Conical pendulum

Problem

An object of mass $m$ hangs from a string of length $\ell = 2\;\text{m}$ and rotates describing a horizontal circle with period $T = 1\;\text{s}$. Find the angle of the string with the vertical.

Solution

Data. $\ell = 2\;\text{m}$, $T = 1\;\text{s}$, $g = 9{,}81\;\text{m/s}^2$.

Geometry. The string makes an angle $\theta$ with the vertical; the radius of the horizontal circle is $r = \ell\sin\theta$

Components of the tension. Towards the centre (horizontal) $t_x = t\sin\theta$; upwards (vertical) $t_y = t\cos\theta$.

Newton’s second law. Horizontal (provides the centripetal force $m\omega^2 r$) and vertical (equilibrium with the weight): $\begin{cases} t\sin\theta = m\omega^2 r = m\omega^2 \ell\sin\theta \\ t\cos\theta = mg \end{cases}$

From the first equation, cancelling $\sin\theta$, we are left with: $t = m\omega^2 \ell$

Substituting into the second: $m\omega^2 \ell\cos\theta = mg \quad\Rightarrow\quad \cos\theta = \frac{g}{\omega^2 \ell}$ (the mass $m$ cancels: the angle does not depend on the mass).

Angular velocity: $\omega = \frac{2\pi}{T} = 2\pi\;\text{rad/s}$

Numerical calculation: $\cos\theta = \frac{9{,}81}{(2\pi)^2 \cdot 2} = \frac{9{,}81}{78{,}96} = 0{,}124 \quad\Rightarrow\quad \theta = \arccos(0{,}124) = 82{,}9^\circ$

Result. $\ev{\cos\theta = \frac{g}{\omega^2 \ell} \;\Rightarrow\; \theta \approx 82{,}9^\circ}$

Links

Topics: Dinamica Concepts: Moto circolare uniforme · Forza centripeta · Tensione · Velocità angolare · Forza peso Skills: Scomposizione vettoriale · Uso della trigonometria Objects: Pendolo conico · Fune ideale