NEET Physics — Circular Motion

Chapter Overview & Weightage

Circular motion is a focused, formula-rich topic that NEET examiners use to test conceptual clarity quickly. Expected weightage: 4-8 marks (1-2 questions every year). The questions are short and punishing for those who confuse centripetal vs centrifugal forces or skip the FBD step.

Year	Questions	Marks
2024	2	8
2023	1	4
2022	2	8
2021	1	4
2020	2	8

Average 6 marks per NEET paper. The chapter is part of “Mechanics” — totally non-negotiable.

Key Concepts You Must Know

Uniform circular motion: speed constant, velocity changes direction.
Centripetal acceleration: $a_c = v^2/r = \omega^2 r$ , directed towards centre.
Centripetal force: net force towards centre, supplied by friction, tension, gravity, etc.
Angular velocity: $\omega = v/r$ , units rad/s.
Angular acceleration: $\alpha = d\omega/dt$ .
Banking of roads: $\tan\theta = v^2/(rg)$ for ideal speed (no friction).
Vertical circular motion: tension at top vs bottom, minimum speed at top $= \sqrt{gr}$ .
Conical pendulum: $\tan\theta = v^2/(rg)$ , $T = 2\pi\sqrt{L\cos\theta / g}$ .

Important Formulas

$a_c = \frac{v^2}{r} = \omega^2 r, \quad F_c = \frac{mv^2}{r}$

For a banked road with friction, the maximum safe speed:

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

When $\mu = 0$ : $v = \sqrt{rg\tan\theta}$ .

At the top of a vertical loop: minimum speed $v_{\text{top, min}} = \sqrt{gr}$ .

At the bottom: $v_{\text{bottom}} = \sqrt{5gr}$ (using energy conservation).

Solved Previous Year Questions

PYQ 1 (NEET 2024)

A car of mass $1500\text{ kg}$ moves on a circular track of radius $50\text{ m}$ at $36\text{ km/hr}$ . Find the centripetal force needed.

Convert: $v = 36\text{ km/hr} = 10\text{ m/s}$ .

$F_c = mv^2/r = 1500 \times 100/50 = 3000\text{ N}$ .

PYQ 2 (NEET 2023)

A stone tied to a string of length $1\text{ m}$ is whirled in a vertical circle. Minimum speed at the top for the string to remain taut?

$v_{\min} = \sqrt{gr} = \sqrt{10 \times 1} = \sqrt{10} \approx 3.16\text{ m/s}$ .

PYQ 3 (NEET 2022)

A turntable rotates at $30\text{ rpm}$ . A coin placed at $0.1\text{ m}$ from the axis. Minimum coefficient of friction for the coin to stay still?

$\omega = 30 \times 2\pi/60 = \pi\text{ rad/s}$ . $a_c = \omega^2 r = \pi^2 \times 0.1 \approx 0.987\text{ m/s}^2$ .

For coin to stay: $\mu g \geq a_c$ , so $\mu \geq a_c/g = 0.987/10 = 0.0987$ .

Difficulty Distribution

Sub-topic	Easy	Medium	Hard
Centripetal basics	60%	35%	5%
Banking	30%	50%	20%
Vertical circles	20%	50%	30%
Conical pendulum	30%	50%	20%

NEET tends to keep things on the easy-to-medium side. Most questions are direct formula application.

Expert Strategy

Always draw an FBD with one axis pointing radially inward (centripetal direction) and one perpendicular. Set $\sum F_{\text{radial}} = mv^2/r$ . This single template solves 90% of NEET circular-motion problems.

For vertical circles, energy conservation links speeds at top and bottom: $v_{\text{bottom}}^2 = v_{\text{top}}^2 + 4gr$ . Use it to skip force analysis.

Memorise three results: $v_{\text{top, min}} = \sqrt{gr}$ , $v_{\text{bottom, min}} = \sqrt{5gr}$ , $\tan\theta_{\text{banking}} = v^2/(rg)$ . Two NEET MCQs every year are solved by these alone.

Common Traps

Treating centrifugal force as a real force in inertial frames. Centrifugal force only exists in rotating (non-inertial) frames. NEET sometimes uses careful wording to test this distinction.

For banking angle, using $\sin\theta$ instead of $\tan\theta$ . The correct relation involves $\tan\theta = v^2/(rg)$ .

Forgetting that the minimum speed at the top of a vertical loop comes from $T = 0$ , not from energy alone. Setting $mg = mv^2/r$ gives the right answer; using energy conservation gives a different (wrong) condition.