Answer to Question #180692 in Mechanics | Relativity for Royce Karam

Question

A simple pendulum consists of a 0.8-kg bob connected to a massless inextensible cord with a length L = 1.4 m. The bob is set into motion and its angular displacement is given by θ(t) = 0.11cos⁡(ωt), where θ is in radians and t is in seconds. Take g = 9.8 m/s^2, determine the mechanical energy of this pendulum.

Answer 1

As the total mechanical energy is determined up to a constant, we can fix it so that the mechanical energy at the lowest point of the trajectory is the kinetic energy (i.e. the potential energy at this point is zero) :

$E_m = E_{k. max}=\max\frac{mv^2}{2}$

As the displacement is angular, we can express the velocity as $v= L\dot\theta=-L\omega \theta_{max} \sin(\omega t)$. Now, using that for a simple pendulum $\omega=\sqrt{\frac{g}{L}}$, we get $v=-\theta_{max}\sqrt{gL} \sin(\omega t)$ and so

$E_m = \max(\frac{mgL \theta^2_{max}}{2} \sin(\omega t))=\frac{mgL\theta^2_{max}}{2}$

$E_m = \frac{0.8 \cdot 9.8\cdot 1.4\cdot0.11^2}{2}\approx 0.066 J = 66mJ$

Another method of calculating the mechanical energy could be calculating the potential energy, which would give

$E_m = mgL(1-\cos \theta_{max})$ with a little bit of trigonometry. This is result is different at first glance, however as $\theta_{max}\ll 1$, developping $\cos$ gives us $E_m \approx mgL \frac{\theta_{max}^2}{2}$ which coincides with the previous result.