Kinematics of Rotational Motion and Angular Velocity – Basic concepts of rotational motion: NEET MCQs with Answers

15 free questions on Kinematics of Rotational Motion and Angular Velocity – Basic concepts of rotational motion for NEET (UG), starting with previous-year NEET and AIPMT questions. Try each one, then tap Show answer for the answer and a short explanation.

Syllabus covered: Basic concepts of rotational motion, rigid body rotation and equations of rotational motion, comparison of linear and rotational motions.

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Question 1

Previous-year question
A flywheel with uniform angular acceleration speeds up from 20π rad/s to 40π rad/s in 10 s. The number of rotations made in this period is [AIPMT 2006]
  1. 100
  2. 150
  3. 200
  4. 50
Show answer
Answer: (B) 150
Explanation: α = (40π − 20π)/10 = 2π rad/s². θ = (ω₂² − ω₁²)/2α = [(40π)² − (20π)²]/(4π) = 1200π²/4π = 300π rad. Number of rotations = 300π/2π = 150.

Question 2

Previous-year question
A fly wheel rotating at 420 rpm is brought to rest by a constant torque in 10 s. The number of rotations made before coming to rest is: [NEET-I 2016]
  1. 35
  2. 70
  3. 140
  4. 280
Show answer
Answer: (A) 35
Explanation: The initial speed is 420 rpm = 7 rev/s and the final speed is zero. Under constant torque the average angular speed is (7 + 0)/2 = 3.5 rev/s, so in 10 s the wheel makes 3.5 × 10 = 35 rotations.

Question 3

Previous-year question
A uniform circular disc of radius 50 cm at rest is free to turn about an axis perpendicular to its plane through its centre. It is subjected to a torque producing a constant angular acceleration of 2.0 rad s⁻². Its net linear acceleration at the rim at the end of 2.0 s is approximately [NEET 2016]
  1. 8.0 m s⁻²
  2. 7.0 m s⁻²
  3. 6.0 m s⁻²
  4. 3.0 m s⁻²
Show answer
Answer: (A) 8.0 m s⁻²
Explanation: After 2 s, ω = αt = 4 rad/s. The radial component is ω²r = 16 × 0.5 = 8 m/s² and the tangential one is αr = 1 m/s², so the resultant is √(64 + 1) ≈ 8.06 ≈ 8.0 m s⁻².

Question 4

Previous-year question
The angular speed of an engine wheel making 90 revolutions per minute is [AIPMT 1995]
  1. 1.5π rad/s
  2. 3π rad/s
  3. 4.5π rad/s
  4. 6π rad/s
Show answer
Answer: (B) 3π rad/s
Explanation: ν = 90/60 = 1.5 rev/s ω = 2πν = 2π × 1.5 = 3π rad/s.

Question 5

Previous-year question
If a flywheel makes 120 revolutions per minute, then its angular speed will be [AIPMT 1996]
  1. 8π rad/s
  2. 6π rad/s
  3. 4π rad/s
  4. 2π rad/s
Show answer
Answer: (C) 4π rad/s
Explanation: ω = 2πν, where ν is the frequency in revolutions per second. ν = 120/60 = 2 rev/s ω = 2π × 2 = 4π rad/s.

Question 6

Previous-year question
What is the value of the linear velocity if r = 3î − 4ĵ + k̂ and ω = 5î − 6ĵ + 6k̂ ? [AIPMT 1999]
  1. 4î − 13ĵ + 6k̂
  2. 18î + 13ĵ − 2k̂
  3. 6î + 2ĵ − 3k̂
  4. 6î − 2ĵ + 8k̂
Show answer
Answer: (B) 18î + 13ĵ − 2k̂
Explanation: v = ω × r = | î ĵ k̂ ; 5 −6 6 ; 3 −4 1 | î-component: (−6)(1) − (6)(−4) = −6 + 24 = 18 ĵ-component: −[(5)(1) − (6)(3)] = −[5 − 18] = 13 k̂-component: (5)(−4) − (−6)(3) = −20 + 18 = −2 v = 18î + 13ĵ − 2k̂.

Question 7

Previous-year question
A constant torque of 1000 N-m turns a wheel of moment of inertia 200 kg-m² about an axis through its centre. Its angular velocity after 3 s is [AIPMT 2001]
  1. 1 rad/s
  2. 5 rad/s
  3. 10 rad/s
  4. 15 rad/s
Show answer
Answer: (D) 15 rad/s
Explanation: α = τ/I = 1000/200 = 5 rad/s². Starting from rest, ω = ω₀ + αt = 0 + 5 × 3 = 15 rad/s.
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Question 8

Previous-year question
A wheel of radius 1 m rolls forward through half a revolution on horizontal ground. The magnitude of the displacement of the point of the wheel that was initially in contact with the ground is [AIPMT 2002]
  1. π
  2. 2π
  3. √(2π)
  4. √(π² + 4)
Show answer
Answer: (D) √(π² + 4)
Explanation: In half a revolution the centre advances a horizontal distance πr = π. The contact point, meanwhile, rises to the top of the wheel, a vertical distance 2r = 2. The two displacements are perpendicular, so the resultant is √((πr)² + (2r)²) = √(π² + 4) with r = 1 m.

Question 9

Previous-year question
A wheel whose moment of inertia about its vertical axis is 2 kg m² rotates about that axis at 60 rpm. The torque that could stop the wheel's rotation in one minute would be [AIPMT 2004]
  1. 2π/15 N-m
  2. π/12 N-m
  3. π/15 N-m
  4. π/18 N-m
Show answer
Answer: (C) π/15 N-m
Explanation: ω = 2π × 60/60 = 2π rad/s. Retardation needed, α = ω/t = 2π/60 = π/30 rad/s². τ = Iα = 2 × π/30 = π/15 N-m.

Question 10

Previous-year question
A wheel has an angular acceleration of 3 rad/s² and an initial angular speed of 2 rad/s. In a time of 2 s, the angle (in radian) through which it has rotated is [AIPMT 2007]
  1. 6
  2. 10
  3. 12
  4. 4
Show answer
Answer: (B) 10
Explanation: Using θ = ω₀t + ½αt² : θ = (2)(2) + ½(3)(2)² = 4 + 6 = 10 rad.

Question 11

Previous-year question
The instantaneous angular position of a point on a rotating wheel is given by θ(t) = 2t³ − 6t². The torque on the wheel becomes zero at [AIPMT 2011]
  1. t = 0.5 s
  2. t = 0.25 s
  3. t = 2 s
  4. t = 1 s
Show answer
Answer: (D) t = 1 s
Explanation: Torque is zero when the angular acceleration is zero, since τ = Iα. ω = dθ/dt = 6t² − 12t α = d²θ/dt² = 12t − 12 Setting α = 0 : 12t − 12 = 0 ⟹ t = 1 s.

Question 12

Previous-year question
An automobile travels along a road at 54 km/h. The radius of its wheels is 0.45 m and the moment of inertia of a wheel about its rotation axis is 3 kg m². If the vehicle is brought to rest in 15 s, the magnitude of the average torque transmitted by the brakes to the wheel is [AIPMT 2015]
  1. 6.66 kg m² s⁻²
  2. 8.58 kg m² s⁻²
  3. 10.86 kg m² s⁻²
  4. 2.86 kg m² s⁻²
Show answer
Answer: (A) 6.66 kg m² s⁻²
Explanation: v = 54 km/h = 54 × 5/18 = 15 m/s. Initial angular speed, ω₀ = v/R = 15/0.45 = 100/3 rad/s. Retardation, α = (ω₀ − 0)/t = (100/3)/15 = 100/45 rad/s². τ = Iα = 3 × 100/45 = 6.66 kg m² s⁻².

Question 13

Previous-year question
A uniform circular disc of radius 50 cm, initially at rest, is free to turn about an axis perpendicular to its plane through its centre. A torque gives it a constant angular acceleration of 2.0 rad/s². Its net acceleration (in m/s²) at the end of 2.0 s is approximately [NEET 2016]
  1. 8.0
  2. 7.0
  3. 6.0
  4. 3.0
Show answer
Answer: (A) 8.0
Explanation: After 2 s, ω = ω₀ + αt = 0 + 2 × 2 = 4 rad/s. For a point on the rim (R = 0.5 m): Radial (centripetal) acceleration, a_c = ω²R = 16 × 0.5 = 8 m/s² Tangential acceleration, a_t = αR = 2 × 0.5 = 1 m/s² These are mutually perpendicular, so a = √(a_c² + a_t²) = √(64 + 1) = √65 ≈ 8.06 ≈ 8.0 m/s².

Question 14

Previous-year question
A particle starting from rest moves along a circle of radius r and attains a velocity of v₀ m/s in the nth round. Its angular acceleration will be [NEET (Odisha) 2019]
  1. v₀²/n rad/s²
  2. v₀²/(2πnr) rad/s²
  3. v₀²/(4πnr²) rad/s²
  4. v₀²/(4πnr) rad/s²
Show answer
Answer: (C) v₀²/(4πnr²) rad/s²
Explanation: Final angular velocity ω = v₀/r, and the angular displacement in n rounds is θ = 2πn. Using ω² = ω₀² + 2αθ with ω₀ = 0 : (v₀/r)² = 2α(2πn) v₀²/r² = 4πnα α = v₀²/(4πnr²) rad/s².

Question 15

Previous-year question
The angular speed of a vehicle's wheel increases from 360 rpm to 1200 rpm in 14 s. Its angular acceleration is [NEET (Oct.) 2020]
  1. 2π rad/s²
  2. 28π rad/s²
  3. 120π rad/s²
  4. 1 rad/s²
Show answer
Answer: (A) 2π rad/s²
Explanation: ω₀ = 2π × 360/60 = 12π rad/s and ω = 2π × 1200/60 = 40π rad/s. α = (ω − ω₀)/t = (40π − 12π)/14 = 28π/14 = 2π rad/s².

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