Conical pendulum — find tension, time period, and angle

Question

A mass of 0.5 kg is attached to a string of length 1 m and moves in a horizontal circle, making an angle of 30° with the vertical (conical pendulum). Find the tension in the string, the speed of the mass, and the time period of revolution.

(JEE Main 2023, similar pattern)

Solution — Step by Step

The mass moves in a horizontal circle. Two forces act on it:

Tension $T$ along the string (at angle $\theta = 30°$ with vertical)
Weight $mg$ downward

The tension has two components:

Vertical: $T\cos\theta = mg$ (balances weight — no vertical acceleration)
Horizontal: $T\sin\theta = \frac{mv^2}{r}$ (provides centripetal force)

From the vertical equation:

T\cos 30° = mg

T = \frac{mg}{\cos 30°} = \frac{0.5 \times 9.8}{\sqrt{3}/2} = \frac{4.9}{0.866} = \mathbf{5.66 \text{ N}}

The radius of circular motion: $r = L\sin\theta = 1 \times \sin 30° = 0.5$ m.

From the horizontal equation:

T\sin 30° = \frac{mv^2}{r}

5.66 \times 0.5 = \frac{0.5 \times v^2}{0.5}

2.83 = v^2

v = \mathbf{1.68 \text{ m/s}}

T_{period} = \frac{2\pi r}{v} = \frac{2\pi \times 0.5}{1.68} = \frac{\pi}{1.68} = \mathbf{1.87 \text{ s}}

Alternatively, using the direct formula:

T_{period} = 2\pi\sqrt{\frac{L\cos\theta}{g}} = 2\pi\sqrt{\frac{1 \times \cos 30°}{9.8}} = 2\pi\sqrt{\frac{0.866}{9.8}} = 2\pi \times 0.297 = 1.87 \text{ s}

Why This Works

A conical pendulum is a beautiful example of uniform circular motion where gravity and tension conspire to produce centripetal acceleration. The vertical component of tension handles gravity, while the horizontal component provides the inward pull needed for circular motion.

Notice the time period formula: $T = 2\pi\sqrt{L\cos\theta/g}$ . It looks similar to a simple pendulum ( $T = 2\pi\sqrt{L/g}$ ) but with an extra $\cos\theta$ factor. As $\theta \to 0$ , the conical pendulum reduces to a simple pendulum.

Interesting: the time period does not depend on mass. A heavier bob at the same angle takes the same time — just like a simple pendulum.

Alternative Method

Divide the two equations: $\frac{T\sin\theta}{T\cos\theta} = \frac{mv^2/r}{mg}$ , giving $\tan\theta = \frac{v^2}{rg}$ . Since $r = L\sin\theta$ : $\tan\theta = \frac{v^2}{gL\sin\theta}$ . This gives $v^2 = gL\sin\theta\tan\theta$ directly without finding $T$ first.

For JEE MCQs, the formula $T_{period} = 2\pi\sqrt{L\cos\theta/g}$ is all you need. No need to find tension or velocity separately. If the question asks “what happens to time period if angle increases?” — since $\cos\theta$ decreases as $\theta$ increases, the time period decreases. The bob goes faster at larger angles.

Common Mistake

The most common error: using $L$ as the radius of the circle instead of $L\sin\theta$ . The mass traces a circle of radius $r = L\sin\theta$ , not $L$ . The string length $L$ is the slant distance, not the horizontal radius. Draw the triangle formed by the string, vertical, and radius to see this clearly.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

Try These Next