Key Fact
Linear–rotational analogy
Translation ↔ Rotation: F ↔ τ, m ↔ I, v ↔ ω, a ↔ α, p = mv ↔ L = Iω, KE = ½mv² ↔ KE_rot = ½ I ω². Newton's 2nd Law analogue: τ = I α.
-- NCERT Class 11 Physics, Ch. 6, p. 119Rotational Motion Basics
8 MCQs9-step worked example
Source: NCERT System of Particles and Rotational MotionPYQ coverage: NEET 2023, 2024Official key: NTA-verifiedLast updated: 7 Oct 2026
Try this first
For a rigid body in pure translational motion, which statement is true at any instant of time?
- A.Particles farther from the centre move faster.
- B.Particles on one line through the body are at rest.
- C.All particles of the body have the same velocity.
- D.Each particle moves in a circle about a common axis.
Tap to see the answer
Answer: C. C is correct: in pure translational motion all particles of the body have the same velocity at any instant (NCERT Class 11 Physics, Chapter 6, page 93).
A is wrong: A is wrong because speed that grows with distance from a point is the pattern of rotation about an axis, not of pure translation, where every particle shares one velocity.
B is wrong: B is wrong because in pure translation no particle is at rest relative to the others; a line of particles at rest belongs to the axis of a rotating body.
D is wrong: D is wrong because circular motion of every particle about a common axis describes rotation about a fixed axis, not pure translation.
Rotational Motion Basics, explained for NEET
A rolling cylinder is not in pure translation, and a spinning body does not give every particle the same speed. Both slips come from skipping the definitions.
In pure translational motion, all particles of a rigid body have the same velocity at any instant (NCERT Class 11 Physics, Chapter 6, page 93). A block sliding down an incline without sidewise movement is the example. A cylinder rolling down the same incline seems to shift from top to bottom, yet its particles do not share one velocity, so NCERT says it is not in pure translational motion. If it rolls without slipping, the velocity of the point of contact is zero at every instant (NCERT Class 11 Physics, Chapter 6, page 93).
To see the "something else", fix the body along a straight line so it cannot translate. Its only possible motion is rotation, and that line is the axis of rotation (NCERT Class 11 Physics, Chapter 6, page 93). Every particle then moves in a circle that lies in a plane perpendicular to the axis, with its centre on the axis. A particle on the axis has r = 0 and stays stationary while the body rotates (NCERT Class 11 Physics, Chapter 6, page 94). Rolling is the combination of rotation about a fixed axis and translation (NCERT Class 11 Physics, Chapter 6, page 95).
The link between the two descriptions is v = ωr. At any given instant this relation applies to every particle of the rigid body, with the same angular velocity ω for all of them, so ω is called the angular velocity of the whole body (NCERT Class 11 Physics, Chapter 6, page 104). The speeds differ because the distances r differ.
Watch-out: "same ω" never means "same speed". Speed grows in proportion to the perpendicular distance from the axis, and it is zero on the axis.
How do you solve a Rotational Motion Basics question? A worked example
- 1
Given
A rigid disc rotates about a fixed axis through its centre, perpendicular to its plane, with angular velocity 10 rad/s. Particle P1 is 0.20 m from the axis, particle P2 is 0.50 m from the axis, and particle P3 lies on the axis.
- 2
Required
The speeds of P1, P2 and P3, and the ratio of the speed of P2 to that of P1.
- 3
Concept
At any instant, v = ωr applies to every particle of the rigid body, with one angular velocity ω for the whole body. A particle on the axis has r = 0 and stays stationary (NCERT Class 11 Physics, Chapter 6, pages 94 and 104).
- 4
Formula
v = ωr for each particle, with the same ω.
- 5
Substitution
v1 = (10 rad/s)(0.20 m); v2 = (10 rad/s)(0.50 m); v3 = (10 rad/s)(0 m).
- 6
Calculation
v1 = 2.0 m/s; v2 = 5.0 m/s; v3 = 0 m/s. The ratio v2 / v1 = 5.0 / 2.0 = 2.5, the same as r2 / r1 = 0.50 / 0.20. No exact constants enter this calculation, so no significant-figure adjustment is needed.
- 7
Final answer
P1 moves at 2.0 m/s, P2 at 5.0 m/s and P3 is stationary. The speed ratio of P2 to P1 is 2.5.
- 8
Common trap
Because all three particles have the same ω, it is tempting to give them the same speed, or to say ω is smaller for the particle nearer the axis. ω is the same for the whole body; only v changes with r.
- 9
Similar NEET-style question
A merry-go-round turns at 2.0 rad/s. A child A sits 1.5 m from the axis and a child B sits 3.0 m from it. What is the speed of each child and the ratio of their speeds? (Answer: v_A = ωr = 2.0 × 1.5 = 3.0 m/s; v_B = 2.0 × 3.0 = 6.0 m/s; v_B / v_A = 6.0 / 3.0 = 2.0, the same as the ratio of the distances.)
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Can you answer these Rotational Motion Basics MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
For a rigid body in pure translational motion, which statement is true at any instant of time?
Show answer and why every option is right or wrong
Answer: C. C is correct: in pure translational motion all particles of the body have the same velocity at any instant (NCERT Class 11 Physics, Chapter 6, page 93).
Why A is wrong: A is wrong because speed that grows with distance from a point is the pattern of rotation about an axis, not of pure translation, where every particle shares one velocity.
Why B is wrong: B is wrong because in pure translation no particle is at rest relative to the others; a line of particles at rest belongs to the axis of a rotating body.
Why D is wrong: D is wrong because circular motion of every particle about a common axis describes rotation about a fixed axis, not pure translation.
MCQ 2Easy RecallPractice
A solid cylinder rolls down an inclined plane without slipping. What is the velocity of the point of the cylinder in contact with the plane at any instant?
Show answer and why every option is right or wrong
Answer: A. A is correct: if the cylinder rolls without slipping, the velocity of the point of contact is zero at any instant (NCERT Class 11 Physics, Chapter 6, page 93).
Why B is wrong: B is wrong because treating the whole cylinder as moving with one velocity assumes pure translation, which NCERT rules out for a rolling cylinder.
Why C is wrong: C is wrong because the point of contact of a cylinder rolling without slipping has zero velocity, not a velocity larger than that of the centre.
Why D is wrong: D is wrong because the contact point of a cylinder rolling without slipping has zero velocity, not the speed of the centre.
MCQ 3Easy RecallPractice
A rigid body rotates about a fixed axis. How does an individual particle of the body move?
Show answer and why every option is right or wrong
Answer: D. D is correct: every particle moves in a circle that lies in a plane perpendicular to the axis and has its centre on the axis (NCERT Class 11 Physics, Chapter 6, page 94).
Why A is wrong: A is wrong because straight-line motion parallel to the axis would be translation, and a fixed axis is exactly what prevents translation.
Why B is wrong: B is wrong because the plane of the circle is perpendicular to the axis; a plane containing the axis would not hold a circle centred on the axis.
Why C is wrong: C is wrong because the centre of each circle is on the axis, not at an arbitrary point inside the body.
MCQ 4Direct ApplicationPractice
A rigid body rotates about a fixed axis with angular velocity 4.0 rad/s. What is the speed of a particle at a perpendicular distance of 0.50 m from the axis?
Show answer and why every option is right or wrong
Answer: B. B is correct: v = ωr applies to every particle of the rigid body, so v = 4.0 rad/s × 0.50 m = 2.0 m/s (NCERT Class 11 Physics, Chapter 6, page 104).
Why A is wrong: A is wrong because 8.0 m/s comes from dividing ω by r (4.0 / 0.50) instead of multiplying.
Why C is wrong: C is wrong because 0.13 m/s comes from dividing r by ω, inverting the relation v = ωr.
Why D is wrong: D is wrong because 4.0 is the angular velocity in rad/s; the speed also depends on the distance 0.50 m.
MCQ 5Direct ApplicationPractice
Two particles of a rigid body rotating about a fixed axis are at perpendicular distances of 0.20 m and 0.60 m from the axis. What is the ratio of their angular velocities, first to second?
Show answer and why every option is right or wrong
Answer: B. B is correct: the same angular velocity ω is used for all particles of the rigid body, so ω is the angular velocity of the whole body (NCERT Class 11 Physics, Chapter 6, page 104). Only their speeds differ, in the ratio 1 : 3.
Why A is wrong: A is wrong because 1 : 3 is the ratio of the speeds (v = ωr with the same ω), not of the angular velocities.
Why C is wrong: C is wrong because it inverts the ratio of the distances and still treats ω as depending on r.
Why D is wrong: D is wrong because 1 : 9 squares the distance ratio, which has no place in v = ωr.
MCQ 6Direct ApplicationPractice
A solid cylinder rolls down an inclined plane. Which description of its motion is correct?
Show answer and why every option is right or wrong
Answer: D. D is correct: the rolling motion of a cylinder down an inclined plane is a combination of rotation about a fixed axis and translation (NCERT Class 11 Physics, Chapter 6, page 95).
Why A is wrong: A is wrong because shifting from top to bottom only seems like translation; the particles do not share one velocity, so the motion is not pure translation (NCERT Class 11 Physics, Chapter 6, page 93).
Why B is wrong: B is wrong because a fixed axis would stop the cylinder from shifting down the incline; the shifting is the translational part.
Why C is wrong: C is wrong because the particles are not at rest while the centre moves; the point of contact has zero velocity only at that instant and the others move in different ways.
MCQ 7CalculationPractice
A rigid body rotates about a fixed axis at 6.0 rad/s. Particle P is at 0.10 m from the axis and particle Q is at 0.40 m from the axis. By how much does the speed of Q exceed the speed of P?
Show answer and why every option is right or wrong
Answer: A. A is correct: with the same ω for both particles, v_Q = 6.0 × 0.40 = 2.4 m/s and v_P = 6.0 × 0.10 = 0.60 m/s, so the difference is 1.8 m/s (NCERT Class 11 Physics, Chapter 6, page 104).
Why B is wrong: B is wrong because 0.30 m is the difference of the distances; it must be multiplied by ω = 6.0 rad/s.
Why C is wrong: C is wrong because 2.4 m/s is the speed of Q alone; the question asks for how much it exceeds the speed of P.
Why D is wrong: D is wrong because 0.60 m/s is the speed of P alone.
MCQ 8CalculationPractice
In a rigid body rotating about a fixed axis, particle P at 0.30 m from the axis has a speed of 1.5 m/s. What is the speed of particle Q at 0.80 m from the axis?
Show answer and why every option is right or wrong
Answer: C. C is correct: ω = v/r = 1.5 / 0.30 = 5.0 rad/s for the whole body, so v_Q = ωr = 5.0 × 0.80 = 4.0 m/s (NCERT Class 11 Physics, Chapter 6, page 104).
Why A is wrong: A is wrong because equal speeds would need equal distances from the axis; the same ω gives different speeds at different r.
Why B is wrong: B is wrong because 0.56 m/s comes from scaling the speed inversely with distance (1.5 × 0.30 / 0.80), but v is proportional to r.
Why D is wrong: D is wrong because 5.0 is the angular velocity in rad/s, found correctly as the intermediate step, not the speed of Q.
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What to remember before solving Rotational Motion Basics questions
7 NCERT lines
Key Fact
Pure translation: every particle has the same velocity; a rolling cylinder is not pure translation
In pure translational motion at any instant of time, all particles of the body have the same velocity. Consider now the rolling motion of a solid metallic or wooden cylinder down the same inclined plane (Fig. 6.2). The rigid body in this problem, namely the cylinder, shifts from the top to the bottom of the inclined plane, and thus, seems to have translational motion. But as Fig. 6.2 shows, all its particles are not moving with the same velocity at any instant. The body, therefore, is not in pure translational motion.
-- NCERT Class 11 Physics, Ch. 6, p. 93In fact, the velocity of the point of contact P3 is zero at any instant, if the cylinder rolls without slipping.
-- NCERT Class 11 Physics, Ch. 6, p. 93The most common way to constrain a rigid body so that it does not have translational motion is to fix it along a straight line. The only possible motion of such a rigid body is rotation. The line or fixed axis about which the body is rotating is its axis of rotation.
-- NCERT Class 11 Physics, Ch. 6, p. 93Key Fact
Every particle moves in a circle perpendicular to the axis; particles on the axis stay stationary
every particle of the body moves in a circle, which lies in a plane perpendicular to the axis and has its centre on the axis. Fig. 6.4 shows the rotational motion of a rigid body about a fixed axis (the z-axis of the frame of reference). Let P1 be a particle of the rigid body, arbitrarily chosen and at a distance r1 from fixed axis. The particle P1 describes a circle of radius r1 with its centre C1 on the fixed axis. The circle lies in a plane perpendicular to the axis. The figure also shows another particle P2 of the rigid body, P2 is at a distance r2 from the fixed axis. The particle P2 moves in a circle of radius r2 and with centre C2 on the axis. This circle, too, lies in a plane perpendicular to the axis. Note that the circles described by P1 and P2 may lie in different planes; both these planes, however, are perpendicular to the fixed axis. For any particle on the axis like P3, r = 0. Any such particle remains stationary while the body rotates.
-- NCERT Class 11 Physics, Ch. 6, p. 94The rolling motion of a cylinder down an inclined plane is a combination of rotation about a fixed axis and translation. Thus, the ‘something else’ in the case of rolling motion which we referred to earlier is rotational motion.
-- NCERT Class 11 Physics, Ch. 6, p. 95We observe that at any given instant the relation v r ω = applies to all particles of the rigid body. Thus for a particle at a perpendicular distance ri from the fixed axis, the linear velocity at a given instant vi is given by i i v r ω = (6.19) The index i runs from 1 to n, where n is the total number of particles of the body. For particles on the axis, 0 = r , and hence v = ω r = 0. Thus, particles on the axis are stationary. This verifies that the axis is fixed. Note that we use the same angular velocity ω for all the particles. We therefore, refer to ω as the angular velocity of the whole body.
-- NCERT Class 11 Physics, Ch. 6, p. 104Rotational Motion Basics: NEET previous year questions (PYQs) with answers
2 questions from NEET 2023, 2024, answers verified against NTA official keys
NEET 2024Revised key
1Point P moves slower than point Q
2Point P moves faster than point Q
3Both the points P and Q move with equal speed
4Point P has zero speed
NTA Answer: Option 2(revised_final)
NEET 2023
The angular acceleration of a body, moving along the circumference of a circle, is
1Along the radius towards the centre
2Along the tangent to its position
3Along the axis of rotation
4Along the radius, away from centre
NTA Answer: Option 3(final)
All 12 past-paper questions from System of Particles and Rotational Motion →
More in System of Particles and Rotational Motion: 7 exam traps and mistakes · 8 formulas · 4 question patterns from its other lessons.
Sources
NCERT refs: Class 11 Physics Chapter 6, p.93 | Class 11 Physics Chapter 6, p.94 | Class 11 Physics Chapter 6, p.95 | Class 11 Physics Chapter 6, p.104
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
Every past-paper question from this chapter:
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