Answer to Question #111420 in Mechanics | Relativity for Brandon Salas

Question

An ideal pendulum of length L = 1.2 m supports a mass of m = 0.5 kg. Initially the pendulum is lifted such that it makes an angle of θ = 21 degrees with respect to the vertical.

part(A) When the pendulum is released from rest, what is the maximum speed, vm in m/s, the mass reaches?

Answer 1

When the pendulum is in lifted state, it has potential energy $mgh$, where $h$ is the height above the origin, and zero kinetic energy.

Pendulum reaches its maximum speed, when going through the equilibrium point ($\theta = 0$).

Lets choose the origin to be located at the equilibrium position of the mass.

Then, in lifted state, the height above the origin will be equal to $h = L - L \cos \theta$ and in maximum speed state $h = 0$.

Hence, total energy in lifted state is just potential energy $E_1 = mgh = mg (L - L\cos \theta)$, and in maximum speed state it is just kinetic energy $E_2 = \frac{m v^2}{2}$.

By the law of conservation of energy, these two energies must be equal, therefore $m g L (1- \cos \theta) = \frac{m v^2}{2}$, from where $v = \sqrt{ 2 g L (1 - \cos \theta)} \approx 4.69\frac{m}{s}$.