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Most students can quote T = 2π√(L/g) for a simple pendulum. Then a JEE Advanced question swings a metre rule from one end, or hangs a ring on a nail, and the formula fails. The fix is not a new formula for each object. It is one torque equation that covers them all. This post derives it, works four problems with numbers, and shows the errors that cost marks.
Is this in the syllabus?
The JEE (Advanced) 2026 brochure lists "Linear and angular simple harmonic motions" along with "Rigid body, moment of inertia, parallel and perpendicular axes theorems". The JEE (Main) 2026 syllabus lists simple harmonic motion and the "simple pendulum: derivation of expression for its time period". Physical pendulums are therefore mainly a JEE Advanced and olympiad topic. Check your exam year's syllabus.
How does NCERT already derive it?
The NCERT Class 11 chapter Oscillations derives the simple pendulum using torque, not force. Taking torque about the support, τ = −L(mg sin θ). With τ = Iα and sin θ ≈ θ for small angles, it gets α = −(mgL/I)θ and so T = 2π√(I/mgL). Only at the last step does it put I = mL² for a point bob on a massless string, which gives the familiar 2π√(L/g).
That middle line is the general result. Replace L with d, the distance from the pivot to the centre of mass, and use the body's real I about the pivot:
T = 2π√(I / mgd)
Any rigid body that obeys τ = −(constant) × θ is in angular SHM, and T = 2π√(I / constant).
How do you find I about the pivot?
NCERT's Table 6.1 in the chapter on rotational motion gives moments of inertia about the centre: ML²/12 for a thin rod about its midpoint, MR² for a ring about its centre, MR²/2 for a disc about its centre. The parallel axes theorem, which the JEE Advanced syllabus lists, shifts these to a pivot a distance d away by adding Md². Our post on moment of inertia and the axis theorems covers it in detail.
Worked example 1: a rod pivoted at one end
A uniform rod of length 1.0 m swings about a horizontal axis through one end. Find T for small oscillations. Take g = 9.8 m/s².
d = L/2 = 0.50 m.
I_pivot = ML²/12 + M(L/2)² = ML²/3.
T = 2π√((ML²/3) / (Mg L/2)) = 2π√(2L/3g) = 2π√(2 × 1.0 / 29.4) = 2π × 0.261 ≈ 1.64 s.
The mass cancels. A simple pendulum with the same period would have length 2L/3 ≈ 0.67 m. This is the equivalent length.
Worked example 2: a ring on a nail
A thin ring of radius 10 cm hangs on a nail through its rim and swings in its own plane. Find T.
d = R = 0.10 m.
I_pivot = MR² + MR² = 2MR².
T = 2π√(2MR² / MgR) = 2π√(2R/g) = 2π√(0.20/9.8) = 2π × 0.143 ≈ 0.90 s.
The equivalent length is 2R, the ring's diameter.
Worked example 3: the pivot that gives the shortest period
A uniform disc of radius R = 20 cm is pivoted at a point a distance d from its centre and swings in its own plane. For which d is T smallest?
I_pivot = MR²/2 + Md².
T = 2π√((R²/2 + d²) / (g d)).
The bracket R²/(2d) + d is smallest when its derivative −R²/(2d²) + 1 = 0, that is, d = R/√2 ≈ 0.141 m.
At d = R/√2: T = 2π√((0.020 + 0.020) / (9.8 × 0.141)) ≈ 1.07 s.
At the rim, d = R = 0.20 m: T = 2π√(0.060/1.96) ≈ 1.10 s.
Near the centre, d = 0.05 m: T = 2π√(0.0225/0.49) ≈ 1.35 s.
As d → 0 the period grows without limit, because gravity's torque vanishes when the pivot is at the centre of mass. This minimum is a favourite JEE Advanced twist.
Worked example 4: a torsion pendulum
A disc of mass 0.50 kg and radius 0.10 m hangs from a wire through its centre. Twisting it by θ gives a restoring torque τ = −κθ, with κ = 0.010 N m/rad. Find T.
I = MR²/2 = 0.50 × 0.010 / 2 = 0.0025 kg m².
T = 2π√(I/κ) = 2π√(0.25) = 2π × 0.5 ≈ 3.14 s.
Gravity plays no part here. The constant in τ = −(constant)θ comes from the wire, so the same angular SHM method applies.
Which mistakes cost marks?
Using I about the centre of mass. Always shift to the pivot.
Using the rod's full length for d. d is pivot to centre of mass, L/2 for an end-pivoted rod.
Forgetting small angles. NCERT notes sin θ ≈ θ holds well up to about 20°. Larger amplitudes are not SHM.
Writing T = 2π√(L/g) for every swinging body. That is only the point-mass limit.
Mixing ω and angular velocity. In angular SHM, ω = √(mgd/I) is a constant of the motion; dθ/dt changes during the swing.
For the linear version of the same method, see our post on simple harmonic motion problems, and for spring systems, springs in series and parallel.
A practice routine
Day 1: Re-derive T = 2π√(I/mgd) from τ = Iα, as NCERT does for the bob.
Day 2: Find T for a rod pivoted at L/4 from one end. (Answer check: d = L/4 and I = 7ML²/48, giving T = 2π√(7L/12g).)
Day 3: Do the disc minimum again for a ring, then compare.
Day 4: Two torsion problems with different wires.
Day 5: Mix five oscillation problems, linear and angular, and solve without headings.
When you set up the torque correctly but your answer still disagrees with the key, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning, so you can find the exact line where your I or d went wrong.
Swing a ruler from one end tonight, time ten swings, and compare the period with 2π√(2L/3g).
