Investigating velocity on a roller coster trail Let's imagine a roller coaster is going to complete a vertical circular motion (so that it will follow the circular track). At the topmost point, we know the velocity is $\sqrt{gr}$ where $r$ is the radius of the track. I saw a problem in which we were asked to find out the velocity of the roller coaster when it reaches the bottom most point from the upper most risky zone point. There they used this equation $v^2=(\sqrt{gr})^2+2g(2r)$. Though it gives the correct answer, I am not sure why it works. This is due to a couple of reasons.
$1)$ When they are using $s=2r$ in the general equation $v^2=u^2+2as$, it means their intention is to consider the vertical motion of the roller coaster. But the velocity of the roller coaster at the topmost point is along the tangent which is directed to the left, and it doesn't have any vertical component. Then how did they use $u=\sqrt{gr}$ in the general equation whereas there is only horizontal component acting in the left direction?
$2)$ Considering point $1$, we would be willing to use $v^2=0^2+2g(2r)$ as the actual equation, but this seems to give some actual value to give rise to a vertical component of velocity in the lowermost point. But we all know that the only velocity in the lower most point will be a tangent vector acting in the rightmost direction (opposite to the direction of uppermost point), so it also shouldn't have any vertical component.
In both the cases, I seem to have developed a contradiction. I know I am doing wrong somewhere since we are dealing with a circular path. Please let me know where the faults are. Thanks in advance.
 A: the the solution given in the book is absolutely wrong .Vertical circular motion is a case of non uniform acceleration .And you can't apply normal /usual equation of kinematics for constant acceleration , in this case.
See its not a particle falling from here .Another force called normal reaction acts when it moves in circular path .Who will consider it . And the reasons told by you are correct . Best way to do this problem is using conservation of energy ,
You can also do it using calculus , find the variable tangential acceleration and use d(speed)/dt = |A tangential|. Benefit here is you won't have to worry about normal acceleration because it provides centripetal acceleration (as it always acts perpendicular to the velocity of particle )
But if you want to analyse it with respect to horizontal and vertical then it will be a tough time .Because normal reaction will be variable dangerously (magnitude + acceleration both ) . It will take time but you will get answer using this too .
So basically your concepts are correct .
A: First, I think your book is not completely wrong. It is just misleading you. That equation is not the usual equation of kinematics. I'll show you. $$v^2=u^2+2as$$ multiplying both sides by $\frac12m$ ($m$ is the mass) $$\frac12mv^2=\frac12mu^2+mas$$
$$\frac12mv^2=\frac12mu^2+mg(2r)$$
$$\frac12mv^2=\frac12mu^2+mgh$$
$$ME_{\text{gained}}=ME_{\text{lost}}$$
You see, this is the equation of energy conservation. That's why your first equation does work, which seems obviously like a equation of kinematics, but actually not.
A: You do it with energy balance, $mv^{2}/2=m(\sqrt{gr}^2)/2+m*g*2r$.
A: The equation $v_2^2 - v_1^2= 2g(h_1-h_2)$ can also be derived using only Newton's second law. It basically boils down to using conservation of energy but it is nice to see that formally you don't have to use it.
Placing the origin of a polar coordinate system at the center of the ring. Let $N$ be the normal force of the ring onto the roller coaster. The forces are
$$m\vec{a} = -mg\hat{y} - N\hat{r} = (-mg\sin\phi - N)\hat{r} + (-mg\cos\phi)\hat{\phi}.$$
On the other hand, plugging in the acceleration in the polar system gives
$$m\vec{a} = m(\ddot{r}- r\dot{\phi}^2)\hat{r} + m(r\ddot{\phi} + 2\dot{r}\dot{\phi})\hat{\phi} = -mR\dot{\phi}^2\hat{r} + mR\ddot{\phi}\hat{\phi}.$$
where $R$ is the radius of the ring. Equating the tangential components gives
$$-mg\cos\phi= mR\ddot{\phi}.$$
Multiply this equation with $\frac{2R\dot{\phi}}{m}$ and integrate from $t_1$ to $t_2$ to obtain
\begin{align}
2g(h_1-h_2) &= 2g(\cos(t_1)-\cos(t_2)) = -2g\cos\phi(t)\Big|_{t_1}^{t_2} = \int_{t_1}^{t_2} -2g(\sin\phi(t))\dot{\phi}(t) \,dt\\
&= \int_{t_1}^{t_2} -2R^2\dot{\phi}(t)\ddot{\phi}(t)\,dt = R^2\dot{\phi}(t)^2\Big|_{t_1}^{t_2} = R^2(\dot{\phi}(t_2) - \dot{\phi}(t_2)) = v_2^2 - v_1^2.
\end{align}
