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I know that I can use conservation of energy to find the velocity of a particle at a point when it's travelling in a vertical circle by saying

$$mgr(1-\cos{\theta})=\frac{1}{2}mv^2$$

then rearranging to get $v=\sqrt{2gr(1-\cos{\theta})}$

But I want to see this done 'the long way' using newtons second law directly and then probably solving a differential equation, but I'm not sure how to do it and nothing I've tried is getting very far.

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For motion in a vertical circle, the velocities need not be unique right. Are you talking about the case where the ody has inimum velocity at the top which equals $\sqrt{gr}$ –  Sandeep Thilakan Mar 3 at 15:33
    
When it has negligible velocity at the top. –  LTS Mar 3 at 16:27

1 Answer 1

I might be able to get you started a direction. Not necessarily the right one or a good one, but a direction.

First, I chose a different place for $\theta=0$. It appears you chose it at the top of the loop, while I chose it on the right side. Oh well. Also, it's likely I have a typo somewhere...

Newton's second law for this problem is

$$m\ddot{x}\hat{x}+m\ddot{y}\hat{y}=-mg\hat{y}-N\hat{R},$$

where $N$ is the normal force provided by the frictionless loop (which will depend on the speed $v$ of the object and maybe even $\theta$), and $\hat{R}=\hat{x}\cos\theta + \hat{y}\sin\theta$ is the outward-pointing normal vector from the center of the loop.

Right now this (vector) equation is in a mix of polar and cartesian coordinates. Let's go to polar $(R, \theta)$. First note $$ \begin{eqnarray} \ddot{x} &=& \frac{d^2}{dt^2}(R\cos\theta) =-R\frac{d}{dt}\dot{\theta}\sin\theta =-R(\ddot{\theta}\sin\theta + \dot{\theta}^2\cos\theta)\\ \ddot{y} &=& \frac{d^2}{dt^2}(R\sin\theta) =R\frac{d}{dt}\dot{\theta}\cos\theta =R(\ddot{\theta}\cos\theta - \dot{\theta}^2\sin\theta) \end{eqnarray} $$

Take the radial component of Newton's second law: $$\begin{eqnarray} m(\ddot{x}\hat{x}\cdot\hat{R} + \ddot{y}\hat{y}\cdot\hat{R})&=&-mg\hat{y}\cdot\hat{R}-N\hat{R}\cdot\hat{R}\\ m\left(-R(\ddot{\theta}\sin\theta\cos\theta + \dot{\theta}^2\cos^2\theta) + R(\ddot{\theta}\cos\theta\sin\theta - \dot{\theta}^2\sin^2\theta)\right)&=&-mg\sin\theta-N\\ -mR\dot{\theta}^2&=&-mg\sin\theta-N.\tag{1} \end{eqnarray}$$

Silly signs. Someone more clever than I could have written this down without going through the song and dance I did. Anyway, Eqn. 1 has square of the speed hiding in the LHS. See it?

Let's now take the tangental component of Newton's second law and see what it gets us. We'll need $\hat{\theta}=-\hat{x}\sin\theta+\hat{y}\cos\theta$.

$$\begin{eqnarray} m(\ddot{x}\hat{x}\cdot\hat{\theta} + \ddot{y}\hat{y}\cdot\hat{\theta})&=&-mg\hat{y}\cdot\hat{\theta}-N\hat{R}\cdot\hat{\theta}\\ m\left( R(\ddot{\theta}\sin^2\theta + \dot{\theta}^2\cos\theta\sin\theta) + R(\ddot{\theta}\cos^2\theta - \dot{\theta}^2\sin\theta\cos\theta) \right)&=&-mg\cos\theta\\ mR\ddot{\theta}&=&-mg\cos\theta\tag{2} \end{eqnarray}$$

Someone else could have written Eqn. 2 down immediately as well. It basically says only the gravitational force can change the speed $R\dot{\theta}$. More importantly, I have no idea how to solve this.

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