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You can know every formula in this chapter and still lose marks, because many rotational motion errors happen in the set-up, not the algebra. Here are five, each with a worked fix.
Which rotational motion topics do JEE and board papers test?
The JEE Main 2025 syllabus published by NTA lists moment of a force, torque, angular momentum and its conservation, moment of inertia, and the parallel and perpendicular axes theorems under Unit 5, Rotational Motion.
For boards, the current NCERT Class 11 chapter, Systems of Particles and Rotational Motion (reprint 2026-27), runs from section 6.1 to 6.12 and has no separate section on rolling. Practise the rolling problems below (mistakes 1, 2 and 5) from another source, such as OpenStax.
What are the five common mistakes in rotational motion problems?
Moments of inertia below come from Table 6.1 of the NCERT chapter: thin rod about its midpoint ML²/12, solid cylinder MR²/2, solid sphere 2MR²/5.
1. Taking torques about the wrong point
The slip: a solid cylinder rolls down an incline of angle θ. You take torques about the contact point, where gravity's lever arm is R, but use I = MR²/2. You get a = 2g sinθ, faster than a frictionless block. Impossible.
The fix: write τ = Iα about the centre of mass or a fixed axis. In rolling without slipping, the contact point works only with I about that point: MR²/2 + MR² = 3MR²/2. Then Mg sinθ × R = (3MR²/2)(a/R), so a = (2/3)g sinθ, which matches the result in OpenStax University Physics, section 11.1.
2. Writing friction as μN in rolling
The slip: writing f = μN, or fixing friction's direction before solving. In rolling without slipping, friction is static: it takes whatever value the motion needs, up to μsN.
The fix: assume a direction, solve for f, then compare it with μsN. For a solid cylinder on an incline, the general result in OpenStax section 11.1 gives f = (1/3)mg sinθ and the condition μs ≥ (1/3)tanθ. Take m = 2 kg, θ = 30°, μs = 0.5 and g = 10 m/s². Friction needed: (1/3)(2)(10)(0.5) ≈ 3.3 N. Maximum available: 0.5 × 2 × 10 × cos30° ≈ 8.7 N. So the cylinder rolls, and friction is 3.3 N, not 8.7 N.
3. Conserving the wrong quantity in collisions and couplings
Angular momentum about a point is conserved only if there is no net external torque about that point, as OpenStax section 11.3 states. When a hinged rod is struck, conserve angular momentum about the hinge, not linear momentum.
The slip: a disc with I = 0.02 kg m² spins at 10 rad/s. An identical disc at rest is dropped onto it and they spin together. Conserving kinetic energy gives a wrong ω = 10/√2 ≈ 7.1 rad/s.
The fix: conserve angular momentum: 0.02 × 10 = 0.04 × ω, so ω = 5 rad/s. Kinetic energy falls from 1 J to 0.5 J; half is lost to friction between the discs.
4. Starting the parallel-axis theorem from the wrong axis
The theorem is I = I_cm + md², where d is the distance from an axis through the centre of mass (OpenStax section 10.5).
The slip: find I for a rod of mass m and length L about a perpendicular axis L/4 from its centre. Starting from the end value mL²/3 and adding m(L/4)² gives 19mL²/48.
The fix: start from the centre of mass: mL²/12 + m(L/4)² = 4mL²/48 + 3mL²/48 = 7mL²/48. From any other axis, shift back to the centre first.
5. Leaving out rotational energy in rolling
The slip: a solid sphere rolls without slipping from rest down a height of 1.4 m. Writing mgh = ½mv² gives v = √(2gh) ≈ 5.2 m/s (g = 9.8 m/s²). Another common slip is subtracting 'work done by friction'.
The fix: in rolling without slipping, static friction does no work because the contact point is at rest, as OpenStax section 11.1 shows. So mgh = ½mv² + ½Iω². With I = 2mR²/5 and ω = v/R, mgh = (7/10)mv², so v² = 10gh/7 = 19.6 and v ≈ 4.4 m/s.
If you cannot find your wrong step, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
How do you check a rotational motion answer before moving on?
Ask these five questions in your next mock:
Is my torque equation about the centre of mass, a fixed axis, or the contact point with the right I?
Did I solve for friction instead of assuming μN?
Is there an external torque about the point where I conserved angular momentum?
Did my parallel-axis shift start from the centre of mass?
Does my energy equation include ½Iω² for every rolling body?
For full derivations, see our posts on rolling without slipping on inclines, the parallel and perpendicular axis theorems and conservation of angular momentum with skaters and discs. If the idea itself feels shaky, the Fermi Concept Library has a lesson for each concept in physics.
