A 0.1 kg ball on a 1 m thread moves in a circle, the thread is 60° from the vertical. Find the tension and the period — Physics problems | PhysLabo
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Conical pendulum: thread tension and period

Medium 10 grade Mechanics / Circular Motion
Problem statement

A ball of mass $m = 0.1$ kg hangs on a thread of length $l = 1$ m and moves uniformly in a circle in a horizontal plane. The thread makes an angle $\alpha = 60°$ with the vertical (see the figure). Find the tension $T$ in the thread and the period $\tau$ of the ball's revolution. Take $g = 10$ m/s².

A conical pendulum: a ball on a thread of length l moves in a circle of radius r, the thread makes angle α with the vertical; mg, T and acceleration a are shown
Given
  • $m = 0.1$ kg
  • $l = 1$ m
  • $\alpha = 60°$
  • $g = 10$ m/s²
  • $T = ?$
  • $\tau = ?$
Hints
Hint 1 / 2
The radius of the circle is $r = l\sin\alpha$. The ball does not accelerate vertically, so the vertical component of the tension balances $mg$.
Hint 2 / 2
The horizontal component of the tension, $T\sin\alpha$, gives the centripetal acceleration: $T\sin\alpha = m\omega^2 r$. Find $\omega$ from it, then the period $\tau = 2\pi/\omega$.
Solution

The ball moves in a circle of radius $r = l\sin\alpha$ with a centripetal acceleration directed horizontally toward the axis. The forces on the ball are gravity and the thread tension.

Force triangle and top view of the circle
Force triangle and top view of the circle

1. There is no vertical acceleration:

$$T\cos\alpha = mg \;\Rightarrow\; T = \frac{mg}{\cos\alpha} = \frac{1}{0.5} = 2 \text{ N}$$

2. Horizontally, a component of the tension provides the centripetal acceleration $a = \omega^2 r$:

$$T\sin\alpha = m\omega^2 r = m\omega^2 l\sin\alpha \;\Rightarrow\; T = m\omega^2 l$$$$\omega^2 = \frac{T}{ml} = \frac{2}{0.1 \cdot 1} = 20 \;\Rightarrow\; \omega \approx 4.47 \text{ rad/s}$$

3. The period:

$$\tau = \frac{2\pi}{\omega} = \frac{2 \cdot 3.14}{4.47} \approx 1.41 \text{ s}$$

The general formula is $\tau = 2\pi\sqrt{\dfrac{l\cos\alpha}{g}}$ — the period does not depend on the mass of the ball.

Answer

$T = 2$ N, $\tau \approx 1.41$ s

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