# A pendulum in a non-inertial frame ?

Nov 12, 2016

Considering the car as a non-inertial reference frame, the pendulum oscilates around the equilibrium position. This occurs with the rope at angle $\phi$ such that

$\phi = \arctan \left(\frac{a}{g}\right)$

because with the rope at this angle, the momentum of $m \vec{\alpha} = m \left(- a \hat{i} - g \hat{j}\right)$ regarding the pendulum fulcrum is null.

Around this point, the small oscillation period must be

$T = 2 \pi \sqrt{\frac{l}{\left\lVert \vec{\alpha} \right\rVert}}$

Note that without additional acceleration should be $2 \pi \sqrt{\frac{l}{g}}$

so

$T = 2 \pi \sqrt{\frac{l}{\sqrt{{g}^{2} + {a}^{2}}}}$