Answer to Question #145428 in Classical Mechanics for alwande hadebe

Question

Solve the equations of motion for the roll pendulum (cp. Example 3 in the
lectures) for small oscillation angle ϕ  1 and initial conditions:
x1(0) = ˙x1(0) = 0 , ϕ(0) = ϕ0 , ϕ˙(0) = 0 .For small angles ϕ  1, use the approximations sin(ϕ) ≈ ϕ, cos ϕ ≈ 1 and ˙ϕ
2
sin ϕ ≈0

Answer 1

Let the mass and length of the string of the simple pendulum is m and L respectively. Which is pivoted at the point P. Now it is displaced small angle and released. Now the pendulum will swing back and forth motion.

Applying newton's second law, for rotational motion

$\tau =I\alpha$

$\Rightarrow -mg\sin\phi L=mL^2\frac{d^2\phi}{dt^2}$

$\Rightarrow \frac{d^2\phi}{dt^2}+\frac{g}{L}\sin\phi =0$

if angle is small, then $\sin \phi = \phi$

Hence, $\Rightarrow \frac{d^2\phi}{dt^2}+\frac{g}{L}\phi =0$