System of Particles and Rotational Motion — Class 11 MCQs with Answers

System of Particles and Rotational Motion — important questions & MCQs with answers (Class 11 Physics)

6 solved questions from this chapter's difficulty levels, each with its answer and explanation. Tap an option to check yourself.

0 of 6 answered

1. Q1 of 6Easy

   Centre of mass of a uniform rod is at:

   AOne endBThe geometric centreC1/3 from one endDOutside the rod
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   Answer: The geometric centre

   Explanation: Uniform rod ⇒ mass symmetric about its midpoint, so centre of mass sits at the geometric centre — e.g. a 2 m rod balances at 1 m from either end. COM is not at an end, and it never lies outside a rod.

2. Q2 of 6Easy

   Torque τ =

   AF·r²BF/rCF + rDr × F
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   Answer: r × F

   Explanation: Torque τ = r×F, the vector cross product of position and force (magnitude rF sinθ, direction by right-hand rule). Adding F+r or dividing F/r ignores that torque has units N·m, not N or 1/m.

3. Q3 of 6Medium

   A force of 10 N acts at a point 2 m from the axis, perpendicular to the position vector of that point. Torque about the axis:

   A10 N·mB20 N·mC2 N·mD100 N·m
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   Answer: 20 N·m

   Explanation: Torque τ = rF sinθ; with the force perpendicular to r, sinθ = sin90° = 1, so τ = 2 m × 10 N = 20 N·m — using sinθ instead of assuming θ=0° (which would wrongly give zero) is the key step.

4. Q4 of 6Medium

   For particle m moving in straight line, angular momentum about any point on its line of motion is:

   AZeroBmvrCMaximumDVariable
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   Answer: Zero

   Explanation: For a particle moving in a straight line, its velocity v is parallel to the position vector r drawn to any point on that line, so sinθ = 0 and L = mvr sinθ = 0 — not mvr, which needs r and v non-parallel.

5. Q5 of 6Hard

   Radius of gyration k for body of moment of inertia I and mass M:

   Ak² = M·IBI = MkCk = √(I/M)Dk = M/I
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   Answer: k = √(I/M)

   Explanation: Radius of gyration k is defined by I = Mk², concentrating all mass at distance k that reproduces the same I — solving gives k = √(I/M), not k² = M·I, which has the wrong dimensions for a length.

6. Q6 of 6Hard

   Solid sphere rolls without slipping at speed v. Fraction of total KE that is rotational:

   A1/3B1/2C2/5D2/7
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   Answer: 2/7

   Explanation: Rolling sphere: KE_rot/KE_total = (⅕v²)/[(½ + ⅕)v²] = (1/5)/(7/10) = 2/7, using I=(2/5)MR² and v_cm=ωR. The (2/5) shows up in the numerator's ⅕ but the total needs BOTH translational and rotational terms.

System of Particles and Rotational Motion notes: key concepts (Class 11 Physics)

Rigid-body mechanics built by analogy with translation: the centre of mass and its motion, torque as the rotational analogue of force, moment of inertia as the analogue of mass, angular momentum and its conservation, the two axis theorems, and the energy accounting of rolling motion.

Centre of mass

The mass-weighted average position of a system, which moves as though the entire mass were concentrated there. For a uniform body it lies at the geometric centre.

Motion of the centre of mass

The centre of mass accelerates only under a net EXTERNAL force. Internal forces, however violent, cannot shift its state of motion — which is why an exploding shell's centre of mass continues on its parabola.

Torque

The turning effect of a force, r × F. Only the component of force perpendicular to the position vector contributes, so a force through the axis produces no torque.

Angular momentum

The rotational analogue of linear momentum, L = r × p, and equal to Iω for a rigid body rotating about a fixed axis.

Conservation of angular momentum

With no net external torque, L stays constant. Reducing I therefore raises ω — the reason a spinning skater speeds up on pulling their arms in.

Moment of inertia

The rotational analogue of mass, I = Σmr². Unlike mass it is not a fixed property of a body: it depends on which axis the rotation is about.

Radius of gyration

The distance from the axis at which the whole mass could be concentrated without changing I. It gives a compact way to compare shapes.

Parallel axis theorem

The moment of inertia about any axis equals that about a parallel axis through the centre of mass plus Md². It only works between PARALLEL axes.

Perpendicular axis theorem

For a PLANAR body only, the moment of inertia about an axis perpendicular to the plane equals the sum of those about two perpendicular axes in the plane.

Rotational kinematics

Angular displacement, velocity and acceleration obey equations identical in form to the linear ones, with θ, ω and α replacing s, v and a.

Rotational kinetic energy

The energy of rotation, ½Iω². A body that both moves and spins carries translational and rotational kinetic energy simultaneously.

Rolling without slipping

The contact point is instantaneously at rest, giving the constraint v = ωR. Friction here does no work, so mechanical energy is still conserved.

Energy split in rolling

Total kinetic energy divides between translation and rotation in a ratio fixed by the body's shape — which is why a ring, a disc and a sphere reach the bottom of an incline in that order, slowest first.

Equilibrium of a rigid body

Requires BOTH the net force and the net torque to vanish. A body can have zero net force and still start spinning, so the second condition is not optional.

Translation-rotation analogy

Every translational quantity has a rotational twin: force to torque, mass to moment of inertia, momentum to angular momentum. Learning the map halves the work.