Conical pendulum

Problem

A mass $m = 0{,}50\ \text{kg}$ hangs from a string of length $L = 1{,}0\ \text{m}$ and moves in uniform horizontal circular motion, with the string making an angle $\theta = 30^\circ$ with the vertical. Calculate the tension in the string and the tangential speed of the mass.

Solution

The mass describes a horizontal circle at constant speed. Two forces act on it: the weight $\vv{P}$ (vertical, downward) and the tension $\vv{T}$ (along the string). Their resultant must be horizontal and directed towards the centre, since it constitutes the centripetal force.

We resolve the tension along the vertical and horizontal directions.

Vertical equilibrium. There is no vertical acceleration, so the vertical component of the tension balances the weight: $T\cos\theta = mg$

From here we obtain the tension: $T = \frac{mg}{\cos\theta} = \frac{0{,}50 \cdot 9{,}81}{\cos 30^\circ} = \frac{4{,}905}{0{,}866}$

$\ev{T \approx 5{,}66\ \text{N}}$

Horizontal direction. The horizontal component of the tension provides the centripetal force: $T\sin\theta = \frac{m v^2}{r}$

The radius of the circle is the horizontal projection of the string: $r = L\sin\theta = 1{,}0 \cdot 0{,}500 = 0{,}50\ \text{m}$

We obtain the speed: $v^2 = \frac{T\sin\theta \cdot r}{m} = \frac{5{,}66 \cdot 0{,}500 \cdot 0{,}50}{0{,}50} = 2{,}83$

$\ev{v \approx 1{,}68\ \text{m/s}}$

As a check, the same speed can be found by dividing the two equations term by term: $\tan\theta = v^2/(g r)$, giving $v = \sqrt{g\,r\tan\theta}$, consistent with the value found.

Links

Topics: Dynamics Concepts: Newton’s second law · Centripetal force · Uniform circular motion · Tension Skills: Vector resolution Objects: Conical pendulum