JEE Mains Physics · Motion in a Plane

Dynamics of Circular Motion: Roads, Strings and Vertical Circles

Circular motion needs a net force mv²/r towards the centre, and some real force has to supply it: friction on a level road, the slope of a banked road, the tension in a string, the pull of a spring or the push of a wall; in a vertical circle gravity joins in and the speed changes from top to bottom.

All Motion in a Plane notes JEE Mains Physics strategy

Why this matters

Thirty-two PYQs, twenty-seven of them multiple choice, and three from 2026: the largest page in the chapter. Nine are about roads, turntables and rotor drums, where friction or banking supplies the force. Thirteen use a string, a spring or a wall in a horizontal circle, including the conical pendulum. Ten are vertical circles, loops and smooth surfaces, where energy conservation and the centripetal equation are used together. The first step is always the same: name the force that points at the centre.

Concept 1 of 3: Friction and banking on a curve

On a level road the only sideways force on a car is friction, so friction has to supply mv²/r, and it cannot give more than μmg. That caps the speed at √(μrg), whatever the mass. Banking the road tilts the normal force so that part of it points at the centre; at one speed this alone is enough and no friction is needed. Friction then lets the car go somewhat faster or slower than that speed.

Definition

Level road or turntable: $\mu mg \ge \dfrac{mv^{2}}{r}$ , so $v_{\max} = \sqrt{\mu rg}$ and $\omega_{\max} = \sqrt{\mu g/r}$ ; mass cancels.
Banked at θ, no friction needed: $\tan\theta = \dfrac{v^{2}}{rg}$ .
Railway track, rail gap l: outer rail raised by $h \approx l\tan\theta = \dfrac{lv^{2}}{rg}$ .
Banked with friction: $v_{\max}^{2} = rg\,\dfrac{\tan\theta + \mu}{1 - \mu\tan\theta}$ , $v_{\min}^{2} = rg\,\dfrac{\tan\theta - \mu}{1 + \mu\tan\theta}$ .
Rotor drum: the wall's normal force $N = m\omega^{2}R$ supplies the centripetal force and friction $\mu N$ holds the weight, so $\omega_{\min} = \sqrt{\dfrac{g}{\mu R}}$ .
A downward aerodynamic force $F_L$ adds to the normal force: $\mu(mg + F_L) = \dfrac{mv^{2}}{R}$ .

Roads and turntables

v_{\max} = \sqrt{\mu rg} \qquad \tan\theta = \frac{v^{2}}{rg} \qquad v_{\max}^{2} = rg\,\frac{\tan\theta + \mu}{1 - \mu\tan\theta}

Worked example

A level curve has radius 80 m and

\mu = 0.5

(g = 10\ \text{m/s}^{2})

. (a) Find the greatest safe speed. (b) If the curve is instead banked at

37^{\circ}

(\tan 37^{\circ} = 0.75)

with a smooth surface, at what speed can it be taken?

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 6 April 2024 · Q97Moderate

Example 1 · Motion in a Plane · Dynamics of Circular Motion: Roads, Strings and Vertical Circles

A car of

800\text{ }kg

is taking turn on a banked road of radius

300\text{ }m

and angle of banking

30^{\circ}

. If coefficient of static friction is 0.2 then the maximum speed with which car can negotiate the turn safely :

\left( g = 10\text{ }m/s^{2},\sqrt{3}= 1.73 \right)

Putting the mass into the safe speed

v_max = √(μrg) has no mass in it: a heavier car needs more force but also gets more friction. A changed mass alone changes nothing.

Friction with the wrong sign on a banked road

At the greatest speed the car tends to slide outwards, so friction acts down the slope: (tan θ + μ)/(1 − μ tan θ). At the least speed it acts up the slope and both signs flip.

Drill 8 more on friction and banking on a curve

Concept 2 of 3: String, spring and wall in a horizontal circle

For a stone whirled on a string on a smooth table, the tension is the centripetal force. Replace the string with a spring and the circle grows, because the spring must stretch to pull; the radius is the natural length plus the stretch. Hang the string from a point instead and it tilts: its vertical part holds up the weight and its horizontal part turns the bob, which is the conical pendulum.

Definition

String on a smooth table: $T = \dfrac{mv^{2}}{L} = m\omega^{2}L$ ; the string breaks when this exceeds its limit.
Spring of natural length $l_0$ : $kx = m\omega^{2}(l_0 + x)$ , so $x = \dfrac{m\omega^{2}l_0}{k - m\omega^{2}}$ .
Conical pendulum, string L at θ to the vertical: $T\cos\theta = mg$ , $T\sin\theta = m\omega^{2}L\sin\theta$ , so $T = m\omega^{2}L$ , $\tan\theta = \dfrac{v^{2}}{rg}$ and $\omega^{2} = \dfrac{g}{L\cos\theta}$ .
A bob hung in a car turning on a curve takes the same tilt: $\tan\theta = v^{2}/rg$ .
A wall or groove pushing inwards: $N = \dfrac{mv^{2}}{r}$ , so $N \propto v^{2}$ .
A liquid of mass M filling a tube of length L rotated about one end presses on the far end with $F = \dfrac{M\omega^{2}L}{2}$ (its centre of mass is at L/2).

Horizontal circles

T = m\omega^{2}L \qquad kx = m\omega^{2}(l_0 + x) \qquad \tan\theta = \frac{v^{2}}{rg}\ (\text{conical pendulum})

Worked example

A 0.5 kg block on a smooth table is tied to a spring of force constant 20 N/m and natural length 30 cm, whose other end is fixed. The block goes round the fixed end at 4 rad/s. Find the stretch, the radius and the spring force.

Practice this conceptself-check · 3 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 6 April 2023 · Q7Moderate

Example 2 · Motion in a Plane · Dynamics of Circular Motion: Roads, Strings and Vertical Circles

A small block of mass

100\text{ }g

is tied to a spring of spring constant

7.5\text{ }N/m

and length

20\text{ }cm

. The other end of spring is fixed at a particular point A. If the block moves in a circular path on a smooth horizontal surface with constant angular velocity

5rad/s

about point

A

, then tension in the spring is

Using the natural length as the radius

A spring stretches to provide the pull, so the radius is l₀ + x. Writing kx = mω²l₀ drops the stretch from the radius and gives the wrong extension.

Radius or string length in a conical pendulum

The bob moves on a circle of radius L sin θ, but the tension comes out as mω²L, with the full string length. Mixing the two gives a tension too small by sin θ.

Drill 12 more on string, spring and wall in a horizontal circle

Concept 3 of 3: Motion in a vertical circle

In a vertical circle the speed is not constant: the body slows as it climbs and speeds up as it falls, so energy conservation links the speeds at different points. At each point the centripetal equation then gives the tension or the normal force. At the bottom the string must hold the weight as well as turn the body; at the top gravity helps, so the tension is least there and the string goes slack first.

Definition

Bottom: $T_b = \dfrac{mv_b^{2}}{r} + mg$ . Top: $T_t = \dfrac{mv_t^{2}}{r} - mg$ . At angle θ from the bottom: $T = \dfrac{mv^{2}}{r} + mg\cos\theta$ .
Energy: $v^{2} = v_b^{2} - 2gr(1 - \cos\theta)$ ; $v_b^{2} = v_t^{2} + 4gr$ ; $T_b - T_t = 6mg$ .
Just completing the circle on a string: $T_t = 0$ , so $v_t = \sqrt{gr}$ and $v_b = \sqrt{5gr}$ .
Slack above the centre, string at φ above the horizontal: $T = 0$ when $mg\sin\varphi = \dfrac{mv^{2}}{r}$ .
Loop of radius R after a smooth slope from height h: $mgh = mg(2R) + \tfrac{1}{2}mv_t^{2}$ and $N_t + mg = \dfrac{mv_t^{2}}{R}$ ; the least height is $\dfrac{5R}{2}$ .
Sliding from the top of a smooth hemisphere: it leaves the surface where $\cos\theta = \dfrac{2}{3}$ , at height $\dfrac{2R}{3}$ above the centre.

Vertical circle on a string

T_b = \frac{mv_b^{2}}{r} + mg,\ T_t = \frac{mv_t^{2}}{r} - mg \qquad v_b^{2} = v_t^{2} + 4gr \qquad v_b \ge \sqrt{5gr}

Worked example

A 0.2 kg ball on a 1 m string moves in a vertical circle with a speed of 8 m/s at the bottom

(g = 10\ \text{m/s}^{2})

. Find the tension at the bottom and at the top, and check that it completes the circle.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 29 Jan 2025 · Q16Moderate

Example 3 · Motion in a Plane · Dynamics of Circular Motion: Roads, Strings and Vertical Circles

A body of mass ' m ' connected to a massless and unstretchable string goes in a vertical circle of radius ' R ' under gravity g . The other end of the string is fixed at the center of circle. If velocity at top of circular path is

n\sqrt{gR}

, where,

n \geq 1

, then ratio of kinetic energy of the body at bottom to that at top of the circle is

Zero speed at the top

On a string, 'just completes the circle' means the tension is zero at the top, which needs a speed of √(gr) there. At zero speed the string would have gone slack well before the top.

Change in velocity as a change in speed

Between the bottom and the point where the string is horizontal, the velocity turns through 90°. The change in velocity is √(v₁² + v₂²), not the difference of the speeds.

Drill 9 more on motion in a vertical circle

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Friction and banking on a curve
Roads and turntables
$v_{\max} = \sqrt{\mu rg} \qquad \tan\theta = \frac{v^{2}}{rg} \qquad v_{\max}^{2} = rg\,\frac{\tan\theta + \mu}{1 - \mu\tan\theta}$
String, spring and wall in a horizontal circle
Horizontal circles
$T = m\omega^{2}L \qquad kx = m\omega^{2}(l_0 + x) \qquad \tan\theta = \frac{v^{2}}{rg}\ (\text{conical pendulum})$
Motion in a vertical circle
Vertical circle on a string
$T_b = \frac{mv_b^{2}}{r} + mg,\ T_t = \frac{mv_t^{2}}{r} - mg \qquad v_b^{2} = v_t^{2} + 4gr \qquad v_b \ge \sqrt{5gr}$

Watch out for (6)

Putting the mass into the safe speed
→ Friction and banking on a curve
Friction with the wrong sign on a banked road
→ Friction and banking on a curve
Using the natural length as the radius
→ String, spring and wall in a horizontal circle
Radius or string length in a conical pendulum
→ String, spring and wall in a horizontal circle
Zero speed at the top
→ Motion in a vertical circle
Change in velocity as a change in speed
→ Motion in a vertical circle

Test yourself on Motion in a Plane

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