Why am I getting this derivation of time period of pendulum in an accelerated frame wrong? [closed]

Question

We are working in the frame of the cart and we are trying to obtain the $\tau=k\theta$ form.

So, let's write the $\tau=I_{axis}\alpha$ first for a small deviation $\theta$ from the vartical.

(The moment of inertia of the bob about the point of suspension would approximately be equal to $ml^2$ as the bob is small.)

Thus,

$mgl\sin\theta - mal\cos\theta=ml^2\alpha$

As $\theta$ is small, $$gl\theta-al=l^2 \alpha$$

Solving this would give $T=2π\sqrt{\frac{l}{g}}$ and not $T=2π\sqrt{\frac{l}{g_{eff}}}$

What am I doing wrong?

Answer 1

Since the equilibrium angle $\theta_0=\arctan\frac{a_0}{g}\neq0$ has now changed, you cannot assume $\theta$ is small. Instead, you can define a shifted angle $\varphi:=\theta-\theta_0$, and use the small-angle approximation on this instead (i.e. Taylor expand around $\theta_0$ rather than $0$).

Answer 2

starting with the free body diagram

obtain the sum of the torques about point A

$$\sum \tau=m\,L^2\ddot\theta+m\,g\,L\,\sin(\theta)- m\,a_a\,L\,\cos(\theta)=0$$

from here with $~\theta=\theta_0+\phi~$

$$\,L^2\ddot\phi+\,g\,L\,\sin(\theta_0+\phi)- \,a_a\,L\,\cos(\theta_0+\phi)=0\tag 2$$

linearized equation (2) with

$$\sin(\theta_0+\phi)=\sin \left( \theta_{{0}} \right) \cos \left( \phi \right) +\cos \left( \theta_{{0}} \right) \sin \left( \phi \right) \\ \cos(\theta_0+\phi)=\cos \left( \theta_{{0}} \right) \cos \left( \phi \right) -\sin \left( \theta_{{0}} \right) \sin \left( \phi \right) \\ \sin(\phi)=\phi~,\cos(\phi)=1 $$

$$L^2\ddot\phi+ \left( g\,L\cos \left( \theta_{{0}} \right) +a_{{0}}L\sin \left( \theta _{{0}} \right) \right) \phi +g\,L\sin \left( \theta_{{0}} \right) -a_{{0}}L\cos \left( \theta_{{0}} \right) =0\tag 3$$

for $~\ddot\phi=0~,\phi=0~$ you obtain $~\theta_0= \arctan\left(\frac{a_0}{g}\right)$

thus equation (3)

$$\ddot\phi+\omega\,\phi=0\quad,\omega=\frac{\sqrt{a_0^2+g^2}}{L}$$