0
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ 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)? $\endgroup$ – BowlOfRed Jun 20 '16 at 21:09
  • $\begingroup$ The acceleration is the vector sum of the radial and tangential velocities. $\endgroup$ – Farcher Jun 20 '16 at 22:44
1
$\begingroup$

$h=R(1-\cos\theta)=R(2\sin^2\frac{\theta}{2})$
$v=\sqrt{2gh}=\sqrt{4gR}\sin\frac{\theta}{2}$
$v$ is a maximum at $\theta=\pi$.

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.