Physics

Physics

Problem

For a simple pendulum of length LL and bob mass mm, the net torque about the pivot is τ=−mgLsin⁡θ\tau = -mgL\sin\theta, where θ\theta is the angular displacement from equilibrium. Derive an expression for the angular acceleration α\alpha of the bob for small angular displacements and identify the constant of proportionality between α\alpha and θ\theta.
Intuitive explanation

The key is recognizing that torque and angular acceleration are related by rotational Newton's second law: τ=Iα\tau = I\alpha. For a point mass at distance LL from the pivot, the moment of inertia is I=mL2I = mL^2. For small angles, sin⁡θ≈θ\sin\theta \approx \theta, which linearizes the equation and reveals simple harmonic motion—the angular acceleration becomes directly proportional to displacement, with a constant that depends on gravity and length.

Intuitive explanation

The key is recognizing that torque and angular acceleration are related by rotational Newton's second law: τ=Iα\tau = I\alpha. For a point mass at distance LL from the pivot, the moment of inertia is I=mL2I = mL^2. For small angles, sin⁡θ≈θ\sin\theta \approx \theta, which linearizes the equation and reveals simple harmonic motion—the angular acceleration becomes directly proportional to displacement, with a constant that depends on gravity and length.

Browse the problem library