What would you say based on your second diagram in the OP? Are all spheres at the same height? Does the answer matter?morpheus343 said:
It's at sphere B which is the midpoint of CD. So i only need the vertical distance from the centre of mass when calculating the potential energy?Orodruin said:
Yes. By definition, the potential energy for any mass distribution is given bymorpheus343 said:
The Lagrangian for a simple pendulum is defined as the difference between the kinetic energy (T) and the potential energy (V) of the system. For a pendulum of length \( l \) and mass \( m \), the Lagrangian \( L \) is given by \( L = T - V \). The kinetic energy \( T \) is \( \frac{1}{2} m l^2 \dot{\theta}^2 \) and the potential energy \( V \) is \( mgl(1 - \cos\theta) \), where \( \theta \) is the angular displacement.
To derive the equation of motion, you first write the Lagrangian \( L = \frac{1}{2} m l^2 \dot{\theta}^2 - mgl(1 - \cos\theta) \). Then, apply the Euler-Lagrange equation \( \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{\theta}} \right) - \frac{\partial L}{\partial \theta} = 0 \). This leads to \( m l^2 \ddot{\theta} + mgl \sin\theta = 0 \). Simplifying, we get the equation of motion \( \ddot{\theta} + \frac{g}{l} \sin\theta = 0 \).
The primary assumptions are that the pendulum is a rigid body with a fixed length, the motion is confined to a plane, and the pivot point is frictionless. Additionally, the mass of the string or rod is negligible compared to the mass of the bob, and air resistance is ignored.
The small-angle approximation assumes that \( \theta \) is small enough such that \( \sin\theta \approx \theta \) (in radians). Under this approximation, the equation of motion \( \ddot{\theta} + \frac{g}{l} \sin\theta = 0 \) simplifies to \( \ddot{\theta} + \frac{g}{l} \theta = 0 \), which is a simple harmonic oscillator equation.
The period \( T \) of a simple pendulum, for small oscillations, is given by \( T = 2\pi \sqrt{\frac{l}{g}} \).