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Most marks in rotational motion are lost to a handful of repeat errors, not to difficult ideas. Here are four, each with a worked example you can redo on paper. The standard results come from OpenStax University Physics, a free textbook, and NCERT's Class 11 chapter on systems of particles and rotational motion covers torque and moment of inertia. Take g = 9.8 m/s² throughout.
Mistake 1: Using a moment of inertia without naming the axis
OpenStax puts it plainly: "the moment of inertia for any object depends on the chosen axis" (OpenStax, calculating moments of inertia). A uniform thin rod has I = (1/12)mL² about its centre but (1/3)mL² about one end.
The parallel-axis theorem, I = I_cm + md², starts from the axis through the centre of mass, where d is the distance from that axis to the new parallel axis. The common slip is to jump between two axes, neither of which passes through the centre of mass.
Worked example. A 2 kg rod is 1.2 m long. Find I about an axis perpendicular to the rod, 0.3 m from one end.
I_cm = (1/12)(2)(1.2)² = 0.24 kg m².
The new axis is 0.6 − 0.3 = 0.3 m from the centre, so I = 0.24 + 2(0.3)² = 0.42 kg m².
The wrong route takes the end value, (1/3)(2)(1.2)² = 0.96 kg m², and adds 2(0.3)² to get 1.14 kg m². That breaks the theorem's condition.
Mistake 2: Multiplying r by F without the angle
Torque is τ = r × F, with magnitude rF sin θ, where θ is the angle between r and F. The quantity r sin θ is the lever arm, the perpendicular distance from the axis to the line of the force, and counterclockwise torque is positive by convention (OpenStax, torque).
Worked example. You pull with 10 N at the end of a 0.5 m spanner, at 30° to the spanner. τ = 0.5 × 10 × sin 30° = 2.5 N m. Writing rF gives 5 N m, double the real value. Quick check: a force along the spanner (θ = 0) must give zero torque.
When several forces act, choose the axis once, give each torque a sign, then add. Changing axis halfway is the rotational version of switching coordinate axes mid-problem, which our post on vector errors in JEE physics covers.
Mistake 3: Writing friction as μN in rolling
For rolling without slipping, v_cm = Rω. OpenStax points out that the bottom of a rolling wheel is at rest relative to the ground, "indicating there must be static friction between the tires and the road surface" (OpenStax, rolling motion). Static friction takes whatever value the motion needs, up to its maximum, so find it from the equations of motion.
Worked example. A 3 kg solid cylinder (I = (1/2)mR²) rolls without slipping down a 30° incline.
Along the incline: mg sin θ − f = ma.
About the centre: fR = Iα = (1/2)mR²(a/R), so f = (1/2)ma.
Add them: mg sin θ = (3/2)ma, so a = (2/3)g sin θ ≈ 3.27 m/s². This matches the OpenStax result for a solid cylinder.
Then f = (1/2)(3)(3.27) ≈ 4.9 N.
OpenStax also points out that this acceleration is independent of the coefficient of static friction. If a question gives μ, use it to check that rolling without slipping is possible (f must not exceed μN), not to compute f.
Mistake 4: Forgetting rotational kinetic energy
In rolling without slipping, the contact point does not slide, so the work done by static friction is zero and, with no other non-conservative forces, the total energy is conserved (OpenStax, rolling motion). But the kinetic energy has two parts: (1/2)mv² + (1/2)Iω².
Worked example. The same cylinder starts from rest 1.5 m above the bottom. Energy: mgh = (1/2)mv² + (1/2)(1/2)mR²(v/R)² = (3/4)mv². So v = √(4gh/3) = √19.6 ≈ 4.43 m/s. Leaving out the rotational term gives √(2gh) ≈ 5.42 m/s, the speed of a block sliding down without friction. If your rolling object arrives as fast as a frictionless block, a term is missing.
A contact point that stays still while the centre moves is hard to picture; the Fermi Simulation Lab has interactive simulations for the ideas that are easier to see than to read. Drawing the forces before writing any equation also helps; see why drawing the diagram first helps.
A five-line checklist before you move on
Every I and every torque has its axis written next to it.
The parallel-axis theorem starts from the centre-of-mass axis.
Torque is force times perpendicular distance, with a sign.
Rolling: v = Rω, friction is static and found from the equations.
Energy includes both translational and rotational terms.
If your answer differs from a worked solution and you cannot see why, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer. For more errors of this kind across mechanics, read five JEE mechanics mistakes and the habits that fix them.
