text{h}\)
    That is practically 1 day, so this is the geostationary orbit.
    \(T \approx 23.7\ \text{h}\) (about 1 day).
WRAP-UP

Challenges

Period and frequency

A fan spins at 1200 rpm. What are its frequency in Hz and its period?

Show solution
\(f = \dfrac{1200}{60} = 20\ \text{Hz}\)
\(T = \dfrac{1}{f} = \dfrac{1}{20} = 0.05\ \text{s}\)
Angular speed

What is the angular speed of the second hand of a clock?

Show solution
\(T = 60\ \text{s}\)
\(\omega = \dfrac{2\pi}{T}\) \(= \dfrac{2\pi}{60} \approx 0.105\ \text{rad/s}\)
Linear speed

A Ferris wheel of radius 10 m makes one turn every 40 s. How fast is someone in a seat moving?

Show solution
\(v = \dfrac{2\pi R}{T} = \dfrac{2 \cdot 3.14 \cdot 10}{40}\)
\(v \approx 1.57\ \text{m/s}\) (\(\approx 5.7\ \text{km/h}\), walking pace)
Centripetal acceleration

A car takes a bend of radius 50 m at 72 km/h. What is its centripetal acceleration?

Show solution
\(72\ \text{km/h} \div 3.6 = 20\ \text{m/s}\)
\(a_c = \dfrac{v^2}{R} = \dfrac{400}{50} = 8\ \text{m/s}^2\)
(almost 1 g, which is why your body ‘goes sideways’)
Transmission

On a bicycle, the chainring has 48 teeth and the rear sprocket 16. Pedalling at 1 turn per second, with a wheel of radius 0.35 m, what is the speed of the bicycle?

Show solution
Teeth \(\propto\) radius \(\Rightarrow f_{\text{sprocket}} = 1 \cdot \dfrac{48}{16} = 3\ \text{Hz}\)
Wheel on the same axle as the sprocket \(\Rightarrow f_{\text{wheel}} = 3\ \text{Hz}\)
\(v = 2\pi R f\) \(= 2 \cdot 3.14 \cdot 0.35 \cdot 3 \approx 6.6\ \text{m/s} \approx 24\ \text{km/h}\)
Think, no maths

One person at the Equator and another in São Paulo turn with the Earth. Who has the greater \(\omega\), and who has the greater \(v\)?

Show solution
\(\omega\) is the same, since both make 1 turn every 24 h.
\(v\) is greater at the Equator, which is farther from the Earth's axis (larger \(R\)), and \(v = \omega\,R\).
Centripetal force

A 0.5 kg stone spins on a 1 m string at 4 m/s. What is the tension in the string?

Show solution
\(F = \dfrac{m\,v^2}{R} = \dfrac{0.5 \cdot 16}{1} = 8\ \text{N}\)
(the tension plays the role of the centripetal force)
Force and friction

The car from the bend challenge has a mass of 1,000 kg. What friction force do the tyres need to provide? With \(\mu = 0.4\) (wet), can they manage?

Show solution
\(F = m\,a_c = 1000 \cdot 8 = 8000\ \text{N}\)
Max friction \(= \mu\,m\,g = 0.4 \cdot 1000 \cdot 9.8 \approx 3920\ \text{N}\)
\(8000\ \text{N} > 3920\ \text{N}\) → it skids and leaves along the tangent

ENEM-style questions

ENEM is Brazil's national secondary school exam, which most students sit to get into university, and the five questions below follow its format, with a short text drawn from everyday life, a question and five options. Each one draws on a different step of the lesson, and at the end of the list we point to real ENEM questions on the same topics, which may well be the best practice for anyone preparing for the exam.

  1. Period and frequency · Step 1

    The manual for a ceiling fan includes a table with the rotation of the blades and the power drawn at each of the three settings on the control. Let us assume that, at each setting, the blades turn at a constant frequency.

    Ceiling fan
    SettingPower (W)Rotation (rpm)
    130120
    245180
    365240

    When we switch the control from setting 1 to setting 3, what happens to the period of rotation of the blades?

    1. It increases, going from 2 s to 4 s.
    2. It decreases, going from 0.5 s to 0.25 s.
    3. It decreases, going from about 0.008 s to about 0.004 s.
    4. It increases, going from 0.25 s to 0.5 s.
    5. It stays the same, because the blades are the same size at all three settings.
    Show solution
    Answer: B.
    Since rpm counts the turns in one minute, we divide by 60 to get the frequency in hertz.
    \(f_1 = \dfrac{120}{60} = 2\ \text{Hz}\), \(T_1 = \dfrac{1}{2} = 0.5\ \text{s}\)
    \(f_3 = \dfrac{240}{60} = 4\ \text{Hz}\), \(T_3 = \dfrac{1}{4} = 0.25\ \text{s}\)
    Doubling the frequency halved the period. Option A swaps frequency for period, C uses 1/rpm, which gives the period in minutes, and D may come from the intuition that spinning faster "takes longer"; the power in the table plays no part in the calculation.
  2. Angular × linear speed · Step 3

    In a music CD, the laser follows the data track at an almost constant linear speed, close to 1.2 m/s, so the disc has to change its rotation as the reading moves on. The track starts about 2.4 cm from the centre and ends about 6.0 cm from it.

    Assuming a linear speed of 1.2 m/s along the whole track, what angular speeds must the disc have when reading the start and the end of the track, in that order?

    1. 0.029 rad/s and 0.072 rad/s.
    2. 0.5 rad/s and 0.2 rad/s.
    3. 20 rad/s and 50 rad/s.
    4. 50 rad/s in both cases, since the disc is rigid.
    5. 50 rad/s and 20 rad/s.
    Show solution
    Answer: E.
    From \(v = \omega\,R\) it follows that \(\omega = \dfrac{v}{R}\), with the radii in metres.
    \(\omega_{\text{start}} = \dfrac{1.2}{0.024} = 50\ \text{rad/s}\)
    \(\omega_{\text{end}} = \dfrac{1.2}{0.060} = 20\ \text{rad/s}\)
    The disc slows down as the reading works its way towards the edge. Option A multiplies v by R instead of dividing, B forgets to convert centimetres to metres and C reverses the order. D would hold for two points on the disc at the same instant, when they all turn together, whereas here the comparison is between two different moments of the reading.
  3. Pulleys and gears · Step 7

    On a road bike, Helena rides with the 44-tooth chainring and the 22-tooth sprocket, keeping a cadence of 90 pedal turns per minute. The rear wheel, tyre included, is 70 cm in diameter. Let us assume that the chain does not slip and that the tyre rolls without sliding.

    Using \(\pi \approx 3.14\), what is the approximate speed of the bike?

    1. 1.8 km/h
    2. 6 km/h
    3. 12 km/h
    4. 24 km/h
    5. 47 km/h
    Show solution
    Answer: D.
    The chainring turns with the pedals, \(f_{\text{chainring}} = \dfrac{90}{60} = 1.5\ \text{Hz}\), and the chain gives both gears the same speed at the rim. Since the number of teeth is proportional to the radius,
    \(f_{\text{sprocket}} = 1.5 \cdot \dfrac{44}{22} = 3\ \text{Hz}\)
    The wheel is on the same axle as the sprocket and also turns at 3 Hz, with radius \(R = 0.35\ \text{m}\).
    \(v = 2\pi R f = 2 \cdot 3.14 \cdot 0.35 \cdot 3\)
    \(v \approx 6.6\ \text{m/s} \approx 24\ \text{km/h}\)
    Inverting the gear ratio leads to 6 km/h, forgetting the gears and letting the wheel turn with the pedals leads to 12 km/h, and using the diameter in place of the radius leads to 47 km/h. Option A divides by 3.6 instead of multiplying.
  4. Centripetal acceleration · Step 5

    In clinical laboratories, a blood sample goes into a centrifuge, which separates the red cells from the plasma by spinning the tubes very fast. In a benchtop centrifuge, the bottom of the tubes sits 10 cm from the axis and the rotor turns at 3000 rpm.

    Using \(\pi \approx 3.14\) and \(g = 10\ \text{m/s}^2\), the centripetal acceleration at the bottom of the tubes is approximately

    1. 3 times g.
    2. 25 times g.
    3. 1,000 times g.
    4. 100,000 times g.
    5. 350,000 times g.
    Show solution
    Answer: C.
    Converting rpm to hertz is the first move, and from it comes ω.
    \(f = \dfrac{3000}{60} = 50\ \text{Hz}\)
    \(\omega = 2\pi f = 2 \cdot 3.14 \cdot 50\)
    \(\omega = 314\ \text{rad/s}\)
    \(a_c = \omega^2 R = 314^2 \cdot 0.10\)
    \(a_c \approx 9860\ \text{m/s}^2\)
    \(\dfrac{a_c}{g} = \dfrac{9860}{10} \approx 986\), close to 1,000 times g.
    Option A forgets to square ω, B leaves out the factor 2π, D uses the radius in centimetres and E keeps the rotation in rpm. A number this large seems to explain why a centrifuge separates in minutes what gravity alone would separate very slowly.
  5. Satellites and Kepler's laws · Steps 6 and 11

    TV satellites sit in a geostationary orbit, in which one trip around the Earth takes about 24 h, so they look fixed in the sky to anyone on the ground. An observation satellite, in a lower circular orbit, goes round once every 3 h.

    In both cases the Earth's gravity plays the role of the centripetal force, and Kepler's 3rd law, \(T^2/r^3 = \text{constant}\), holds for all of the Earth's satellites. How many times larger is the radius of the geostationary orbit than the radius of the observation satellite's orbit?

    1. 2
    2. 4
    3. 8
    4. 23
    5. 64
    Show solution
    Answer: B.
    Since the constant is the same for both satellites, we can set the ratios equal.
    \(\dfrac{T_G^2}{r_G^3} = \dfrac{T_O^2}{r_O^3}\), \(\left(\dfrac{r_G}{r_O}\right)^3 = \left(\dfrac{T_G}{T_O}\right)^2\)
    \(\left(\dfrac{r_G}{r_O}\right)^3 = \left(\dfrac{24}{3}\right)^2\) \(= 64\)
    \(\dfrac{r_G}{r_O} = \sqrt[3]{64} = 4\)
    Option C assumes the radius grows in the same proportion as the period, and E stops at \(8^2\) without taking the cube root. A takes the cube root of 8 without squaring the period, and D swaps the exponents in the law.

Real ENEM questions on this topic

The official papers and answer keys are on the INEP website, where each question can be found by year, day, booklet colour and number.

  • ENEM 2013, Day 1, blue booklet, question 66. A butcher chooses between two ways of mounting the pulleys of a band saw, and the task is to decide which one makes the blade slower by comparing frequencies and rim speeds.
  • ENEM 2014, Day 1, blue booklet, question 82. Starting from a comic strip on a satellite in orbit, the question asks for the tangential acceleration of a character moving round at constant speed.
  • ENEM 2016, Day 1, blue booklet, question 66. A watchmaker builds a stopwatch from gears, and the task is to follow the motor's frequency, pair by pair, all the way to the hand.
  • ENEM 2018, Day 2, blue booklet, question 122. Swapping a car's wheels for ones of larger diameter affects its stability and makes the speedometer read wrongly, and the question asks for these two consequences.
  • ENEM 2019, Day 2, blue booklet, question 109. Starting from fragments of a Soviet rocket that fell into the Atlantic in 1978, the question compares the rocket's angular speed with the Earth's.

ENEM papers and answer keys on the INEP website (in Portuguese) →

Cheat sheet

Period and frequency\(T = 1/f\)
Angle in radians\(\theta = s/R \quad 2\pi = 360^\circ\)
Angular speed\(\omega = 2\pi/T = 2\pi f\)
Linear speed\(v = \omega R = 2\pi R/T\)
Centripetal acceleration\(a_c = v^2/R = \omega^2 R\)
Belt / teeth\(f_A R_A = f_B R_B\)
Same axle\(\omega_A = \omega_B\)
Uniformly accelerated circular motion\(\omega = \omega_0 + \alpha t \quad a_t = \alpha R\)
Globe of death\(v_{\text{min}} = \sqrt{gR}\)
Kepler's 3rd law\(T^2/a^3 = \text{const.}\)
Shadows of a spin\(x = R\cos\omega t \quad y = R\,\sin\omega t\)
Centripetal force\(F_c = m v^2/R\)
UnitsHz · rad/s · m/s · m/s² · N