68 free MCQs on System of Particles and Rotational Motion with worked answers and explanations. Moment of inertia, torque, angular momentum, rolling motion. Very high JEE weightage.
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Torque about axis O equals the force F multiplied by the perpendicular distance r⊥ from the axis to the line of action of the force, producing rotation.
Q1.
What is the moment of inertia of a solid sphere of mass M and radius R about its diameter?
Answer: A. 2MR²/5
Why: For a solid sphere about its diameter, I = 2MR²/5. The hollow sphere gives 2MR²/3.
Q2.
The angular equivalent of Newton's second law (F = ma) is:
Answer: A. τ = Iα
Why: Torque = moment of inertia x angular acceleration, just as force = mass x linear acceleration.
Q3.
A ring, a disk, and a solid sphere of equal mass and radius roll down an incline. Which reaches the bottom first?
Answer: A. Solid sphere
Why: The solid sphere has the smallest I/mR² ratio (2/5), so it has the greatest acceleration down the incline.
Q4.
The SI unit of torque is:
Answer: A. N·m
Why: Torque = r x F, so the unit is newton-metre (N·m). Although numerically equal to joules, torque is a vector quantity.
Q5.
When a spinning ice skater pulls in her arms, her angular velocity:
Answer: A. Increases
Why: Angular momentum L = Iω is conserved. Pulling arms in reduces I, so ω must increase to keep L constant.
Q6.
The moment of inertia of a uniform rod of mass M and length L about one end is:
Answer: A. ML²/3
Why: About one end, I = ML²/3. About the centre, I = ML²/12 (use parallel axis theorem to verify).
Q7.
Which theorem relates the moment of inertia about an axis through the centre of mass to that about a parallel axis?
Answer: A. Parallel axis theorem
Why: Parallel axis theorem: I = I<sub>cm</sub> + Md², where d is the distance between the two parallel axes.
Q8.
For a body rolling without slipping on a surface, which condition must hold?
Answer: A. v = ωR
Why: In pure rolling, the contact point is instantaneously at rest. This gives v<sub>cm</sub> = ωR.
Q9.
Angular momentum L of a rotating body is given by:
Answer: A. L = Iω
Why: L = Iω (angular momentum = moment of inertia x angular velocity). It is conserved when net torque is zero.
Q10.
A disk has moment of inertia I about its axis. What is the moment of inertia of a ring of same mass and radius about its axis?
Answer: A. 2I
Why: Disk: I<sub>disk</sub> = MR²/2. Ring: I<sub>ring</sub> = MR² = 2I<sub>disk</sub>. So the ring has twice the MI of the disk.
Q11.
Torque is defined as:
Answer: A. r × F (cross product of position and force)
Why: Torque τ = r × F. Its magnitude is rF sinθ, where θ is the angle between r and F.
Q12.
The rotational kinetic energy of a body with moment of inertia I rotating at angular speed ω is:
Answer: A. (1/2)Iω²
Why: KE_rot = (1/2)Iω², the rotational analogue of (1/2)mv² for translational KE.
Q13.
The perpendicular axis theorem applies only to:
Answer: A. Planar (flat) bodies
Why: Perpendicular axis theorem (I<sub>z</sub> = I<sub>x</sub> + I<sub>y</sub>) only applies to flat, planar (2D) objects like disks, rings, and thin rods.
Q14.
If a sphere rolls down without slipping, what fraction of its total KE is rotational?
Answer: A. 2/7
Why: For a solid sphere, KE_rot/KE_total = (1/2)Iω² / [(1/2)mv² + (1/2)Iω²] = (2/5)/(1 + 2/5) = 2/7.
Q15.
What is the SI unit of angular velocity?
Answer: A. rad/s
Why: Angular velocity is measured in radians per second (rad/s). It is related to frequency by ω = 2πf.
Q16.
A torque does zero work on a rotating body when the force is:
Answer: A. Directed toward the axis of rotation
Why: If force is directed toward the rotation axis (radial), it is perpendicular to displacement, so work = 0.
Q17.
Which of the following has the largest moment of inertia for same mass M and radius R, about the central axis?
Answer: D. Thin ring
Why: Ring: I = MR² (all mass at maximum distance). Hollow cylinder: I = MR². Solid cylinder: MR²/2. Solid sphere: 2MR²/5. Ring and hollow cylinder tie at MR².
Q18.
What is the angular acceleration α if a torque of 10 N·m acts on a body with I = 5 kg·m²?
Answer: A. 2 rad/s²
Why: α = τ/I = 10/5 = 2 rad/s². This is the rotational analogue of a = F/m.
Q19.
The linear velocity of a point on a rotating body at distance r from the axis is:
Answer: A. v = ωr
Why: v = ωr (tangential velocity = angular velocity x radius). Points farther from the axis move faster.
Q20.
Which conservation law explains why a planet moves faster when closer to the Sun?
Answer: A. Conservation of angular momentum
Why: As the planet gets closer to the Sun, r decreases, so v must increase to keep L = mvr constant (Kepler's second law).
Q21.
A solid cylinder of mass 2 kg and radius 0.1 m rolls down a 30° incline without slipping. What is its acceleration?
Answer: A. (2/3)g sin30°
Why: For rolling without slipping, a = g sinθ/(1 + I/mR²). For solid cylinder, I = mR²/2, so a = g sinθ/(1 + 1/2) = (2/3)g sin30°.
Q22.
A wheel of moment of inertia 2 kg·m² is rotating at 10 rad/s. A tangential force applies a torque of 4 N·m for 5 s. What is the final angular velocity?
Answer: A. 20 rad/s
Why: α = τ/I = 4/2 = 2 rad/s². ω_f = ω_i + αt = 10 + 2 x 5 = 20 rad/s.
Q23.
Using the parallel axis theorem, the MI of a rod of mass M, length L about one end is ML²/3. What is the MI about the centre?
Answer: A. ML²/12
Why: I<sub>end</sub> = I<sub>cm</sub> + M(L/2)². So ML²/3 = I<sub>cm</sub> + ML²/4 → I<sub>cm</sub> = ML²/3 - ML²/4 = ML²/12.
Q24.
A torque of 20 N·m acts on a rigid body for 4 s. If the body starts from rest, what is the angular momentum after 4 s?
Answer: A. 80 kg·m²/s
Why: Angular impulse = τ × t = ΔL. L = 20 × 4 = 80 kg·m²/s (starting from rest).
Q25.
A turntable rotating at 60 rpm has moment of inertia 4 kg·m². A person of mass 60 kg sits at the edge (r = 1 m). What is the new angular velocity? (Assume no friction)
Answer: A. 60π/32 rad/s
Why: L conserved: Iω = (I + mr²)ω'. ω_i = 2π rad/s. 4 × 2π = (4+60)ω' → ω' = 8π/64 = π/8 rad/s.
Q26.
For a disk rolling without slipping at speed v, the speed of the topmost point is:
Answer: A. 2v
Why: In rolling without slipping, the top point has velocity 2v (translational v plus rotational ωR = v), while the contact point has velocity 0.
Q27.
The radius of gyration k of a body is defined by:
Answer: A. I = Mk²
Why: Radius of gyration k is defined so that I = Mk². It represents the rms distance of mass from the axis.
Q28.
A solid sphere is rotating at ω. If its radius is doubled keeping mass constant, and angular momentum is conserved, what is the new angular velocity?
Answer: A. ω/4
Why: L = Iω = (2/5)MR²ω. New I = (2/5)M(2R)² = 4 × (2/5)MR². For L conservation: ω' = ω/4.
Q29.
A rigid body is in rotational equilibrium. This means:
Answer: A. Net torque acting on it is zero
Why: Rotational equilibrium requires Στ = 0. For complete (static) equilibrium, both ΣF = 0 and Στ = 0 are needed.
Q30.
A particle of mass m moves in a circle of radius r. The angular momentum about the centre is:
Answer: A. mvr
Why: L = r × p = mvr (when v is perpendicular to r, as in circular motion). Also L = Iω = mr² × (v/r) = mvr.
Q31.
The moment of inertia of a rectangular plate of mass M, dimensions a × b about an axis through its centre perpendicular to its plane is:
Answer: A. M(a²+b²)/12
Why: I<sub>z</sub> = M(a² + b²)/12. By perpendicular axis theorem: I<sub>z</sub> = I<sub>x</sub> + I<sub>y</sub> = Mb²/12 + Ma²/12.
Q32.
An object rolls down without slipping. The ratio of translational KE to rotational KE for a solid disk is:
Answer: A. 2:1
Why: For disk: I = mR²/2. KE_trans = (1/2)mv², KE_rot = (1/2)(mR²/2)(v/R)² = mv²/4. Ratio = 2:1.
Q33.
The work done by a torque τ through an angular displacement θ is:
Answer: A. τθ
Why: W = τ × θ (rotational analogue of W = F × d for translational work).
Q34.
A flywheel has I = 8 kg·m². It is brought to rest from 100 rad/s in 20 s. What is the braking torque?
Answer: A. 40 N·m
Why: α = (0 - 100)/20 = -5 rad/s². τ = I|α| = 8 × 5 = 40 N·m.
Q35.
For the perpendicular axis theorem, if I<sub>x</sub> = I<sub>y</sub> for a symmetric disk, then I<sub>z</sub> equals:
Answer: A. 2I<sub>x</sub>
Why: By perpendicular axis theorem: I<sub>z</sub> = I<sub>x</sub> + I<sub>y</sub>. If I<sub>x</sub> = I<sub>y</sub>, then I<sub>z</sub> = 2I<sub>x</sub>.
Q36.
Two disks of equal mass are joined coaxially. Their MIs about their common axis are 3 kg·m² and 5 kg·m². What is the combined MI?
Answer: A. 8 kg·m²
Why: For bodies on the same axis, moments of inertia simply add: I<sub>total</sub> = I<sub>1</sub> + I<sub>2</sub> = 3 + 5 = 8 kg·m².
Q37.
The kinetic energy of a rolling object can be written as (1/2)mv²(1 + k²/R²) where k is radius of gyration. For a ring, k²/R² = 1. So the KE is:
Answer: A. mv²
Why: KE = (1/2)mv²(1 + 1) = mv². For a ring, half is translational and half is rotational.
Q38.
A diver tucks in during a dive. What happens to their angular velocity?
Answer: A. Increases
Why: Tucking in reduces the moment of inertia. By conservation of angular momentum (no external torque during dive), ω increases.
Q39.
A point mass m is attached to the end of a string and rotates in a horizontal circle of radius r. The angular momentum about the centre is L = mvr. If r is halved and v is doubled, L becomes:
Answer: A. L
Why: L' = m(2v)(r/2) = mvr = L. Angular momentum remains the same.
Q40.
The moment of inertia of a thin ring about a diameter (not central axis) is:
Answer: A. MR²/2
Why: By perpendicular axis theorem for ring: I<sub>z</sub> = MR² = I<sub>x</sub> + I<sub>y</sub>. By symmetry I<sub>x</sub> = I<sub>y</sub>, so each = MR²/2.
Q41.
A solid cylinder of mass M and radius R is free to rotate about its horizontal axis. A string wound around it carries mass m. What is the acceleration of the hanging mass?
Answer: A. 2mg/(M+2m)
Why: Net torque on cylinder = mR - 0 = (MR²/2)α. Also a = Rα and T = m(g-a). Solving: T = mRg/(2m+M) x M/R, and a = 2mg/(M+2m).
Q42.
A rod of length L stands vertically on a frictionless floor and falls. What is the angular velocity when it hits the floor?
Answer: A. √(3g/L)
Why: Energy conservation: mg(L/2) = (1/2)(ML²/3)ω². Solving: ω = √(3g/L). The CM falls by L/2.
Q43.
A wheel of radius R rolls without slipping. What is the acceleration of the topmost point?
Answer: A. 2a (horizontal, direction of motion)
Why: The top point has velocity 2v. Its acceleration has both centripetal (ω²R = a upward) and translational (a forward) components. The horizontal component is 2a (both a<sub>cm</sub> and centripetal contribute horizontally).
Q44.
Two particles of masses m and 2m are attached to the ends of a uniform rod of mass M and length L. What is the MI about the centre of the rod?
Answer: A. (3m+M/3)L²/4
Why: I = ML²/12 (rod about centre) + m(L/2)² + 2m(L/2)² = ML²/12 + mL²/4 + mL²/2 = ML²/12 + 3mL²/4.
Q45.
A spinning top (gyroscope) precesses when tilted. The precession angular velocity is:
Answer: A. Ω = Mgr/Lω (where Lω = Iω is spin angular momentum)
Why: Gyroscopic precession: Ω = τ/L = Mgr/(Iω). The torque from gravity causes the angular momentum vector to precess.
Q46.
A ball is thrown with backspin onto a rough floor. Initially v (forward) and ωR > v. The friction force on the ball is:
Answer: A. Forward (in direction of motion)
Why: Since ωR > v, the contact point moves backward relative to the floor. Kinetic friction acts forward on the ball (opposing relative slip), accelerating it and decelerating its spin until rolling condition v = ωR is met.
Q47.
The angular momentum of a system is conserved when:
Answer: A. The net external torque about the reference point is zero
Why: Conservation of L requires Στ_ext = dL/dt = 0. Forces can still act as long as their torques cancel.
Q48.
A uniform disk of mass M and radius R has a hole of radius R/2 cut from it, centred at R/2 from the centre. What is the new moment of inertia about the disk centre?
Answer: A. 13MR²/32
Why: Mass of hole = M/4 (proportional to area R²/4). I<sub>hole</sub> about disk centre = (1/2)(M/4)(R/2)² + (M/4)(R/2)² = MR²/32 + MR²/16 = 3MR²/32. I<sub>new</sub> = MR²/2 - 3MR²/32 = 13MR²/32.
Q49.
A solid sphere rolls without slipping up an incline, starting with translational speed v. How high does it go?
Answer: A. 7v²/10g
Why: Total initial KE = (1/2)mv² + (1/2)(2/5)mR²(v/R)² = (7/10)mv². Setting equal to mgh: h = 7v²/10g.
Q50.
A particle of mass m moves in a plane. Its position vector is r = (t², 3t). What is the angular momentum about the origin at t = 1?
Answer: A. 3m - 6m = -3m (in z-direction)
Why: L = m(r × v). v = (2t, 3). At t=1: r = (1,3), v = (2,3). L<sub>z</sub> = m(rx vy - ry vx) = m(1×3 - 3×2) = m(3-6) = -3m.
Q51.
For a conical pendulum, the tension T in the string provides centripetal force and supports weight. The angular velocity is:
Answer: A. ω = √(g/L cosθ)
Why: T cosθ = mg and T sinθ = mω²r where r = L sinθ. Dividing: tanθ = ω²r/g = ω² L sinθ/g → ω = √(g/L cosθ).
Q52.
A uniform rod of length L and mass M can rotate about one end. It is held horizontal and released. What is the velocity of the free end just as the rod becomes vertical?
Answer: A. √(3gL)
Why: Energy conservation: Mg(L/2) = (1/2)(ML²/3)ω² → ω = √(3g/L). Speed of free end = ωL = √(3gL).
Q53.
Two equal masses m are at the ends of a rod of negligible mass, length 2a. The rod rotates about its centre at ω. One mass suddenly falls off. What happens to ω?
Answer: A. ω doubles
Why: When a mass falls off, external torque is impulsive for an instant. Immediately after, I = ma² (one mass). By conservation of angular momentum: 2ma² ω = ma² ω' → ω' = 2ω.
Q54.
A horizontal disk rotating at ω has a coefficient of friction μ between it and a small block placed at distance r from centre. The block will start sliding if:
Answer: A. ω > √(μg/r)
Why: Static friction provides centripetal force: μmg ≥ mω²r. Sliding occurs when ω²r > μg, i.e., ω > √(μg/r).
Q55.
The spin angular momentum of Earth is approximately 7 × 10³³ kg·m²/s. If Earth suddenly contracted to half its radius, its day would be:
Answer: A. 6 hours
Why: I = (2/5)MR². If R halves, I becomes (1/4) of original. L = Iω is conserved, so ω quadruples and T = 24/4 = 6 hours.
Q56.
A disk of moment of inertia I and angular velocity ω is dropped onto another stationary disk with the same I. They eventually reach the same speed due to friction. Final ω is:
Answer: A. ω/2
Why: By conservation of angular momentum (friction is internal between the two disks): Iω = 2Iω_f → ω_f = ω/2.
Q57.
The precession rate of a gyroscope increases if:
Answer: A. The spin speed decreases
Why: Precession rate Ω = Mgr/(Iω). Decreasing spin speed ω (denominator) increases Ω.
Q58.
A particle of mass m undergoes uniform circular motion at radius r with speed v. Its angular momentum magnitude about a point P that is at distance d from the centre (coplanar) is:
Answer: A. mvr (independent of d for uniform circular motion)
Why: For a particle in circular motion, L about the centre = mvr. About any other point, L depends on geometry. However, for the centre specifically, L = mvr. For point P offset by d, L = mvr (it is constant and equal to mvr for any point on the rotation axis line in 2D uniform circular motion: this is Kepler's 2nd law applied).
Q59.
A solid ball rolling without slipping on a curved bowl oscillates about the bottom. The effective restoring force gives period T = 2π√(7R/5g) for a bowl of radius R. This is analogous to a pendulum of effective length:
Answer: A. 7R/5
Why: For a simple pendulum, T = 2π√(L<sub>eff</sub>/g). Comparing, L<sub>eff</sub> = 7R/5. The rolling condition adds the factor of 7/5 compared to a sliding particle.
Q60.
The moment of inertia of a uniform solid sphere of mass M and radius R about a diameter is:
Answer: A. (2/5)MR²
Why: For a solid sphere about its diameter, I = (2/5)MR².
Q61.
The moment of inertia of a solid sphere of mass M and radius R about a tangent line is:
Answer: B. 7MR²/5
Why: By the parallel-axis theorem: I = (2/5)MR² + MR² = 7MR²/5.
Q62.
For a ring rolling without slipping, the fraction of its total kinetic energy that is rotational is:
Answer: A. 1/2
Why: For a ring I = MR², so rotational KE = translational KE, each being half the total.
Q63.
A uniform solid disc rolls without slipping down an incline of angle θ. Its linear acceleration is:
Answer: C. 2g sinθ/3
Why: a = g sinθ/(1 + I/MR²) = g sinθ/(1 + 1/2) = 2g sinθ/3.
Q64.
A skater spinning with her arms out pulls them in so her moment of inertia halves. Her rotational kinetic energy:
Answer: C. doubles
Why: L = Iω conserved, so ω doubles; KE = (1/2)Iω² = (1/2)(I/2)(2ω)² = 2× the original - it doubles.
Q65.
Two point masses, 2 kg at the origin and 3 kg at x = 5 m, form a system. The x-coordinate of the centre of mass is:
Answer: B. 3 m
Why: x<sub>cm</sub> = (2·0 + 3·5)/(2 + 3) = 15/5 = 3 m.
Q66.
A solid cylinder and a hollow cylinder of the same mass and radius are released from rest at the top of the same incline. Which reaches the bottom first?
Answer: A. the solid cylinder
Why: The solid cylinder has smaller I/MR², so larger acceleration, and reaches the bottom first.
Q67.
A torque of 2 N·m acts on a disc of moment of inertia 0.5 kg·m². Its angular acceleration is:
Answer: C. 4 rad/s²
Why: α = τ/I = 2/0.5 = 4 rad/s².
Q68.
A force of 10 N is applied perpendicular to a wrench at 0.5 m from the pivot. The torque produced is:
Answer: B. 5 N·m
Why: τ = rF sin90° = 0.5·10 = 5 N·m.