Chapter 3 ยท Motion in a Plane

3.4 Uniform Circular Motion

A body going round a circle at constant speed is accelerating all the time โ€” towards the centre, with magnitude $a_c=\dfrac{v^2}{r}=\omega^2r$. Angular speed, period, frequency and centripetal acceleration.

Whirl a stone round your head on a string, then let go. The stone does not fly outward along the string; it shoots off along the tangent, in the direction it was moving at that instant. While you held the string it was being pulled, at every moment, away from that straight line and towards your hand. Even though its speed never changed, its velocity did โ€” and a changing velocity means an acceleration.

The same idea runs through a surprising amount of physics: the blades of a ceiling fan, a car on a roundabout, a sample in a laboratory centrifuge, a satellite in orbit, and the Moon going round the Earth. Isaac Newton compared the Moon's acceleration towards the Earth with that of a falling apple, and found the clue that led to the law of gravitation (Chapter 7). In this section you will describe motion on a circle with angles, angular speed, period and frequency, and derive the most important result of the chapter: a body moving uniformly on a circle accelerates towards the centre with $a_c=v^2/r$.

Describing motion on a circle

๐Ÿ“˜ Definition 3.10 โ€” Uniform circular motion

A body is in uniform circular motion when it moves along a circle of radius $r$ with constant speed $v$. Its velocity is always tangent to the circle, so although the speed is constant, the direction of the velocity changes continuously.

Position on a circle is easiest to describe by an angle. Take the origin at the centre $O$; the position vector $\mathbf{r}$ has constant length $r$ and turns as the body moves.

๐Ÿ“˜ Definition 3.11 โ€” Angular displacement, angular speed, period and frequency

If the position vector turns through an angle $\Delta\theta$ (in radians) in a time $\Delta t$, $\Delta\theta$ is the angular displacement; the arc covered is $\Delta s=r\,\Delta\theta$ (the definition of the radian, Section 1.1).

The angular speed is $\omega=\dfrac{\Delta\theta}{\Delta t}$ (instantaneously, $\omega=\dfrac{d\theta}{dt}$), in $\text{rad/s}$; in uniform circular motion it is constant.

The period $T$ is the time for one complete revolution, and the frequency $\nu=\dfrac1T$ is the number of revolutions per second (unit $\text{s}^{-1}$ or hertz, $\text{Hz}$). Engineers also use revolutions per minute: $1\ \text{rpm}=\tfrac1{60}\ \text{rev/s}$.

๐Ÿ“ Result 3.10 โ€” Linear and angular speed

For uniform circular motion with radius $r$:

$$v=\omega r,\qquad \omega=\frac{2\pi}{T}=2\pi\nu,\qquad v=\frac{2\pi r}{T}.$$

Proof. In time $\Delta t$ the arc covered is $\Delta s=r\,\Delta\theta$, so $v=\dfrac{\Delta s}{\Delta t}=r\dfrac{\Delta\theta}{\Delta t}=r\omega$. One revolution is $2\pi$ radians in time $T$, so $\omega=\dfrac{2\pi}{T}$. These hold only with $\theta$ in radians.

โœ๏ธ Example 1 โ€” A ceiling fan
The tips of a ceiling fan's blades are $0.60\ \text{m}$ from the axis and the fan turns at $300\ \text{rpm}$. Find the frequency, the period, the angular speed and the speed of a blade tip.
Frequency $\nu=\dfrac{300}{60}=5.0\ \text{Hz}$; period $T=\dfrac1\nu=0.20\ \text{s}$.
Angular speed $\omega=2\pi\nu=2\pi\times5.0=31.4\ \text{rad/s}$.
Tip speed $v=\omega r=31.4\times0.60=18.8\ \text{m/s}$ โ€” about $68\ \text{km/h}$.
Check $v=\dfrac{2\pi r}{T}=\dfrac{2\pi(0.60)}{0.20}=18.8\ \text{m/s}$ โœ“. Every point of the blade has the same $\omega$, but the speed grows with $r$.
โœ๏ธ Example 2 โ€” Riding on the spinning Earth
The Earth turns once in $24\ \text{h}$ (to the accuracy we need) and its radius is $R_E=6.37\times10^6\ \text{m}$. Find (a) its angular speed, (b) the speed of a point on the equator, (c) the speed of a person in Delhi, latitude $28.6^\circ$ N, who moves on a circle of radius $R_E\cos28.6^\circ$.
(a) $T=24\times3600=86\,400\ \text{s}$; $\ \omega=\dfrac{2\pi}{86\,400}=7.27\times10^{-5}\ \text{rad/s}$.
(b) $v=\omega R_E=7.27\times10^{-5}\times6.37\times10^6=463\ \text{m/s}$ โ€” faster than a jet airliner.
(c) $r=6.37\times10^6\cos28.6^\circ=5.59\times10^6\ \text{m}$; $v=\omega r=407\ \text{m/s}$.
Check Everyone on the Earth has the same $\omega$; speed falls to zero at the poles, where $r=0$.

Centripetal acceleration

Let the body be at $P$ at time $t$ and at $P'$ a short time $\Delta t$ later, the radius having turned through $\Delta\theta$. Its velocities $\mathbf{v}$ at $P$ and $\mathbf{v}'$ at $P'$ have the same magnitude $v$, and each is perpendicular to its radius, so $\mathbf{v}'$ is $\mathbf{v}$ turned through the same angle $\Delta\theta$.

With calculus the same result appears in two lines. Measure $\theta=\omega t$ from the $x$-axis, so $\mathbf{r}=r\cos\omega t\,\hat{\mathbf{i}}+r\sin\omega t\,\hat{\mathbf{j}}$. Differentiating (you will prove the rates of change of $\sin$ and $\cos$ in maths), $\mathbf{v}=\dfrac{d\mathbf{r}}{dt}=-\omega r\sin\omega t\,\hat{\mathbf{i}}+\omega r\cos\omega t\,\hat{\mathbf{j}}$ and $\mathbf{a}=\dfrac{d\mathbf{v}}{dt}=-\omega^2r\cos\omega t\,\hat{\mathbf{i}}-\omega^2r\sin\omega t\,\hat{\mathbf{j}}=-\omega^2\mathbf{r}$: magnitude $\omega^2r$, pointing opposite to $\mathbf{r}$, i.e. at the centre.

๐Ÿ“ Result 3.11 โ€” Centripetal acceleration

A body in uniform circular motion (radius $r$, speed $v$, angular speed $\omega$, period $T$, frequency $\nu$) has an acceleration directed towards the centre of the circle, of magnitude

$$a_c=\frac{v^2}{r}=\omega^2r=4\pi^2\nu^2r=\frac{4\pi^2r}{T^2}.$$

It is called the centripetal acceleration ("centre-seeking"). Its magnitude is constant but its direction turns with the body, so it is not a constant vector: the equations of Result 3.7 do not apply to circular motion. The acceleration is always perpendicular to the velocity, which is why the speed does not change.

Interactive 1 โ€” Velocity and Acceleration Going Round

Left: a body in uniform circular motion with its velocity $\mathbf{v}$ (blue, tangent) and acceleration $\mathbf{a}$ (red, towards the centre). Right: the same velocity vectors drawn from one point โ€” their tips go round a circle of radius $v$, once per period, at speed $\omega v$; that rate of change of velocity is the acceleration, $a=\omega v=\dfrac{v^2}{r}$. Change the radius and speed, and press Cut the string to see where the body goes when the inward pull stops.

1.5 m3.0 m/s
angular speed $\omega=v/r$โ€”
period $T=2\pi r/v$โ€”
$a_c=v^2/r$โ€”
angle between $\mathbf{v}$ and $\mathbf{a}$โ€”
โœ๏ธ Example 3 โ€” How hard does a fan blade pull?
Find the centripetal acceleration of the blade tip in Example 1 and compare it with $g$.
Using v $a_c=\dfrac{v^2}{r}=\dfrac{18.85^2}{0.60}=592\ \text{m/s}^2$.
Using ฯ‰ $a_c=\omega^2r=31.42^2\times0.60=592\ \text{m/s}^2$ โœ“.
Compare $\dfrac{592}{9.8}=60$: the tip accelerates at about $60g$ towards the axis.
Meaning Anything fixed to the tip โ€” a loose screw, a lump of dirt โ€” needs an inward force about $60$ times its own weight to keep it on the circle. That is why fan blades must be firmly fixed and balanced.
โœ๏ธ Example 4 โ€” Newton's Moon test
The Moon orbits the Earth in a nearly circular path of radius $3.84\times10^8\ \text{m}$ with a period of $27.3$ days. (a) Find its centripetal acceleration. (b) The Moon is about $60$ Earth radii away. Compare your answer with $\dfrac{g}{60^2}$.
Period $T=27.3\times86\,400=2.36\times10^6\ \text{s}$.
(a) $a_c=\dfrac{4\pi^2r}{T^2}=\dfrac{4\pi^2(3.84\times10^8)}{(2.359\times10^6)^2}=2.72\times10^{-3}\ \text{m/s}^2$, directed towards the Earth.
(b) $\dfrac{9.8}{3600}=2.72\times10^{-3}\ \text{m/s}^2$ โ€” the same.
Meaning The Moon "falls" towards the Earth with the acceleration an apple would have if gravity weakened as $\dfrac{1}{r^2}$. This was Newton's evidence for the inverse-square law of gravitation (Chapter 7).
โœ๏ธ Example 5 โ€” Average versus instantaneous acceleration
A car goes round a roundabout of radius $25\ \text{m}$ at a steady $15\ \text{m/s}$. (a) Find its centripetal acceleration. (b) Find the magnitude of its average acceleration over a quarter of a turn, and compare.
(a) $a_c=\dfrac{15^2}{25}=9.0\ \text{m/s}^2$.
ฮ”v Over a quarter turn $\mathbf{v}$ turns through $90^\circ$: $|\Delta\mathbf{v}|=\sqrt{15^2+15^2}=21.2\ \text{m/s}$.
ฮ”t Arc $=\tfrac14(2\pi\times25)=39.3\ \text{m}$, so $\Delta t=\dfrac{39.3}{15}=2.62\ \text{s}$; average acceleration $=\dfrac{21.2}{2.62}=8.1\ \text{m/s}^2$.
Compare $\dfrac{8.1}{9.0}=0.90$ (exactly $\tfrac{2\sqrt2}{\pi}$). The average over a big arc is smaller than $v^2/r$; Interactive 2 shows it climbing to $v^2/r$ as the arc shrinks.
โœ๏ธ Example 6 โ€” Designing a centrifuge
A laboratory centrifuge spins test tubes whose bottoms are $10\ \text{cm}$ from the axis. At what rate (in rpm) must it turn to give an acceleration of $1000g$ at the bottom of the tubes?
Required ฯ‰ $\omega^2r=1000g$: $\ \omega=\sqrt{\dfrac{1000\times9.8}{0.10}}=313\ \text{rad/s}$.
Frequency $\nu=\dfrac{\omega}{2\pi}=49.8\ \text{rev/s}$.
Answer $49.8\times60=2.99\times10^3\ \text{rpm}$, i.e. about $3000\ \text{rpm}$.
Check $v=\omega r=31.3\ \text{m/s}$ and $\dfrac{v^2}{r}=\dfrac{980}{0.10}=9.8\times10^3\ \text{m/s}^2=1000g$ โœ“.
โš ๏ธ Common mistakes

1. "Constant speed, so no acceleration." In circular motion the direction of $\mathbf{v}$ changes, so $\mathbf{a}\ne\mathbf{0}$. Constant speed is not constant velocity.

2. "The acceleration points outward." Seen from the ground, the acceleration of a body moving on a circle points inward, towards the centre. The outward "fling" you feel in a turning car is your body trying to continue in a straight line (Chapter 4).

3. Using the constant-acceleration equations. $\mathbf{a}$ keeps turning, so $v=u+at$ and its relatives do not apply to circular motion.

4. Units. $v=\omega r$ and $a=\omega^2r$ need $\omega$ in $\text{rad/s}$: convert rpm with $\omega=2\pi\times\dfrac{\text{rpm}}{60}$.

5. Squares. $a_c\propto v^2$: doubling the speed on the same circle quadruples the acceleration. At fixed $v$, a tighter circle (smaller $r$) means a larger acceleration; at fixed $\omega$, a larger $r$ does.

If the speed is also changing, the body has a second, tangential component of acceleration along $\mathbf{v}$ as well as the centripetal one; that non-uniform case is met with rotation in Chapter 6.

Interactive 2 โ€” Shrinking the Arc: Where $v^2/r$ Comes From

A body moves at $v=3.0\ \text{m/s}$ on a circle of radius $r=1.5\ \text{m}$, so $\dfrac{v^2}{r}=6.0\ \text{m/s}^2$. It goes from $P$ to $P'$ through the angle $\Delta\theta$. On the right, $\mathbf{v}$ and $\mathbf{v}'$ are drawn tail to tail and $\Delta\mathbf{v}=\mathbf{v}'-\mathbf{v}$ joins their heads. Press Play (or drag the slider) to shrink $\Delta\theta$: the average acceleration $|\Delta\mathbf{v}|/\Delta t$ rises to exactly $v^2/r$, and its direction (red arrow drawn at $P$) swings round to point at the centre.

90ยฐ
$|\Delta\mathbf{v}|=2v\sin\tfrac{\Delta\theta}{2}$โ€”
$\Delta t=r\,\Delta\theta/v$โ€”
$|\Delta\mathbf{v}|/\Delta t$โ€”
ratio to $v^2/r$โ€”

Exercises

I
Angular speed, period and frequency.
  • 1 A turntable rotates at $120\ \text{rpm}$. Find its frequency, angular speed and period.
  • 2 A wheel turns at $25\ \text{rad/s}$. Find its frequency and period.
  • 3 Find the angular speeds of the second, minute and hour hands of a clock.
  • 4 The second hand is $2.0\ \text{cm}$ long. Find the speed of its tip.
  • 5 A point on the rim of a wheel of radius $0.30\ \text{m}$ goes round the axle $15$ times a second. Find its speed relative to the axle in $\text{m/s}$ and $\text{km/h}$.
  • 6 Using Example 2, find the speed due to the Earth's rotation at latitude $60^\circ$.
  • 7 The Earth orbits the Sun on a nearly circular path of radius $1.50\times10^{11}\ \text{m}$ in $365.25$ days. Find its orbital speed.
1 $\nu=\tfrac{120}{60}=2.0\ \text{Hz}$; $\omega=2\pi\nu=12.6\ \text{rad/s}$; $T=0.50\ \text{s}$.
2 $\nu=\dfrac{25}{2\pi}=3.98\ \text{Hz}$; $T=0.251\ \text{s}$.
3 Second hand $\dfrac{2\pi}{60}=0.105\ \text{rad/s}$; minute hand $\dfrac{2\pi}{3600}=1.75\times10^{-3}\ \text{rad/s}$; hour hand $\dfrac{2\pi}{43\,200}=1.45\times10^{-4}\ \text{rad/s}$.
4 $v=\omega r=0.105\times0.020=2.1\times10^{-3}\ \text{m/s}$ ($2.1\ \text{mm/s}$).
5 $v=2\pi(15)(0.30)=28.3\ \text{m/s}=102\ \text{km/h}$.
6 $r=R_E\cos60^\circ=3.19\times10^6\ \text{m}$; $v=7.27\times10^{-5}\times3.19\times10^6=232\ \text{m/s}$ (half the equatorial speed).
7 $T=365.25\times86\,400=3.156\times10^7\ \text{s}$; $v=\dfrac{2\pi(1.50\times10^{11})}{3.156\times10^7}=2.99\times10^4\ \text{m/s}$, about $30\ \text{km/s}$.
II
Centripetal acceleration.
  • 1 A stone on a $0.80\ \text{m}$ string is whirled in a horizontal circle at $2.0$ revolutions per second. Find its speed and centripetal acceleration.
  • 2 In part 1 the stone's speed is doubled (same string). What is the new acceleration?
  • 3 Two points on the same rotating disc are at $r$ and $2r$ from the axis. Which has the larger centripetal acceleration, and by what factor? Now compare two cars on circles of radius $r$ and $2r$ at the same speed.
  • 4 Using Exercise I.7, find the Earth's centripetal acceleration towards the Sun.
  • 5 Find the centripetal acceleration of a point on the equator due to the Earth's rotation, as a percentage of $g$.
  • 6 A space station orbits at radius $6.77\times10^6\ \text{m}$ with a period of $92.4\ \text{min}$. Find its speed and centripetal acceleration.
  • 7 In a simple model of the hydrogen atom an electron moves at $2.2\times10^6\ \text{m/s}$ on a circle of radius $5.3\times10^{-11}\ \text{m}$. Find its centripetal acceleration.
  • 8 A racing car can take a sideways (centripetal) acceleration of at most $1.2g$. What is its greatest speed on a bend of radius $80\ \text{m}$?
1 $\omega=4\pi=12.57\ \text{rad/s}$; $v=\omega r=10.1\ \text{m/s}$; $a_c=\omega^2r=126\ \text{m/s}^2$.
2 $a_c\propto v^2$: $4\times126=505\ \text{m/s}^2$.
3 Same disc: same $\omega$, $a_c=\omega^2r$, so the outer point has twice the acceleration. Same speed: $a_c=v^2/r$, so the car on radius $2r$ has half the acceleration. Which formula to use depends on what is held fixed.
4 $a_c=\dfrac{v^2}{r}=\dfrac{(2.99\times10^4)^2}{1.50\times10^{11}}=5.95\times10^{-3}\ \text{m/s}^2$.
5 $a_c=\omega^2R_E=(7.27\times10^{-5})^2(6.37\times10^6)=0.0337\ \text{m/s}^2$, i.e. $0.34\%$ of $g$.
6 $T=5544\ \text{s}$; $v=\dfrac{2\pi r}{T}=7.67\times10^3\ \text{m/s}$; $a_c=\dfrac{v^2}{r}=8.69\ \text{m/s}^2$ โ€” gravity is still about $89\%$ of its surface value up there.
7 $a_c=\dfrac{(2.2\times10^6)^2}{5.3\times10^{-11}}=9.1\times10^{22}\ \text{m/s}^2$.
8 $v=\sqrt{1.2\times9.8\times80}=30.7\ \text{m/s}$, about $110\ \text{km/h}$.
III
Vectors in circular motion. A particle moves at a steady $4.0\ \text{m/s}$ on a circle of radius $2.0\ \text{m}$.
  • 1 Find the magnitude of its displacement and of its average velocity over a quarter revolution. Compare with its average speed.
  • 2 Repeat for half a revolution.
  • 3 Find the magnitude of its average acceleration over half a revolution and compare it with $a_c$.
  • 4 Is uniform circular motion "motion with constant acceleration"? Explain.
  • 5 What is the angle between $\mathbf{v}$ and $\mathbf{a}$ in uniform circular motion, and why does that keep the speed constant?
  • 6 The string holding a whirled stone breaks when the stone is at the top of its (vertical-plane) circle, moving east. In which direction does the stone set off?
1 Displacement $=r\sqrt2=2.83\ \text{m}$; time $=\dfrac{\pi r/2}{v}=0.785\ \text{s}$; average velocity $3.6\ \text{m/s}$, less than the average speed $4.0\ \text{m/s}$.
2 Displacement $=2r=4.0\ \text{m}$ in $\dfrac{\pi r}{v}=1.57\ \text{s}$: average velocity $2.5\ \text{m/s}$ ($=\tfrac2\pi v$).
3 $\mathbf{v}$ reverses: $|\Delta\mathbf{v}|=8.0\ \text{m/s}$; $\dfrac{8.0}{1.57}=5.1\ \text{m/s}^2$, compared with $a_c=\dfrac{16}{2.0}=8.0\ \text{m/s}^2$ (ratio $\tfrac2\pi$).
4 No. The magnitude $v^2/r$ is constant but the direction keeps turning, so the acceleration vector is not constant.
5 $90^\circ$. An acceleration perpendicular to the velocity only turns it; no part of it lies along $\mathbf{v}$ to speed the body up or slow it down.
6 Along the tangent at that point: horizontally east (and then it falls as a projectile, Section 3.3).
IV
Applications.
  • 1 A centrifuge must produce $2000g$ at $8.0\ \text{cm}$ from its axis. Find the required rotation rate in rpm.
  • 2 A giant wheel of radius $30\ \text{m}$ makes one turn every $2.0\ \text{min}$. Find a rider's speed and centripetal acceleration.
  • 3 An astronaut-training centrifuge has an arm of $8.0\ \text{m}$. What period gives the trainee a centripetal acceleration of $9g$? How many rpm is that?
  • 4 A washing-machine drum of radius $0.25\ \text{m}$ spins at $1200\ \text{rpm}$. Find the centripetal acceleration of the clothes against the drum, in units of $g$.
  • 5 (Modelling) A ring-shaped space station of radius $100\ \text{m}$ spins so that people standing on the rim have a centripetal acceleration of $9.8\ \text{m/s}^2$ ("artificial gravity"). Find its period. Why would a much smaller station be uncomfortable?
1 $\omega=\sqrt{\dfrac{2000\times9.8}{0.080}}=495\ \text{rad/s}$; $\ \dfrac{495\times60}{2\pi}=4.7\times10^3\ \text{rpm}$.
2 $v=\dfrac{2\pi(30)}{120}=1.57\ \text{m/s}$; $a_c=\dfrac{v^2}{r}=0.082\ \text{m/s}^2$ โ€” less than $1\%$ of $g$.
3 $\omega=\sqrt{\dfrac{9\times9.8}{8.0}}=3.32\ \text{rad/s}$; $T=\dfrac{2\pi}{\omega}=1.89\ \text{s}$, i.e. $31.7\ \text{rpm}$.
4 $\omega=2\pi\times20=125.7\ \text{rad/s}$; $a_c=\omega^2r=3.95\times10^3\ \text{m/s}^2\approx400g$ โ€” which is how a spin cycle wrings water out.
5 $T=2\pi\sqrt{\dfrac rg}=2\pi\sqrt{\dfrac{100}{9.8}}=20\ \text{s}$. In a small station the spin would be fast, and $a_c=\omega^2r$ would differ noticeably between a person's head and feet, causing dizziness.
๐ŸŽฏ Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board ยท 3 marks Show that a particle moving with constant speed $v$ on a circle of radius $r$ has an acceleration of magnitude $\dfrac{v^2}{r}$ directed towards the centre.

Figure Positions $P$, $P'$ at $t$, $t+\Delta t$; radii $\mathbf{r}$, $\mathbf{r}'$ at angle $\Delta\theta$; velocities $\mathbf{v}$, $\mathbf{v}'$ (equal magnitude $v$), each perpendicular to its radius, so also at angle $\Delta\theta$. [1]
Similar triangles Triangle $(\mathbf{r},\mathbf{r}',\Delta\mathbf{r})$ and triangle $(\mathbf{v},\mathbf{v}',\Delta\mathbf{v})$ are isosceles with equal apex angles: $\dfrac{|\Delta\mathbf{v}|}{v}=\dfrac{|\Delta\mathbf{r}|}{r}$. [1]
Limit $a=\displaystyle\lim_{\Delta t\to0}\frac{|\Delta\mathbf{v}|}{\Delta t}=\frac vr\lim_{\Delta t\to0}\frac{|\Delta\mathbf{r}|}{\Delta t}=\frac vr\cdot v=\frac{v^2}{r}$; as $\Delta\theta\to0$, $\Delta\mathbf{v}\perp\mathbf{v}$, i.e. along the radius towards the centre. [1]
Exam tip The direction argument is often left out โ€” and it carries a mark.

E2. NEET For a particle in uniform circular motion, which quantity is constant? (a) velocity (b) acceleration (c) speed (d) displacement from the starting point

Idea Velocity and acceleration have constant magnitudes but rotating directions; the displacement keeps changing.
Answer (c).
Exam tip "Uniform" refers to the speed only. (Kinetic energy, a scalar, is also constant.)

E3. NEET A body moves on a circle of radius $1.0\ \text{m}$ at $120\ \text{rpm}$. Its centripetal acceleration is about (a) $158\ \text{m/s}^2$ (b) $39.5\ \text{m/s}^2$ (c) $12.6\ \text{m/s}^2$ (d) $1.44\times10^4\ \text{m/s}^2$

Convert $\omega=2\pi\times\tfrac{120}{60}=4\pi=12.57\ \text{rad/s}$.
Compute $a_c=\omega^2r=158\ \text{m/s}^2$.
Answer (a).
Exam tip (b) takes $\omega=2\pi\ \text{rad/s}$, forgetting that the frequency is $2.0\ \text{rev/s}$; (c) is $\omega$ itself; (d) squares the rpm without converting.

E4. JEE Main Two particles $A$ and $B$ move on concentric circles of radii $2.0\ \text{m}$ and $3.0\ \text{m}$. (i) If they have the same period, the ratio $a_A:a_B$ of their centripetal accelerations is (ii) if they have the same speed, it is โ€” (a) $2:3$ and $3:2$ (b) $3:2$ and $2:3$ (c) $4:9$ and $9:4$ (d) $2:3$ and $9:4$

(i) Same $T$ means same $\omega$; $a=\omega^2r\propto r$: $2:3$.
(ii) Same $v$; $a=\dfrac{v^2}{r}\propto\dfrac1r$: $3:2$.
Answer (a).
Exam tip Pick the form of $a_c$ in which the shared quantity appears: $\omega^2r$ for equal periods, $v^2/r$ for equal speeds.

E5. JEE Advanced (One or more correct) A particle moves uniformly with speed $v$ on a circle of radius $r$. Which statements are correct? (A) The magnitude of its average velocity over half a revolution is $\dfrac{2v}{\pi}$. (B) The magnitude of its average acceleration over half a revolution is $\dfrac{2v^2}{\pi r}$. (C) Its average acceleration over one full revolution is $\dfrac{v^2}{r}$. (D) Its velocity and acceleration are perpendicular at every instant.

(A) Displacement $2r$ in time $\dfrac{\pi r}{v}$: $\dfrac{2r}{\pi r/v}=\dfrac{2v}{\pi}$ โœ“.
(B) $|\Delta\mathbf{v}|=2v$ in $\dfrac{\pi r}{v}$: $\dfrac{2v^2}{\pi r}$ โœ“.
(C) After a full turn $\mathbf{v}$ is back to its starting value: $\Delta\mathbf{v}=\mathbf{0}$, so the average acceleration is zero โœ—.
(D) $\mathbf{v}$ is tangential, $\mathbf{a}$ radial โœ“.
Answer (A), (B), (D).

E6. Board ยท Case study A giant wheel at a fair has a radius of $15\ \text{m}$ and turns steadily, making one revolution every $40\ \text{s}$. (a) Find its angular speed and the speed of a rider. [2] (b) Find the rider's centripetal acceleration. [1] (c) At the lowest point of the ride, in which direction is the rider's acceleration? [1]

(a) $\omega=\dfrac{2\pi}{40}=0.157\ \text{rad/s}$; $v=\omega r=2.36\ \text{m/s}$. [2]
(b) $a_c=\omega^2r=0.370\ \text{m/s}^2$. [1]
(c) Towards the centre โ€” vertically upward at the lowest point. [1]
Exam tip "Towards the centre" changes meaning round the circle: up at the bottom, down at the top. Chapter 4 uses this to explain why you feel heavier at the bottom.
๐Ÿ”ฌ Researchโ€“Teaching NexusResearch-tutored ยท optional, about 40 minutes

Research question. Can a smartphone in a salad spinner verify $a_c=\omega^2r$ โ€” and find where inside the phone its accelerometer is?

Why it matters. Every smartphone contains a tiny accelerometer and a gyroscope (which measures $\omega$). Physics-education researchers have built free apps that turn phones into laboratory instruments, so that students anywhere can do quantitative experiments cheaply. The same sensors stabilise drones, count steps, rotate your screen and trigger airbags; engineers calibrate them in exactly this way, on a turntable at known $\omega$.

What to do.

  1. Install a free sensor app that records the gyroscope (angular speed) and the accelerometer together (some apps have a ready-made "centripetal acceleration" experiment).
  2. Fix the phone flat in a salad spinner (or on a rotating stool), with padding so it cannot move. Spin at several different rates and record $\omega$ and the acceleration along the radius.
  3. Plot $a$ against $\omega^2$. Is it a straight line through the origin, as Result 3.11 predicts? The slope is $r$, the distance of the accelerometer chip from the axis.
  4. Measure where the axis is relative to the phone and estimate where the chip sits inside the phone.

Tools: Smartphone with a sensor app, salad spinner or swivel chair, ruler, spreadsheet.

Hand in: Your $a$โ€“$\omega^2$ graph with a best-fit line, the value of $r$, and a short discussion of the sources of scatter.

Going further. Spin at a steady rate while the phone is placed at two different distances from the axis. Does the ratio of accelerations equal the ratio of radii, as $a=\omega^2r$ (same $\omega$) predicts?

Instructor note: typical salad spinners reach $2$โ€“$4\ \text{rev/s}$; with the chip $r=0.08\ \text{m}$ from the axis, $2.0\ \text{rev/s}$ gives $\omega=12.6\ \text{rad/s}$ and $a=12.6\ \text{m/s}^2$, so the readings are comfortably above sensor noise. Good data give a straight line with intercept close to zero; a non-zero intercept usually means the phone's acceleration axis was not exactly radial.

๐Ÿงช Self-check quiz

0/5 correct

Q1. In uniform circular motion the acceleration points

Q2. A car's speed on a roundabout doubles. Its centripetal acceleration becomes

Q3. A body goes round a circle once every $0.50\ \text{s}$. Its angular speed is

Q4. The angle between the velocity and the acceleration of a body in uniform circular motion is

Q5. A cyclist rides round a circular track at a constant $8\ \text{m/s}$. Which statement is true?

โ† 3.3 Motion in a Plane with Constant Acceleration and Projectile MotionChapter 4 โ€” Laws of Motion โ†’