Question

Pendulum. We discussed in class the equation of motion for the simple pendulum:

$ml\ddot{\Theta } = -mgsin\Theta$.

Here m is the mass of the bob, $l$ is the length of the arm, and $g$ is the acceleration of gravity,
and $\Theta$ is the angle of the arm from away from the vertical.

The total energy of the pendulum is a sum of the kinetic and potential terms:

$E = \frac{1}{2}m(l\dot{\Theta })^2$  $E = -mglcos\Theta$

a. Draw a picture of the pendulum that shows all of the parameters.

b. Show that the equation of motion implies that $E$ is a constant, i.e. $\dot{E} = 0$ [Hint: use the chain
rule to compute $\dot{E}$ directly in terms of $\Theta$, $\dot{\Theta }$, and $\ddot{ \Theta }$ , and then use the equation of motion to eliminate the $\ddot{\Theta }$ term.]

c. Suppose we start the pendulum at $t = 0$ with initial conditions $\Theta (0) = 0$ and $\dot{\Theta } = \sigma$ , where $\sigma > 0$ is a constant with units of angular velocity (i.e. frequency). What is the minimum value for $\sigma$ in order for the pendulum to reach $\Theta = \pi$?

Answer 1

a) mg me Econstan C) Eequating enuey at A and B