Speed of a conical pendulum

Problem

A small sphere is tied to a string of length $\ell = 0{,}80\;\mathrm{m}$ and rotates in a horizontal circle (conical pendulum). If the angle of the string with the vertical is $\theta = 30^\circ$, what is the speed of the sphere?

Solution

Geometry. The string traces a cone of half-angle $\theta$. The radius of the horizontal circle is the projection of the string: $r = \ell\sin\theta = 0{,}80\cdot\sin 30^\circ = 0{,}80\cdot 0{,}50 = 0{,}40\;\mathrm{m}$

Free-body diagram. Only the weight $mg$ (vertical, downward) and the tension $T$ along the string act on the sphere. The tension splits into:

• a vertical component $T\cos\theta$, which balances the weight;
• a horizontal component $T\sin\theta$, which provides the centripetal force.

Newton’s equations. $\begin{aligned} T\cos\theta &= mg \\ T\sin\theta &= \frac{m v^2}{r} \end{aligned}$

Eliminating $T$. Dividing the second equation by the first term by term ($T$ and $m$ cancel): $\tan\theta = \frac{v^2}{g\,r} \;\Rightarrow\; v = \sqrt{g\,r\,\tan\theta}$

Numerical substitution. $v = \sqrt{9{,}8\cdot 0{,}40\cdot\tan 30^\circ} = \sqrt{9{,}8\cdot 0{,}40\cdot 0{,}577} = \sqrt{2{,}26}\;\mathrm{m/s}$

$\ev{v \approx 1{,}5\;\mathrm{m/s}}$

Note: the speed does not depend on the mass of the sphere (it cancels out), only on the length of the string and the angle.

Links

Topics: Dinamica Concepts: Forza centripeta · Tensione · Moto circolare uniforme · Forza peso Skills: Diagramma di corpo libero · Scomposizione vettoriale · Uso della trigonometria · Applicazione delle leggi di Newton Methods: Metodo del diagramma di corpo libero Objects: Pendolo conico · Fune ideale