Equation of Motion

Given the angle, what is the angular acceleration?

Two results are now in hand:

• The torque gravity applies: \(\tau = -mg\frac{L}{2}\sin\theta\)
• The rod’s moment of inertia: \(I = \frac{mL^2}{3}\)

Newton’s second law, for rotation

For something sliding, a force produces an acceleration, \(a = F/m\). Turning works the same way, with torque in place of force and moment of inertia in place of mass:

\[\alpha = \frac{\tau}{I}\]

Putting it together

\[\alpha = \frac{-mg\frac{L}{2}\sin\theta}{\frac{mL^2}{3}} = -\frac{3g}{2L}\sin\theta\]

Writing \(\alpha\) as the second derivative of the angle gives the rod’s equation of motion:

\[\ddot{\theta} = -\frac{3g}{2L}\sin\theta\]

What it tells us

• The mass cancels. A heavier rod feels more torque but is equally harder to spin, so the two cancel exactly. Only the length and gravity set the motion.
• The minus sign means the acceleration always points back towards hanging straight down, so the rod swings back and forth around \(\theta = 0\) instead of running away.
• Shorter rods swing faster. \(L\) sits in the denominator.

For the 0.3 m rod with \(g = 9.81 \ \text{m/s}^2\):

\[\ddot{\theta} \approx -49\,\sin\theta \ \ \text{rad/s}^2\]

This one equation is the model. Given the angle at any instant, it gives the angular acceleration.

Stepping this equation forward in time predicts the whole swing: see The Simulation.