# How do we analyze a pendulum from an Inertial frame?

When we analyze a system of the pendulum and a car in which both are moving with an acceleration $$a_0$$.

We can easily analyze the situation from inside the Non Inertial frame, the pendulum is at rest (" Inside the Non-Inertial frame ") , we may find the angle which the pendulum makes with the vertical as $$\displaystyle\frac{a_0}{g}$$. The problem i'm facing is that if we do all the measurements outside the Non Inertial frame than how can we write the equations since then there is no force pseudo-force that counts while choosing the Inertial frame as the system. We will only have the gravitational force and the tension then how can we account the acceleration of the pendulum. I knwo im wrong somewhere.

• Possible duplicate here. Jun 1, 2021 at 7:41
• Yes, It is almost same but have a different problem than mine Jun 2, 2021 at 2:25

I think it's worth looking at this in some detail as working in non-inertial frames can be a dangerous thing to do. It's easy to find yourself making mistakes because assumptions you are used to making in inertial frames no longer apply. It may seem that the calculation is harder in the ground (inertial) frame but it is just a matter of applying the second law.

In this case we have an unknown tension $$T$$ in the string, and an unknown angle, $$\theta$$. However we can use these two unknowns to write two equations for the horizontal and vertical acceleration of the bob.

Consider the vertical acceleration first. We know this is zero so the net vertical force on the bob must be zero. There is a downwards gravitational force $$mg$$ and an upwards force due to the tension of $$T\cos\theta$$, so our first equation is:

$$T\cos\theta = mg \tag{1}$$

Now the horizontal acceleration. The bob is accelerating at the same acceleration as the car so the horizontal acceleration is $$a_0$$ and therefore the net horizontal force is $$ma_0$$. The only horizontal force is $$T\sin\theta$$ so we get our second equation:

$$T\sin\theta = ma_0 \tag{2}$$

So we have two simultaneous equations in the two unknowns $$T$$ and $$\theta$$, and solving the equations gives us their values. In this case we can simply divide (2) by (1) to get:

$$\tan\theta = \frac{a_0}{g}$$

And that's our solution.

• You solution just opens my mind, Thank you Jun 1, 2021 at 17:56

you can see it also from those equations:

$$\mathbf R=\left[ \begin {array}{c} -x \left( t \right) +L\sin \left( \varphi \right) \\ L\cos \left( \varphi \right) \end {array} \right]$$

from here with the kinetic and potential energy you obtain the equation of motion $$\ddot{\varphi}+\frac{g}{L}\,\sin(\varphi)-\frac{a_0}{L}\,\cos(\varphi)=0\tag 1$$ where $$~a_0=\ddot{x}(t)=$$constant

with $$\ddot{\varphi}=0$$ you obtain the equilibrium position $$\varphi_0$$

$$\varphi_0=\arctan \left( {\frac {a_{{0}}}{g}} \right)$$

the equation of motion with $$\varphi\mapsto\varphi+\varphi_0$$ is now

$$\ddot{\varphi}+\frac{g}{L}\,\sin(\varphi+\varphi_0)-\frac{a_0}{L}\,\cos(\varphi+\varphi_0)=0$$

take the taylor series for small $$\varphi$$ you get:

$$\ddot\varphi+\omega^2\varphi=0$$

where

$$\omega^2=\frac{\sqrt{g^2+a_0^2}}{L}$$