A stone tied to string whirled in horizontal circle — find tension

Question

A stone of mass 0.5 kg is tied to a string of length 1.5 m and whirled in a horizontal circle at a speed of 6 m/s. Find the tension in the string.

Solution — Step by Step

When an object moves in a circle, it needs a centripetal force directed toward the centre. For a stone whirling on a string in a horizontal plane, the only horizontal force acting on the stone is the tension in the string — and it points toward the centre (inward along the string).

Therefore: Tension = Centripetal force required.

F_c = \frac{mv^2}{r}

Where:

$m$ = mass of stone = 0.5 kg
$v$ = speed = 6 m/s
$r$ = radius = length of string = 1.5 m

T = \frac{mv^2}{r} = \frac{0.5 \times (6)^2}{1.5}

= \frac{0.5 \times 36}{1.5} = \frac{18}{1.5} = 12 \text{ N}

The tension in the string is 12 N.

If the string can withstand at most 12 N, this is right at the breaking limit. Increasing the speed further would require more tension, and the string would snap — the stone would then fly off tangentially.

Why This Works

In uniform circular motion, the speed is constant but the direction continuously changes. This change in direction means the velocity is changing — so there IS acceleration, directed toward the centre. By Newton’s 2nd law, there must be a net force in that direction: $F = ma_c = mv^2/r$ .

The string can only pull (not push), so tension always acts inward. This is exactly the centripetal direction. No other horizontal force acts on the stone (ignoring air resistance), so tension equals the centripetal force needed.

The formula can also be written as $T = m\omega^2 r$ where $\omega$ is the angular velocity. Since $v = \omega r$ , the two are equivalent: $mv^2/r = m(\omega r)^2/r = m\omega^2 r$ . In JEE, you’ll often be given angular velocity instead of linear speed — use the second form directly.

Alternative — Using Angular Velocity

If the stone completes 2 revolutions per second, then $\omega = 2\pi \times 2 = 4\pi$ rad/s.

T = m\omega^2 r = 0.5 \times (4\pi)^2 \times 1.5 = 0.5 \times 16\pi^2 \times 1.5 \approx 118.4 \text{ N}

This shows how dramatically tension increases with angular velocity — the $\omega^2$ term means doubling frequency quadruples the tension.

Common Mistake

A very common error is thinking “centripetal force” is a separate, additional force acting on the object. There is no “centripetal force” as a new force — it’s a role played by an existing force (here, tension). The tension IS the centripetal force. If you draw a free body diagram and add “centripetal force” as a separate arrow alongside tension, you’ve double-counted. Free body diagrams should show only real forces: tension and gravity (weight). Centripetal force is the net inward component of those real forces.

Question

Solution — Step by Step

Why This Works

Alternative — Using Angular Velocity

Common Mistake

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