Acceleration in full-swing pendulum

Consider a $360^{0}$ swing pendulum with a massless rod.

If I start it from the position of unstable balance, at the very top of the circumference, according to the conservation of energy, its velocity should be $v=\sqrt{2gh}$, with $h=R(1-\cos(x))$, $x$ being the angle measured from the initial position.

Now if I plot this, I get the maximum acceleration at the starting point (curve is steepest) and the change in velocity is greater between 0 and $\pi/4$ than between $\frac{\pi}{4}$ and $\frac{\pi}{2}$. However, if we consider the tangential component of $g$ $(g\sin(x))$, shouldn't the acceleration peak at $\frac{\pi}{2}$? And therefore should I not see an inflexion point in the plot at $\frac{\pi}{2}$, as in a sinusoidal curve? I cannot reconcile the two.

• You show an equation where $v$ is a function of angle. How do you get from that to a function of time (necessary to calculate acceleration)? – BowlOfRed Jun 20 '16 at 21:09
• The acceleration is the vector sum of the radial and tangential velocities. – Farcher Jun 20 '16 at 22:44

$h=R(1-\cos\theta)=R(2\sin^2\frac{\theta}{2})$
$v=\sqrt{2gh}=\sqrt{4gR}\sin\frac{\theta}{2}$
$v$ is a maximum at $\theta=\pi$.
$\frac{dv}{dt}=\sqrt{4gR}\cos\frac{\theta}{2}.\frac12\frac{d\theta}{dt}=\sqrt{4gR}\cos\frac{\theta}{2}.\frac12.\frac{1}{R}\sqrt{4gR}\sin\frac{\theta}{2}=g\sin\theta$
as expected. This is a maximum at $\theta=\frac12 \pi$.
(Note that I have used $\frac{d\theta}{dt}=\frac{v}{R}$.)