# Vertical Circular Motion

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.

• 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 '14 at 15:33
• When it has negligible velocity at the top. – user85798 Mar 3 '14 at 16:27

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.

Consider at any instant t, the forces acting on the body. Towards the centre, two forces act: Tension ($\vec{T}$) and radial component of weight($\vec{w}$),i.e, $|\vec{w}|\,cos\theta \,\mathbf{r}$ where $\theta$ is the angle made by the string with the vertical and $\mathbf{r}$ is the radius unit vector. These radial forces do not change the magnitude of the velocity of the particle in UCM (Uniform Circular Motion).

The tangential forces, which change the magnitude of velocity is $|\vec{w}|sin\theta$ ,obviously directed along the tangent at that point.Assuming the velocity at topmost point to be $v$, we get

$$a_{t}=\frac{d|\vec{v}|}{dt}=\frac{dv}{dt}=v\frac{dv}{dx}$$ $$\implies a_{t}=v\frac{dv}{r.d\theta}$$ $$\implies \int_v^{v'} vdv=gr.\int_0^{\theta} sin\theta\,d\theta$$ $$\implies v'^2-v^2=2gr(1-cos\theta)$$ If you use the fact that minimum velocity at the top is $\sqrt{gr}$, we get $v^2=gr$. Therefore, $$v'^2=gr(3-2cos\theta) \implies v'=\sqrt{gr(3-2cos\theta)}$$ I believe that you got the expressions before that since your change in kinetic energy was taken to be wrong. The actual change in kinetic energy would be $\frac12mv'^2-\frac12mv^2$ and not $\frac12mv^2$