Equilibrium — Question 4

Question 4

A turning point in a pendulum is not a static state

A simple pendulum of length $1.0\,\mathrm m$ and bob mass $0.50\,\mathrm{kg}$ is instantaneously at rest at $30^\circ$ from downward vertical. Use $g=10\,\mathrm{m/s^2}$; the cord is taut.

1. Find tangential acceleration and tension at that instant.

2. Why is its zero speed not evidence of equilibrium?

Question 4 - Solution

A turning point in a pendulum is not a static state

(a) Derivation and calculation

Resolve gravity into radial and tangential components. Toward the bottom, $$\boxed{|a_t|=g\sin30^\circ=5\ \mathrm{m/s^2}}.$$ The inward radial equation at $v=0$ is $$T-mg\cos30^\circ=mv^2/L=0,$$ $$\boxed{T=(.50)(10)\cos30^\circ=\frac{5\sqrt3}{2}\ \mathrm N\simeq4.330\ \mathrm N}.$$

(b) Result and interpretation

Taking increasing angular displacement away from the bottom, $$\tau=-mgL\sin\theta,\qquad \ddot\theta=-\frac gL\sin\theta.$$ At $30^\circ$, $$\boxed{\ddot\theta=-5\ \mathrm{rad/s^2}\ne0}.$$ The cord balances only radial gravity; the tangential component reverses the motion.

Physical check

Tension is positive, consistent with the taut-cord assumption, but the net force is not zero.