Conical Pendulum: Tension, Angle and the Minimum Speed
A small ball of mass 0.4 kg is attached to a light inextensible string of length 1.2 m whose other end is fixed. The ball is set moving in a horizontal circle so that the string traces a cone of semi-vertical angle θ with the vertical. The string can withstand a maximum tension of 25 N. A student wants to spin the ball as fast as possible and assumes that a larger angle of the string automatically means a larger tension, so the string will break at a small angle. You are asked to test that assumption using the conical pendulum model with g = 9.8 m/s². The only forces on the ball are the tension along the string and its weight: the vertical components balance and the horizontal component supplies the centripetal force. Consider very low angular speeds as well as high ones, and decide whether there is a minimum angular speed below which steady conical motion is impossible.
1. Show that the tension is given by T = mω²L and use it to find the maximum angular speed before the string breaks.
Read the worked answer
Resolve the forces. Vertically: T cosθ = mg. Horizontally the tension component supplies the centripetal force: T sinθ = mω²r, where r = L sinθ.
Substituting r:
T sinθ = mω²L sinθ → T = mω²L (for θ ≠ 0).
Comment on the student's assumption. The tension depends only on m, ω and L - it does not contain θ at all. So at a fixed angular speed, letting the ball swing out to a larger angle does not increase the tension. The angle adjusts to the speed; the tension does not depend on the angle.
Maximum ω. Breaking occurs at T = 25 N:
25 = 0.4 × ω² × 1.2 = 0.48ω² → ω² = 52.08 → ω = 7.22 rad/s.
At that speed cosθ = mg/T = (0.4 × 9.8)/25 = 0.157, so θ = 81°. The string breaks when the string is almost horizontal, not when θ is small - the student's intuition is exactly backwards.
2. At ω = 5 rad/s, find the tension, the semi-vertical angle and the radius of the circle, and verify consistency.
Read the worked answer
Tension. T = mω²L = 0.4 × 25 × 1.2 = 12 N, well below the 25 N limit, so the string is safe.
Angle. From T cosθ = mg: cosθ = 3.92/12 = 0.327, so θ = 70.9°.
Radius. r = L sinθ = 1.2 × 0.945 = 1.13 m.
Checks. Vertical balance: T cosθ = 12 × 0.327 = 3.92 N = mg ✓. Horizontal: mω²r = 0.4 × 25 × 1.13 = 11.3 N and T sinθ = 12 × 0.945 = 11.3 N ✓. Both original equations are satisfied, confirming that T = mω²L is a valid shortcut and that the angle is determined by ω through cosθ = g/(ω²L).
3. Find the minimum angular speed for steady conical motion, and describe what happens below it. State the full safe operating range.
Read the worked answer
Minimum speed. Combining T cosθ = mg with T = mω²L gives
cosθ = g/(ω²L).
Since cosθ ≤ 1, steady conical motion requires ω² ≥ g/L, i.e.
ω ≥ √(9.8/1.2) = 2.86 rad/s.
Below the minimum. If ω < 2.86 rad/s the equation would require cosθ > 1, which is impossible. Physically the ball cannot maintain a horizontal circle: it hangs vertically (θ = 0), where there is no horizontal force to provide the centripetal acceleration. At exactly ω = 2.86 rad/s the limiting case θ = 0 occurs with T = mg = 3.92 N.
Safe range. 2.86 rad/s ≤ ω ≤ 7.22 rad/s. Below the lower limit the motion collapses to a vertical hang; above the upper limit the string breaks at T = 25 N with θ ≈ 81°. The student's fear of breaking the string at a small angle is misplaced - small angles correspond to low speeds and low tension.