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A small bob on a string, swinging so that it traces a horizontal circle while the string sweeps out a cone, is one of the most common circular motion set-ups in JEE papers. It looks simple, but students often lose marks by taking the wrong radius or adding a force that is not there. This post gives the two equations that solve every conical pendulum problem, three worked examples and a short practice plan.
Which forces act on the bob?
Only two: the weight mg straight down, and the tension T along the string. There is no separate "centripetal force". The NCERT Class 11 chapter Laws of Motion (section 4.10) explains that the centripetal force mv²/R is not a new force but the net force that real forces must provide. In a conical pendulum the horizontal part of the tension does that job.
The Syllabus for JEE (Main) 2026 lists "dynamics of uniform circular motion, centripetal force and its applications" under Laws of Motion. Check the syllabus for your own exam year.
What are the two equations?
Let the string have length L and make angle θ with the vertical. The bob moves in a circle of radius r = L sin θ, not L.
Vertical (no acceleration): T cos θ = mg.
Horizontal (towards the centre): T sin θ = m ω² r = m ω² L sin θ.
From equation 2, the sin θ cancels and T = m ω² L. Dividing 2 by 1 gives tan θ = ω² r / g, or ω² = g / (L cos θ). So the period is
T_period = 2π √(L cos θ / g) = 2π √(h / g),
where h = L cos θ is the height of the suspension point above the plane of the circle. The period depends only on h, not on the mass or the length alone.
Worked example 1: find tension, period and speed
A 0.20 kg bob on a 1.0 m string moves in a horizontal circle with the string at 60° to the vertical. Take g = 9.8 m/s².
Tension: T = mg / cos 60° = (0.20 × 9.8) / 0.50 = 3.92 N.
Height: h = L cos 60° = 0.50 m.
Period: 2π √(0.50 / 9.8) = 2π × 0.226 ≈ 1.42 s.
Radius: r = 1.0 × sin 60° ≈ 0.866 m.
Speed: from tan θ = v² / (r g), v² = 9.8 × 0.866 × 1.732 ≈ 14.7, so v ≈ 3.83 m/s.
Check: ω = 2π / 1.42 ≈ 4.43 rad/s and ωr ≈ 4.43 × 0.866 ≈ 3.83 m/s. The two routes agree.
Worked example 2: find the angle from the spin rate
A 0.10 kg bob on a 0.50 m string is whirled at 2.0 revolutions per second. Find the angle and the tension.
ω = 2π × 2.0 = 4π ≈ 12.57 rad/s, so ω² ≈ 157.9 s⁻².
cos θ = g / (ω² L) = 9.8 / (157.9 × 0.50) ≈ 0.124, so θ ≈ 82.9°.
T = m ω² L = 0.10 × 157.9 × 0.50 ≈ 7.9 N.
Check: T cos θ ≈ 7.9 × 0.124 ≈ 0.98 N, which equals mg. At high spin rates the string is nearly horizontal but never exactly, because some tension must always hold up the weight.
There is also a minimum spin rate. Since cos θ cannot exceed 1, conical motion needs ω² > g/L. For L = 0.50 m that is ω > √19.6 ≈ 4.4 rad/s. Spin slower and the bob simply hangs.
Worked example 3: when does the string break?
A string of length 1.0 m can bear at most 5.0 N. A 0.20 kg bob is spun faster and faster. At what ω and angle does it break?
T = m ω² L, so 5.0 = 0.20 × ω² × 1.0, giving ω² = 25 and ω = 5.0 rad/s.
cos θ = mg / T = 1.96 / 5.0 = 0.392, so θ ≈ 66.9°.
Exam questions like this test whether you know that T = mω²L, which avoids having to find θ first.
Which mistakes are most common?
Using r = L. Always r = L sin θ.
Adding a centrifugal force in the ground frame. In the ground frame there are only two forces. A centrifugal force belongs only to a rotating frame, and then you must leave out the mv²/r on the other side.
Writing T = mg. The tension is always larger than mg, because T cos θ = mg and cos θ < 1.
Thinking the period depends on L. It depends on h = L cos θ. Two pendulums of different lengths hanging from the same ceiling with bobs at the same height have the same period.
Mixing rev/s and rad/s. Multiply revolutions per second by 2π first.
The same tan θ = v²/(rg) appears in a vehicle on a frictionless banked road, where the normal force plays the role of the tension. Our guide to circular motion and banking problems shows the comparison.
Does solving more problems fix the concept?
Not on its own. Kim and Pak (2002) surveyed students who estimated they had solved between 300 and 2,900 textbook physics problems, about 1,500 on average. The students handled formulas and mathematics well but still had many of the well-known conceptual difficulties in basic mechanics, with little correlation between the number of problems solved and conceptual understanding. Quantity of practice is not the same as checking your force diagram.
A practice routine
Free-body diagram first, always, with just two arrows. Our post on free-body diagrams and Newton's second law has a checklist.
Solve three problems three ways: given θ find ω; given ω find θ; given breaking tension find both.
Do one limit check per problem. As ω grows, θ should approach 90° and T should grow as ω².
If you get stuck at the set-up, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning, so you can compare your force equations with a full solution.
Next time you practise, solve one conical pendulum problem without the angle given, using only T = mω²L and T cos θ = mg.
