Chapter 4 · Laws of Motion

4.4 Circular Motion and Solving Problems in Mechanics

Turning needs a net force towards the centre, $mv^2/r$, supplied by friction, tension or the tilt of a banked road. Then the full method: free-body diagrams for every body, linked by strings, pulleys and lifts.

A car takes a bend on a wet road a little too fast. The driver turns the wheel, but the car carries straight on towards the outside of the curve. What went wrong? Section 3.4 showed that anything moving on a circle has an acceleration $v^2/r$ towards the centre. Newton's second law now adds the missing piece: that acceleration needs a net force towards the centre, of size $mv^2/r$. On a level road the only thing that can supply it is friction between the tyres and the road — and friction has a limit. Above a certain speed the road simply cannot push the car sideways hard enough. That is why race tracks and highway curves are tilted, or banked: tilting the road lets the normal force help to turn the car.

The second half of this section puts the whole chapter to work. Real problems often involve several bodies at once — two blocks tied by a string over a pulley, a lift and its passenger, a train of wagons. The method is always the same: one free-body diagram for each body, Newton's second law for each, and the conditions that link them (the same string, the same acceleration). With that method you can solve almost any problem in Class 11 mechanics.

Dynamics of circular motion

📘 Definition 4.8 — Centripetal force

The centripetal force on a body moving in a circle is the net force on it directed towards the centre of the circle. It is not a new kind of force: it is the resultant of the real forces acting (tension, gravity, friction, normal force, …), or of their components towards the centre.

📐 Result 4.9 — Dynamics of uniform circular motion

A body of mass $m$ moving uniformly in a circle of radius $r$ with speed $v$ (angular speed $\omega$) needs a net force towards the centre

$$F_c=\frac{mv^2}{r}=m\omega^2r .$$

This follows from $\mathbf{F}_\text{net}=m\mathbf{a}$ with the centripetal acceleration of Result 3.11. If the available inward force becomes smaller than $mv^2/r$, the body moves off along a wider path; if the inward force vanishes, it moves off along the tangent (first law).

So in a circular-motion problem, do not add an extra force called "centripetal force" to the free-body diagram. Draw the real forces, choose one axis pointing towards the centre, and write $\sum F_\text{towards centre}=\dfrac{mv^2}{r}$. Nor is there an outward "centrifugal force" in the ground frame: the feeling of being thrown outward on a bend is your body's inertia — it tends to go straight on while the car turns under it.

✏️ Example 1 — A stone on a string

A $0.50\ \text{kg}$ stone is whirled in a horizontal circle of radius $1.2\ \text{m}$ on a smooth table, tied to a string fixed at the centre. It makes $2.0$ revolutions per second. (a) Find the tension in the string. (b) The string breaks at $150\ \text{N}$. What is the greatest number of revolutions per second possible?

💡 Show solution▶

FBD Vertical: weight and the table's normal force cancel. Horizontal: only the tension, towards the centre — it alone provides $mv^2/r$.

(a) $\omega=2\pi\nu=4.0\pi=12.6\ \text{rad/s}$; $T=m\omega^2r=0.50\times(12.57)^2\times1.2=94.7\ \text{N}$.

(b) $\omega_\text{max}=\sqrt{\dfrac{T_\text{max}}{mr}}=\sqrt{\dfrac{150}{0.60}}=15.8\ \text{rad/s}$, so $\nu_\text{max}=\dfrac{15.8}{2\pi}=2.5\ \text{rev/s}$.

Check $T\propto\nu^2$: $94.7\times\left(\tfrac{2.52}{2.0}\right)^2=150\ \text{N}$ ✓. When the string breaks the stone flies off along the tangent, not outward along the string.

Cars on level and banked curves

Level road. On a flat curve the forces on a car are its weight $mg$ down, the normal force $N=mg$ up, and friction from the road on the tyres. Only friction is horizontal, so friction must provide $mv^2/r$. The tyres are not sliding sideways, so this is static friction, at most $\mu_sN=\mu_smg$. The car can follow the curve only if $\dfrac{mv^2}{r}\le\mu_smg$.

Banked road. Now let the road be tilted at angle $\theta$, with its outer edge higher. The normal force is perpendicular to the road, so it leans towards the centre and its horizontal component $N\sin\theta$ helps to turn the car. At one particular speed this is enough on its own and no friction is needed at all. Above that speed friction acts down the slope (the car tends to slide outward and up); below it, friction acts up the slope.

To find the greatest speed, take friction at its limit, $\mu_sN$, down the slope. Resolve horizontally (towards the centre) and vertically:

$$N\sin\theta+\mu_sN\cos\theta=\frac{mv^2}{r},$$ $$N\cos\theta-\mu_sN\sin\theta=mg .$$

Dividing the first equation by the second removes $N$ and $m$: $\dfrac{v^2}{rg}=\dfrac{\sin\theta+\mu_s\cos\theta}{\cos\theta-\mu_s\sin\theta}=\dfrac{\tan\theta+\mu_s}{1-\mu_s\tan\theta}$.

📐 Result 4.10 — Speed limits on level and banked curves

For a car on a curve of radius $r$ with coefficient of static friction $\mu_s$ between tyres and road:

Level road:

$$v_\text{max}=\sqrt{\mu_s\,r\,g}.$$

Road banked at angle $\theta$: the ideal (optimum) speed, needing no friction, is

$$v_0=\sqrt{r\,g\tan\theta},$$

and the safe range of speeds is $v_\text{min}\le v\le v_\text{max}$ with

$$v_\text{max}=\sqrt{r\,g\,\frac{\mu_s+\tan\theta}{1-\mu_s\tan\theta}},\qquad v_\text{min}=\sqrt{r\,g\,\frac{\tan\theta-\mu_s}{1+\mu_s\tan\theta}}\ \ (v_\text{min}=0\text{ if }\tan\theta\le\mu_s).$$

None of these depends on the mass of the car. (If $\mu_s\tan\theta\ge1$ there is no upper limit.)

✏️ Example 2 — A flat roundabout

A car rounds a level roundabout of radius $50\ \text{m}$. (a) If $\mu_s=0.60$ (dry road), what is the greatest safe speed? (b) What is it on a wet road with $\mu_s=0.30$? (c) Would a heavier lorry be able to go faster?

💡 Show solution▶

(a) $v_\text{max}=\sqrt{0.60\times50\times9.8}=\sqrt{294}=17.1\ \text{m/s}$, about $62\ \text{km/h}$.

(b) $\sqrt{0.30\times50\times9.8}=12.1\ \text{m/s}$ ($44\ \text{km/h}$): halving $\mu_s$ cuts the safe speed by a factor $\sqrt2$, not 2.

(c) No: the mass cancels. A heavier vehicle needs more inward force, but its greater weight presses it harder onto the road, giving proportionally more friction.

Check At $17.1\ \text{m/s}$ the needed acceleration is $\dfrac{294}{50}=5.88\ \text{m/s}^2=0.60g$ ✓.

✏️ Example 3 — A banked curve

A highway curve of radius $100\ \text{m}$ is banked at $15^\circ$, and $\mu_s=0.30$ between tyres and road. Find (a) the ideal speed, (b) the greatest safe speed, (c) the least safe speed.

💡 Show solution▶

(a) $\tan15^\circ=0.268$: $v_0=\sqrt{100\times9.8\times0.268}=16.2\ \text{m/s}$ ($58\ \text{km/h}$).

(b) $v_\text{max}=\sqrt{980\times\dfrac{0.30+0.268}{1-0.30\times0.268}}=\sqrt{980\times0.6175}=24.6\ \text{m/s}$ ($89\ \text{km/h}$).

Check On the same curve unbanked, $v_\text{max}=\sqrt{0.30\times980}=17.1\ \text{m/s}$; banking raises the limit by $43\%$. These are the default settings of Interactive 1.

A cross-section of a curve, with the centre of the circle to the left. On the $1000\ \text{kg}$ car act its weight (black), the normal force (green) and static friction (blue), drawn to scale; their sum, the dashed red arrow, must point horizontally towards the centre with size $mv^2/r$. The bar above shows the safe range of speeds and the ideal speed $v_0$. Set the bank angle to $0^\circ$ for a level road; press Play to sweep the speed and watch friction change direction at $v_0$ and run out at the limits.

radius:100 mbank $\theta$:15°$\mu_s$:0.30speed:20 m/s

ideal speed $v_0$—

safe range—

friction needed / limit—

result—

Solving problems: connected bodies, pulleys and lifts

🧭 Problem-solving strategy — Systems of bodies

Draw a separate free-body diagram for each body (or for any group of bodies that moves together). A string's tension appears on both bodies it connects, pulling each towards the string.
Write the constraints. Bodies joined by a taut, inextensible string have accelerations of equal size; over an ideal pulley the tension is the same on both sides.
Choose a positive direction for each body consistent with the motion — for example, "down" for the heavier hanging mass and "up" for the lighter one — so that one symbol $a$ serves for all.
Apply $\sum F=ma$ to each body along its direction of motion, giving as many equations as unknowns. Add them to eliminate the tension.
Check a special case (equal masses, zero friction, $\theta=0$) and that tensions come out positive.

The simplest pulley system is the Atwood machine: two masses $m_1\gt m_2$ hung on either side of an ideal pulley. Taking $m_1$ moving down: $m_1g-T=m_1a$ and $T-m_2g=m_2a$. Adding eliminates $T$.

📐 Result 4.11 — Atwood machine

For masses $m_1\gt m_2$ connected by an ideal string over an ideal pulley, the heavier mass descends with acceleration

$$a=\frac{m_1-m_2}{m_1+m_2}\,g,$$

and the tension in the string is

$$T=\frac{2m_1m_2}{m_1+m_2}\,g .$$

$T$ lies between $m_2g$ and $m_1g$: less than the weight of the falling mass (so it accelerates down) and more than that of the rising one (so it accelerates up).

✏️ Example 4 — An Atwood machine

Masses of $3.0\ \text{kg}$ and $2.0\ \text{kg}$ hang over a light frictionless pulley and are released from rest. Find (a) the acceleration, (b) the tension, (c) the time for each mass to move $1.0\ \text{m}$, (d) the force on the pulley's support.

💡 Show solution▶

(a) $a=\dfrac{3.0-2.0}{5.0}\times9.8=1.96\ \text{m/s}^2$ (the $3.0\ \text{kg}$ mass down).

(b) $T=\dfrac{2\times3.0\times2.0}{5.0}\times9.8=23.5\ \text{N}$.

(d) The string pulls down on the pulley on both sides: $2T=47.0\ \text{N}$ — less than the total weight $49\ \text{N}$, because the system's centre is accelerating downward.

Check $3.0\,\text{kg}$: $29.4-23.5=5.9=3.0\times1.96$ ✓; $2.0\,\text{kg}$: $23.5-19.6=3.9=2.0\times1.96$ ✓.

✏️ Example 5 — A block on a table pulled by a hanging mass

A $4.0\ \text{kg}$ block on a horizontal table is tied by a light string, passing over a light pulley at the table's edge, to a $1.0\ \text{kg}$ mass hanging freely. For the block and table $\mu_s=0.22$ and $\mu_k=0.20$. (a) Does the system move from rest? (b) Find the acceleration and the tension.

💡 Show solution▶

(a) To stay at rest, friction on the block would have to equal the hanging weight, $9.8\ \text{N}$. The limit is $\mu_sN=0.22\times39.2=8.6\ \text{N}\lt9.8\ \text{N}$, so it moves.

Equations Block: $T-\mu_kN=4.0\,a$ with $\mu_kN=0.20\times39.2=7.84\ \text{N}$. Hanging mass: $9.8-T=1.0\,a$.

(b) Adding: $9.8-7.84=5.0\,a$, so $a=0.392\ \text{m/s}^2$; then $T=9.8-0.392=9.41\ \text{N}$.

Check $T\lt mg$ for the hanging mass, as it must be for it to accelerate down. With a smooth table, $a=\tfrac{9.8}{5.0}=1.96\ \text{m/s}^2$ — try both in Interactive 2.

✏️ Example 6 — Apparent weight in a lift

A $60\ \text{kg}$ student stands on a bathroom scale in a lift. What does the scale read (in newtons, and as the "mass" it displays) when the lift (a) moves up at constant speed, (b) accelerates upward at $2.0\ \text{m/s}^2$, (c) accelerates downward at $2.0\ \text{m/s}^2$, (d) falls freely after its cable snaps?

💡 Show solution▶

FBD On the student: weight $588\ \text{N}$ down, normal force $N$ from the scale up. The scale reads $N$ (by the third law, she presses on it with $N$). With up positive: $N-mg=ma$, so $N=m(g+a)$.

(a), (b) $a=0$: $N=588\ \text{N}$ ($60\ \text{kg}$). $a=+2.0$: $N=60\times11.8=708\ \text{N}$ (displays $72.2\ \text{kg}$).

(d) $a=-g$: $N=0$. She and the scale fall together and the scale reads zero — she feels "weightless", though gravity still acts on her.

Check Only the acceleration matters, not the velocity: a lift moving down but slowing down has $a$ upward and the scale reads more than $588\ \text{N}$.

Choose a set-up, set the masses (and the friction coefficient $\mu$ of the table, taken here as both static and kinetic), and press Play. On each mass the forces are drawn to scale: weight (black), tension (purple) and friction (blue). The acceleration and tension come from $\sum F=ma$ written for each body, exactly as in Examples 4 and 5. Look for the case where nothing moves, and notice that the tension is never equal to the hanging weight while things accelerate.

set-up:$m_1$:3 kg$m_2$:2 kg$\mu$:0.20

acceleration—

tension—

friction on $m_1$—

time to move 1 m—

⚠️ Common mistakes

1. Adding a "centripetal force" or "centrifugal force" to the diagram. The centripetal force is the net inward force of the real forces; in the ground frame there is no outward force on a body going round.

2. Using $N=mg$ on a banked road. On a banked curve $N$ is found from the two equations together; it is larger than $mg\cos\theta$ when friction acts down the slope.

3. Taking the tension equal to the hanging weight. $T=mg$ only if the hanging mass is not accelerating. In an Atwood machine $T$ lies strictly between the two weights.

4. Inconsistent signs for connected bodies. Choose each body's positive direction along the way the system moves, so one $a$ serves for all; a negative answer then means the system moves the other way.

5. Confusing velocity and acceleration in lifts. The scale reading depends only on the acceleration: $N=m(g+a)$ with $a$ measured upward.

Exercises

Take $g=9.8\ \text{m/s}^2$. Strings and pulleys are light and frictionless unless stated otherwise.

Dynamics of circular motion.

1 A $0.20\ \text{kg}$ ball on a string moves in a horizontal circle of radius $0.80\ \text{m}$ on a smooth table at $4.0\ \text{m/s}$. Find the tension.
2 A string breaks at a tension of $50\ \text{N}$. What is the greatest speed at which a $0.50\ \text{kg}$ mass can be whirled on it in a horizontal circle of radius $1.0\ \text{m}$ (smooth table)?
3 A $1000\ \text{kg}$ car rounds a level curve of radius $40\ \text{m}$ at $12\ \text{m/s}$. Find the friction force needed and the least $\mu_s$ that provides it.
4 Find the greatest safe speed on a level curve of radius $80\ \text{m}$ with $\mu_s=0.50$.
5 A coin lies on a turntable $10\ \text{cm}$ from the axis; $\mu_s=0.30$. Find the greatest angular speed (in rad/s and rpm) at which it does not slide.
6 Passengers in a car taking a sharp left bend feel pushed to the right. Explain this in terms of inertia.
7 A $1500\ \text{kg}$ car goes over the top of a hump-backed bridge (a circular arc of radius $20\ \text{m}$) at $10\ \text{m/s}$. Find the normal force on it at the top, and the speed at which it would just lose contact.

💡 Show solutions▶

1 $T=\dfrac{mv^2}{r}=\dfrac{0.20\times16}{0.80}=4.0\ \text{N}$.

2 $v=\sqrt{\dfrac{Tr}{m}}=\sqrt{\dfrac{50\times1.0}{0.50}}=10\ \text{m/s}$.

3 $f=\dfrac{1000\times144}{40}=3.6\times10^3\ \text{N}$; $\mu_s\ge\dfrac{3600}{9800}=0.37$.

4 $v=\sqrt{0.50\times80\times9.8}=19.8\ \text{m/s}$.

5 $m\omega^2r\le\mu_smg$: $\omega_\text{max}=\sqrt{\dfrac{0.30\times9.8}{0.10}}=5.4\ \text{rad/s}$, i.e. $\dfrac{5.42\times60}{2\pi}=52\ \text{rpm}$.

6 Their bodies tend to keep moving in a straight line (first law) while the car turns left beneath them; the seat and door must push them leftward to make them turn. No outward force acts.

7 At the top the centre is below: $mg-N=\dfrac{mv^2}{r}$, so $N=1500(9.8-5.0)=7.2\times10^3\ \text{N}$. Contact is lost when $N=0$: $v=\sqrt{rg}=\sqrt{196}=14\ \text{m/s}$.

Banked roads.

1 A curve of radius $60\ \text{m}$ is banked at $12^\circ$. Find the speed at which no friction is needed.
2 At what angle should a curve of radius $100\ \text{m}$ be banked for an ideal speed of $20\ \text{m/s}$?
3 A curve of radius $80\ \text{m}$ is banked at $10^\circ$ with $\mu_s=0.25$. Find the greatest safe speed.
4 For the curve in part 3, find the least safe speed.
5 Give two reasons why highway curves are banked instead of relying on friction alone.
6 On a railway curve of radius $500\ \text{m}$ the outer rail is raised so that a train at $20\ \text{m/s}$ needs no sideways push from the rails. The rails are $1.5\ \text{m}$ apart. By how much is the outer rail raised?

💡 Show solutions▶

1 $v_0=\sqrt{60\times9.8\times\tan12^\circ}=\sqrt{125}=11.2\ \text{m/s}$.

2 $\tan\theta=\dfrac{v^2}{rg}=\dfrac{400}{980}=0.408$: $\theta=22.2^\circ$.

3 $\tan10^\circ=0.176$: $v_\text{max}=\sqrt{784\times\dfrac{0.25+0.176}{1-0.25\times0.176}}=\sqrt{350}=18.7\ \text{m/s}$.

4 $\tan10^\circ\lt\mu_s$, so $v_\text{min}=0$: a car at rest on it does not slide down.

5 Friction is unreliable (wet or icy roads, worn tyres) and relying on it wears tyres; banking lets the normal force supply most or all of the inward force, so curves are safe at higher speeds.

6 $\tan\theta=\dfrac{400}{500\times9.8}=0.0816$, $\theta=4.67^\circ$; height $=1.5\sin\theta=0.12\ \text{m}$ ($12\ \text{cm}$).

III

Connected bodies, pulleys and lifts.

1 Masses of $5.0\ \text{kg}$ and $3.0\ \text{kg}$ hang over a pulley (Atwood machine). Find the acceleration and the tension.
2 In part 1, starting from rest, find the speed of the masses after $2.0\ \text{s}$ and the distance each has moved.
3 Blocks of $2.0\ \text{kg}$ and $3.0\ \text{kg}$ on a smooth table are joined by a string. A $10\ \text{N}$ horizontal force pulls the $3.0\ \text{kg}$ block. Find the acceleration and the tension.
4 A $6.0\ \text{kg}$ block on a smooth table is tied over a pulley at the edge to a $2.0\ \text{kg}$ hanging mass. Find the acceleration and the tension.
5 Repeat part 4 with $\mu_k=0.10$ between block and table (the system moves).
6 A $50\ \text{kg}$ person stands on a scale in a lift accelerating downward at $1.8\ \text{m/s}^2$. What does the scale read?
7 An $800\ \text{kg}$ lift accelerates upward at $1.2\ \text{m/s}^2$. Find the tension in its cable.
8 A $2.0\ \text{kg}$ block on a smooth $30^\circ$ incline is tied by a string, parallel to the incline and over a pulley at its top, to a $3.0\ \text{kg}$ mass hanging vertically. Find the acceleration and the tension.

💡 Show solutions▶

1 $a=\dfrac{2.0}{8.0}\times9.8=2.45\ \text{m/s}^2$; $T=\dfrac{2\times5.0\times3.0}{8.0}\times9.8=36.8\ \text{N}$.

2 $v=at=4.9\ \text{m/s}$; $s=\tfrac12at^2=4.9\ \text{m}$.

3 $a=\dfrac{10}{5.0}=2.0\ \text{m/s}^2$; the string pulls the $2.0\ \text{kg}$ block: $T=4.0\ \text{N}$.

4 $a=\dfrac{2.0\times9.8}{8.0}=2.45\ \text{m/s}^2$; $T=6.0\times2.45=14.7\ \text{N}$.

5 $a=\dfrac{19.6-0.10\times58.8}{8.0}=\dfrac{13.7}{8.0}=1.72\ \text{m/s}^2$; $T=2.0(9.8-1.72)=16.2\ \text{N}$.

6 $N=m(g-a)=50\times8.0=400\ \text{N}$ (it displays $40.8\ \text{kg}$).

7 $T=m(g+a)=800\times11.0=8.8\times10^3\ \text{N}$.

8 The hanging mass descends: $29.4-T=3.0a$ and $T-2.0\times9.8\sin30^\circ=2.0a$. Adding: $29.4-9.8=5.0a$, $a=3.92\ \text{m/s}^2$; $T=3.0(9.8-3.92)=17.6\ \text{N}$.

Harder problems and modelling.

1 In an Atwood machine one mass is twice the other. Show that the acceleration is $g/3$ and the tension is $\tfrac43$ of the weight of the lighter mass.
2 Three blocks of $1.0$, $2.0$ and $3.0\ \text{kg}$ lie in a row on a smooth table, joined by strings. A $12\ \text{N}$ force pulls the $3.0\ \text{kg}$ block along the line. Find the acceleration and the tension in each string.
3 A uniform rope of mass $2.0\ \text{kg}$ pulls an $8.0\ \text{kg}$ block across a smooth floor; the free end is pulled with $20\ \text{N}$. Find the acceleration and the tension at the middle of the rope.
4 A $60\ \text{kg}$ climber goes up a rope with an acceleration of $1.0\ \text{m/s}^2$. Find the tension. If the rope can bear at most $700\ \text{N}$, what is the greatest upward acceleration she can safely have?
5 A $1200\ \text{kg}$ car takes the curve of Example 3 at $22\ \text{m/s}$. Find the normal force and the friction (magnitude and direction), and check that it does not slide.
6 (Modelling) A jet pulls out of a dive along a vertical circle of radius $500\ \text{m}$ at $100\ \text{m/s}$. At the lowest point, how many times her normal weight does the $60\ \text{kg}$ pilot's seat push on her?

💡 Show solutions▶

1 $m_1=2m$, $m_2=m$: $a=\dfrac{m}{3m}g=\dfrac g3$; $T=\dfrac{2(2m)(m)}{3m}g=\tfrac43mg$.

2 $a=\dfrac{12}{6.0}=2.0\ \text{m/s}^2$. The string between the $3.0$ and $2.0\ \text{kg}$ blocks pulls $3.0\ \text{kg}$: $6.0\ \text{N}$; between the $2.0$ and $1.0\ \text{kg}$ blocks: $2.0\ \text{N}$.

3 $a=\dfrac{20}{10}=2.0\ \text{m/s}^2$; the middle of the rope pulls the block and half the rope: $T=9.0\times2.0=18\ \text{N}$. (The tension varies along a rope that has mass.)

4 $T=60(9.8+1.0)=648\ \text{N}$; $a_\text{max}=\dfrac{700}{60}-9.8=1.87\ \text{m/s}^2$.

5 With $f$ down the slope: $N=m\left(g\cos\theta+\tfrac{v^2}{r}\sin\theta\right)=1.29\times10^4\ \text{N}$, $f=m\left(\tfrac{v^2}{r}\cos\theta-g\sin\theta\right)=2.6\times10^3\ \text{N}$ down the slope. $f/N=0.20\lt0.30$: no sliding (as expected, $22\lt24.6\ \text{m/s}$).

6 $N-mg=\dfrac{mv^2}{r}$: $N=m(9.8+20)=29.8m$, i.e. $\dfrac{29.8}{9.8}=3.0$ times her weight ("$3g$").

🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board · 5 marks A car moves on a road banked at angle $\theta$, round a curve of radius $r$; the coefficient of static friction is $\mu_s$. Draw the free-body diagram of the car moving at its greatest safe speed and derive an expression for that speed. Deduce the speed at which no friction is needed.

💡 Show solution▶

FBD Weight $mg$ down; normal force $N$ perpendicular to the road; friction $\mu_sN$ along the road, down the slope (the car tends to slide outward). [1]

Equations Vertical: $N\cos\theta=mg+\mu_sN\sin\theta$. Horizontal (towards the centre): $N\sin\theta+\mu_sN\cos\theta=\dfrac{mv^2}{r}$. [2]

Result Dividing: $\dfrac{v^2}{rg}=\dfrac{\sin\theta+\mu_s\cos\theta}{\cos\theta-\mu_s\sin\theta}$, so $v_\text{max}=\sqrt{rg\,\dfrac{\mu_s+\tan\theta}{1-\mu_s\tan\theta}}$. [1]

No friction Put $\mu_s=0$: $v_0=\sqrt{rg\tan\theta}$. [1]

Exam tip Mention that the result is independent of the mass, and that with $\theta=0$ it reduces to $\sqrt{\mu_srg}$ for a level road — a quick check examiners appreciate.

E2. NEET The greatest speed at which a car can round a level curve of radius $20\ \text{m}$ if $\mu_s=0.50$ ($g=9.8\ \text{m/s}^2$) is (a) $7.0\ \text{m/s}$ (b) $9.9\ \text{m/s}$ (c) $14\ \text{m/s}$ (d) $98\ \text{m/s}$

💡 Show solution▶

Formula $v_\text{max}=\sqrt{\mu_srg}=\sqrt{0.50\times20\times9.8}=\sqrt{98}=9.9\ \text{m/s}$.

Answer (b).

Exam tip (c) is $\sqrt{rg}$ (forgetting $\mu_s$); (d) forgets the square root.

E3. JEE Main NEET Masses of $5\ \text{kg}$ and $3\ \text{kg}$ are tied to the ends of a light string passing over a light frictionless pulley. Taking $g=10\ \text{m/s}^2$, the acceleration of the masses and the tension in the string are (a) $2.5\ \text{m/s}^2$, $37.5\ \text{N}$ (b) $2.5\ \text{m/s}^2$, $50\ \text{N}$ (c) $5.0\ \text{m/s}^2$, $37.5\ \text{N}$ (d) $2.5\ \text{m/s}^2$, $30\ \text{N}$

💡 Show solution▶

Acceleration $a=\dfrac{5-3}{5+3}\times10=2.5\ \text{m/s}^2$.

Tension $T=\dfrac{2\times5\times3}{8}\times10=37.5\ \text{N}$ (check: $3(10+2.5)=37.5$ ✓).

Answer (a).

Exam tip The tension must lie between the two weights, $30$ and $50\ \text{N}$ — which rules out (b) and (d) at a glance.

E4. JEE Main (Numerical value) A $60\ \text{kg}$ man stands on a weighing machine in a lift that is moving downward and speeding up at $2.0\ \text{m/s}^2$. Taking $g=10\ \text{m/s}^2$, find the reading of the machine in newtons.

💡 Show solution▶

Direction Moving down and speeding up: the acceleration is downward, $a=-2.0\ \text{m/s}^2$ (up positive).

Equation $N=m(g+a)=60(10-2.0)=480\ \text{N}$.

Answer $480$.

Exam tip "Moving down" alone tells you nothing; "moving down and slowing down" would mean an upward acceleration and a reading of $720\ \text{N}$.

E5. JEE Advanced (One or more correct) A curve of radius $100\ \text{m}$ is banked with $\tan\theta=0.20$, and $\mu_s=0.20$ between tyres and road. Take $g=10\ \text{m/s}^2$. Which statements are correct? (A) At about $14\ \text{m/s}$ a car needs no friction to take the curve. (B) A car can stand still on the curve without sliding down. (C) The greatest safe speed is about $20\ \text{m/s}$. (D) At speeds above the ideal speed, friction on the tyres acts up the slope.

💡 Show solution▶

(A) $v_0=\sqrt{100\times10\times0.20}=\sqrt{200}=14.1\ \text{m/s}$ ✓.

(B) $\tan\theta=0.20\le\mu_s=0.20$, so $v_\text{min}=0$: the car is just able to stand still ✓.

(D) Above $v_0$ the car tends to slide outward, up the slope, so friction acts down the slope ✗.

Answer (A), (B), (C).

E6. Board · Case study A $4.0\ \text{kg}$ block on a rough horizontal table ($\mu_k=0.20$) is connected by a light string over a smooth pulley at the edge of the table to a $2.0\ \text{kg}$ mass hanging freely. The system is released and moves. (i) Find the acceleration. (2 marks) (ii) Find the tension in the string. (1 mark) (iii) What is the least coefficient of static friction for which the system would stay at rest? (1 mark)

💡 Show solution▶

(i) Block: $T-\mu_k(4.0)(9.8)=4.0a$; hanging mass: $19.6-T=2.0a$ [1]. Adding: $19.6-7.84=6.0a$, $a=1.96\ \text{m/s}^2$ [1].

(ii) $T=2.0(9.8-1.96)=15.7\ \text{N}$. [1]

(iii) At rest $T=19.6\ \text{N}$ and friction must balance it: $\mu_s\times39.2\ge19.6$, so $\mu_s\ge0.50$. [1]

Exam tip Write one equation per body and say which way each is taken as positive — that alone usually earns the first mark.

🔬 Research–Teaching NexusResearch-tutored · optional, about 45 minutes

Research question. How tight can a highway curve be for a given speed limit — and why do road engineers never rely on all the friction a tyre can give?

Why it matters. Highway design codes around the world (for example those used by national road agencies in India, Europe and North America) set a minimum curve radius for each design speed. They combine the bank angle — written as the "superelevation" $e=\tan\theta$, usually capped at about $0.07$ so that slow vehicles and trucks do not slide inward on icy or wet days — with a "side-friction factor" $f$ of only about $0.12$–$0.16$, far below a dry tyre's $\mu_s\approx0.8$. The margin keeps curves safe on wet roads, with worn tyres and for drivers who brake while turning.

What to do.

For small angles, Result 4.10 gives approximately $\dfrac{v^2}{rg}=e+f$. Show this by putting $\mu_s\tan\theta\approx0$ in the formula for $v_\text{max}$.
Find the minimum radius for design speeds of $50$, $80$ and $100\ \text{km/h}$, with $e=0.07$ and $f=0.14$.
Recalculate the $80\ \text{km/h}$ radius using a dry-road $\mu_s=0.80$ instead of $f$. By what factor is the code more cautious?
Find a sharp curve near you (on an online map, measure the radius by fitting a circle to three points). What speed does the formula allow? Compare with the posted limit.

Tools: Spreadsheet; an online map with a distance-measuring tool.

Hand in: A table of design speed against minimum radius, your check on a real curve, and a paragraph explaining to a town council why the speed limit on that curve is (or is not) appropriate.

Going further. Racing circuits bank some curves at over $30^\circ$. Use Interactive 1 to find how fast a car can go on a $30^\circ$ curve of radius $100\ \text{m}$ with racing tyres ($\mu_s=1.0$). What happens to the formula when $\mu_s\tan\theta\ge1$?

Instructor note: with $e+f=0.21$, $r_\text{min}=\dfrac{v^2}{0.21g}$ gives $94\ \text{m}$ ($50\ \text{km/h}$), $240\ \text{m}$ ($80\ \text{km/h}$) and $375\ \text{m}$ ($100\ \text{km/h}$). With $\mu_s=0.80$ and $e=0.07$ the full formula gives about $55\ \text{m}$ for $80\ \text{km/h}$, so the code is roughly four times more cautious. For the racing curve, $\mu_s\tan30^\circ=0.577\lt1$ and $v_\text{max}=\sqrt{980\times1.577/0.423}=60\ \text{m/s}$ ($217\ \text{km/h}$); when $\mu_s\tan\theta\ge1$ friction and banking together could supply any inward force — there is no speed limit in this model (real cars then hit other limits).

🧪 Self-check quiz

0/5 correct

Q1. A car rounds a level curve at constant speed. The force towards the centre of the curve is provided by

Q2. A road of radius $r$ is banked so that a car at speed $v$ needs no friction. The bank angle satisfies

Q3. In an Atwood machine the two masses are equal. If the system is given a small push, the masses

Q4. A person stands on a scale in a lift whose cable has snapped, so that the lift falls freely. The scale reads

Q5. A car's speed on a level curve is doubled. The friction force needed to keep it on the curve becomes

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