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We have the equation of motion in polar coordinates:

$$\frac{d^{2}\vec r}{dt^2} = (\frac{d^2 |\vec r|}{dt^2} - |\vec r|\cdot (\frac{d\theta}{dt})^2)\hat r + (|\vec r|\cdot \frac{d^2\theta}{dt^2}+2\frac{d\theta}{dt}\cdot\frac{d|\vec r|}{dt})\hat \theta.$$

For example, we have a pendulum that is consisted by a rigid rod and mass and we wish to find its velocity at a certain angle after we release from and initial angle $\theta_0$.

We could solve the problem using the Work-Energy theorem but say we wish to solve it by the equation of motion. We would have to integrate it but I find some difficulties doing so, like in in the radial direction where you have to integrate the term $(\frac{d\theta}{dt})^2$. So how do we integrate it?

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  • $\begingroup$ You cannot have a perfect solution for this. When you try to find the time period of pendulum (with oscillations of high $\theta$) the result you get is a second order differential equation which is unsolvable by current knowledge, although an approximation can be made. You can watch youtube.com/watch?v=p_di4Zn4wz4 and the proceeding episode for an insight. $\endgroup$
    – Dorothea
    Commented Aug 1, 2020 at 7:51
  • $\begingroup$ OP's eq. is strictly speaking not an EOM, it's a kinematic identity. $\endgroup$
    – Qmechanic
    Commented Aug 1, 2020 at 8:14
  • $\begingroup$ If you fix your origin at the pivot, then $r$ doesn't change, so $\dot{r}$ and $\ddot{r}$ are zero. That should simplify your problem a bit. Write down the forces in the radial and tangential directions and equate the components. You'll get two differential equations. $\endgroup$ Commented Aug 1, 2020 at 8:46

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