Physics
Laws of Motion
Circular Motion Dynamics – Centripetal Force

Laws of Motion · JEE & NEET Physics

Circular Motion Dynamics – Centripetal Force: notes and previous year questions

The centripetal force and what supplies it: strings, conical pendulums, flat and banked roads, vertical circles and the well of death.

34 JEE Main questions (2002–2026)

Circular Motion Dynamics – Centripetal Force in short

Circular motion needs a net inward (centripetal) force mv²/r.
The centripetal force is supplied by existing forces: tension, friction, gravity or the normal force.
Conical pendulum: tan θ = v²/(rg) and cos θ = g/(Lω²); faster spin, wider cone.
On a flat bend friction supplies the force, and v_max = √(μrg) whatever the mass.

1Centripetal force

A body moving in a circle accelerates toward the centre with $a_c = v^2/r = \omega^2 r$ . By the second law, the net force on it must point to the centre too:

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

Always toward the centre, at right angles to the velocity.

A 0.5 kg stone on a 2 m string at 4 m/s needs $T = 0.5 \times 16/2 = 4$ N. If the string breaks above $T_{\max}$ , the top speed is $v_{\max} = \sqrt{rT_{\max}/m}$ : with 50 N, $\sqrt{200} \approx 14.1$ m/s. (A real whirled string sags a little below the horizontal: that is the conical pendulum.)

2The conical pendulum

A bob on a string of length $L$ moves in a horizontal circle while the string makes angle $\theta$ with the vertical. Only the weight and the tension act. The vertical part of $T$ holds the bob up; its horizontal part pulls it to the centre:

T\cos\theta = mg,\quad T\sin\theta = \frac{mv^2}{r}

Radius of the circle r = L sin θ.

\tan\theta = \frac{v^2}{rg}

\cos\theta = \frac{g}{L\omega^2}

Faster spin, smaller cos θ, wider cone. It never reaches 90°.

T_{\text{period}} = 2\pi\sqrt{\frac{L\cos\theta}{g}}

Also v² = Lg sin θ tan θ.

3Turning on a flat road

On a flat bend only sideways static friction can pull the car inward: $f = mv^2/r$ . At most $f = \mu_s mg$ , so the car holds the bend only if $mv^2/r \le \mu_s mg$ :

v_{\max} = \sqrt{\mu_s rg}

The mass cancels: a heavy truck and a light car have the same top speed.

Examples: $r = 50$ m, $\mu = 0.4$ : $v_{\max} = \sqrt{200} \approx 14.1$ m/s. $r = 40$ m, $\mu = 0.4$ : $\sqrt{160} \approx 12.6$ m/s. Better tyres (bigger $\mu$ ) or a gentler bend (bigger $r$ ) allow more speed. Too fast, and the car skids off along the tangent.

4Banked roads

Tilt the road by $\theta$ . The normal force is perpendicular to the road, so it leans toward the centre. With no friction:

N\cos\theta = mg,\quad N\sin\theta = \frac{mv^2}{r}

\tan\theta = \frac{v^2}{rg}

Design (optimum) speed: v = √(rg tan θ). At that speed no friction is needed.

Examples: a 100 m bend for 20 m/s: $\tan\theta = 400/1000 = 0.4$ , $\theta \approx 21.8°$ . A 40 m bend for 20 m/s: $\tan\theta = 1$ , $\theta = 45°$ . A 90 m bend banked at 45° suits $\sqrt{900} = 30$ m/s. Banking makes turns safer at speed and saves tyre wear.

5Banking with friction

Speed	The car tends to	Friction acts	Effect
Too fast	slide up and out	down the slope	adds to the inward force
Design speed	stay put	not at all	N alone is enough
Too slow	slide down and in	up the slope	reduces the inward force

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

From N sin θ + f cos θ = mv²/r and N cos θ = mg + f sin θ, with f = μN.

v_{\min} = \sqrt{rg\,\frac{\tan\theta - \mu}{1 + \mu\tan\theta}}

Example: $r = 40$ m, $\theta = 15°$ ( $\tan\theta \approx 0.268$ ), $\mu = 0.2$ : design speed $\sqrt{400 \times 0.268} \approx 10.35$ m/s; $v_{\max} = \sqrt{400 \times 0.468/0.946} \approx 14.06$ m/s; $v_{\min} \approx 5.1$ m/s.

6Vertical circles

Swing a stone on a string in a vertical circle of radius r. Gravity now acts along or against the centre direction, and the speed changes round the loop: fastest at the bottom, slowest at the top.

Position	Equation	Tension
Bottom	$T - mg = mv^2/r$	$T = mg + mv^2/r$ (largest)
Side	$T = mv^2/r$	in between
Top	$T + mg = mv^2/r$	$T = mv^2/r - mg$ (smallest)

To get round, the string must stay tight at the top: $T \ge 0$ . At the limit $T = 0$ , gravity alone supplies the force, so $v_{\text{top}} = \sqrt{gr}$ . Energy conservation (next chapter) links the two ends: $v_{\text{bottom}}^2 = v_{\text{top}}^2 + 4gr$ .

v_{\text{top}} \ge \sqrt{gr},\qquad v_{\text{bottom}} \ge \sqrt{5gr}

Remember "1 and 5". The same holds for a car looping a track, with the normal force in place of T.

Inside a smooth bowl or sphere, measure $\theta$ from the lowest point: $N - mg\cos\theta = mv^2/r$ , with $v^2 = u^2 - 2gr(1 - \cos\theta)$ . The body can lose contact ( $N = 0$ ) only above the level of the centre, where $\cos\theta < 0$ .

7The well of death and centrifugal force

A rider circles on the inside of a vertical drum. The wall's normal force points to the axis and is the centripetal force: $N = mv^2/r$ . Friction on the wall, at most $\mu N$ , holds the rider up: $\mu N \ge mg$ .

v_{\min} = \sqrt{\frac{rg}{\mu}}

r = 5 m, μ = 0.5: v_min = √100 = 10 m/s. Slower, and the rider slides down.

Summary

Key ideas

Circular motion needs a net inward (centripetal) force mv²/r.
The centripetal force is supplied by existing forces: tension, friction, gravity or the normal force.
Conical pendulum: tan θ = v²/(rg) and cos θ = g/(Lω²); faster spin, wider cone.
On a flat bend friction supplies the force, and v_max = √(μrg) whatever the mass.
A road banked at tan θ = v²/(rg) needs no friction at that speed.
On a banked road, friction acts down the slope when too fast and up the slope when too slow.
In a vertical circle tension is largest at the bottom and smallest at the top.
To loop, a body needs at least √(gr) at the top and √(5gr) at the bottom.
In the well of death the wall's normal force is centripetal and friction holds the rider up: v ≥ √(rg/μ).
Centrifugal force is a pseudo force of the rotating frame.

Every equation

Centripetal force: $F_c = \frac{mv^2}{r} = m\omega^2 r = mv\omega$
Centripetal acceleration: $a_c = \frac{v^2}{r} = \omega^2 r$
String breaking speed: $v_{\max} = \sqrt{rT_{\max}/m}$
Conical pendulum: $\tan\theta = \frac{v^2}{rg}$
Conical pendulum (ω): $\cos\theta = \frac{g}{L\omega^2}$
Conical pendulum period: $T = 2\pi\sqrt{\frac{L\cos\theta}{g}}$
Flat bend: $v_{\max} = \sqrt{\mu rg}$
Banking angle: $\tan\theta = \frac{v^2}{rg}$
Design speed: $v = \sqrt{rg\tan\theta}$
Banked, fastest: $v_{\max} = \sqrt{rg\frac{\tan\theta + \mu}{1 - \mu\tan\theta}}$
Banked, slowest: $v_{\min} = \sqrt{rg\frac{\tan\theta - \mu}{1 + \mu\tan\theta}}$
Bottom of a vertical circle: $T = mg + \frac{mv^2}{r}$
Top of a vertical circle: $T = \frac{mv^2}{r} - mg$
Top and bottom speeds: $v_b^2 = v_t^2 + 4gr$
Least speeds: $v_t = \sqrt{gr},\ v_b = \sqrt{5gr}$
Inside a sphere: $N - mg\cos\theta = \frac{mv^2}{r}$
Well of death: $v_{\min} = \sqrt{rg/\mu}$

Previous year questions with solutions

Real JEE and NEET questions on circular motion dynamics – centripetal force. Try each one before you open the solution.

Q1JEE Main 2026One correct option

A car moving with a speed of $54 \mathrm{km}/h$ takes a turn of radius 20 m . A simple pendulum is suspended from the ceiling of the car. Determine the angle made by the string of the pendulum with the vertical during the turning. (Take $g=10 m/s^{2}$ )

A ${\tan}^{-1}(0.5)$
B ${\tan}^{-1}(0.75)$
C ${\tan}^{-1}(1.125)$
D ${\tan}^{-1}(0.25)$

Show answer and solution

Answer: Option C

Dividing the two equations, $\dfrac{T\sin\theta}{T\cos\theta} = \dfrac{mv^{2}/r}{mg}$ , so $\tan\theta = \dfrac{v^{2}}{rg}$ . First put the speed into metres per second: $54\ \mathrm{km/h} = 54 \times \dfrac{5}{18} = 15\ \mathrm{m/s}$ . Then $\tan\theta = \dfrac{15^{2}}{20 \times 10} = \dfrac{225}{200} = 1.125$ , so $\theta = \tan^{-1}(1.125)$ . The conversion is where this one is lost — left in $\mathrm{km/h}$ the speed is squared along with its wrong units, and the answer comes out about thirteen times too big.

Q2JEE Main 2026One correct option

A 0.5 kg mass is in contact against the inner wall of a cylindrical drum of radius 4 m rotating about its vertical axis. The minimum rotational speed of the drum to enable the mass to remain stuck to the wall (without falling) is 5 rad/s. The coefficient of friction between the drum’s inner wall surface and mass is _________. (Take $g=10 m/s^{2}$ )

A0.1
B0.5
C0.7
D0.3

Show answer and solution

Answer: Option A

The same drum read for $\mu$ instead of $\omega$ . At the minimum speed the inequality is an equality, so $\mu = \dfrac{g}{\omega^{2}R} = \dfrac{10}{5^{2} \times 4} = \dfrac{10}{100} = 0.1$ . The $0.5\ \mathrm{kg}$ never enters — it cancelled between the weight friction has to hold and the normal reaction that supplies the grip. A wall as slippery as this still works because $\omega^{2}R = 100\ \mathrm{m/s^{2}}$ is ten times $g$ , so a tenth of it is enough.

Q3JEE Main 2025One correct option

A car of mass ' $m$ ' moves on a banked road having radius ' $r$ ' and banking angle $\theta$ . To avoid slipping from banked road, the maximum permissible speed of the car is $v_{0}$ . The coefficient of friction $\mu$ between the wheels of the car and the banked road is

A $\mu =\frac{v_{0}^{2}+rg\tan \theta }{rg+v_{0}^{2}\tan \theta }$
B $\mu =\frac{v_{0}^{2}-rg\tan \theta }{\mathrm{rg}-v_{o}^{2}\tan \theta }$
C $\mu =\frac{v_{0}^{2}-rg\tan \theta }{rg+v_{0}^{2}\tan \theta }$
D $\mu =\frac{v_{o}^{2}+rg\tan \theta }{rg-v_{o}^{2}\tan \theta }$

Show answer and solution

Answer: Option C

The same formula with $\mu$ as the unknown instead of the speed. From $v_{0}^{2} = rg\dfrac{\tan\theta + \mu}{1 - \mu\tan\theta}$ , multiply out: $v_{0}^{2} - \mu v_{0}^{2}\tan\theta = rg\tan\theta + \mu rg$ . Collect the $\mu$ terms on one side, $v_{0}^{2} - rg\tan\theta = \mu\left(rg + v_{0}^{2}\tan\theta\right)$ , so $\mu = \dfrac{v_{0}^{2} - rg\tan\theta}{rg + v_{0}^{2}\tan\theta}$ . Two options go on sight: if $v_{0}^{2} = rg\tan\theta$ the car is at the design speed and needs no friction, so the right expression must give $\mu = 0$ there, and only the ones with $v_{0}^{2} - rg\tan\theta$ on top do.

Practice questions, easy to hard

Three questions from the circular motion dynamics – centripetal force practice ladder: one easy, one medium, one hard.

Q4Numerical answer

Put $r = L\sin\theta$ into $\tan\theta = \dfrac{\omega^{2}r}{g}$ and the $\sin\theta$ cancels off both sides, leaving $\omega^{2} = \dfrac{g}{L\cos\theta}$ . One circuit therefore takes $T_{p} = 2\pi\sqrt{\dfrac{L\cos\theta}{g}}$ , where $T_{p}$ is the period — kept apart from $T$ , which in this section always means the tension.

A bob on a string of length $0.8\ \mathrm{m}$ sweeps its cone at $60^{\circ}$ to the vertical, with $g = 10\ \mathrm{m/s^{2}}$ . What is $\omega$ , in $\mathrm{rad/s}$ ?

Show answer and solution

Answer: 5

$\omega^{2} = \dfrac{g}{L\cos\theta} = \dfrac{10}{0.8 \times 0.5} = 25$ , so $\omega = 5\ \mathrm{rad/s}$ . What sets the period is $L\cos\theta$ , the depth of the bob below its support — two pendulums of different lengths swung at different angles keep the same time if their bobs sweep at the same depth.

Q5One correct option

A bead is threaded on a smooth circular wire of radius $r$ held in a vertical plane, and the wire is spun about its vertical diameter at angular speed $\omega$ . The bead settles at rest on the wire, at an angle $\theta$ from the lowest point as seen from the centre of the ring. Because the wire is smooth it can only push at right angles to itself — that is, along the radius of the ring towards the centre — so the pair becomes $N\cos\theta = mg$ and $N\sin\theta = m\omega^{2}(r\sin\theta)$ .

What does the second of those give for $N$ ?

A $m\omega^{2}r$
B $mg$
C $m\omega^{2}r\sin\theta$
D $\dfrac{mg}{\sin\theta}$

Show answer and solution

Answer: Option A

The bead's own circle has radius $r\sin\theta$ , not $r$ — that is the step this question is built around — so the right-hand side is $m\omega^{2}r\sin\theta$ and the $\sin\theta$ cancels against the left: $N = m\omega^{2}r$ . Divide the two equations instead and the same cancellation leaves $\omega^{2} = \dfrac{g}{r\cos\theta}$ , which is the conical pendulum's $\omega^{2} = \dfrac{g}{L\cos\theta}$ with the ring's radius doing the work of the string's length.