Problem — Speed of a pendulum

Pendulum from a height. A pendulum of length $L = 1\;\mathrm{m}$ is pulled to $60^\circ$ from the vertical and released. Calculate the speed at the lowest point. Does the mass matter?

Solution

Data: $L = 1\;\mathrm{m}$, initial angle $\theta = 60^\circ$ from the vertical, starting from rest.

Height difference. In the initial position the mass is higher than the lowest point by

$h = L(1 - \cos\theta) = 1\cdot(1 - \cos 60^\circ) = 1\cdot(1 - 0{,}5) = 0{,}5\;\mathrm{m}$

Conservation of mechanical energy (ideal string, no friction). All the potential energy at the starting point becomes kinetic energy at the bottom:

$mgh = \tfrac{1}{2}mv^2 \quad\Rightarrow\quad v = \sqrt{2gh} = \sqrt{2\cdot 9{,}81\cdot 0{,}5} = \sqrt{9{,}81} \approx 3{,}13\;\mathrm{m/s}$

$\ev{v \approx 3{,}13\;\mathrm{m/s}}$

Does the mass matter?

$\ev{\text{No: } m \text{ cancels out}}$

In the equation $mgh = \tfrac12 mv^2$ the mass appears on both sides and cancels: $v = \sqrt{2gh}$ depends only on $g$ and the height difference. It is the same principle as free fall: all bodies fall with the same acceleration, regardless of mass.

Collegamenti

Argomenti: Work and energy Concetti: Conservation of mechanical energy · Gravitational potential energy Competenze: Conservation of energy Oggetti: Simple pendulum