If we have a body like the one below , What will be the minimum initial velocity $V_0$ to complete one revolution, My assumption was that it has to reach $\theta=180$ ,But how do I describe this mathematically and why?

enter image description here

  • $\begingroup$ A fee body follows a straight line, Newtons first law. en.wikipedia.org/wiki/Newton%27s_laws_of_motion a force is needed to make a circular path, $\endgroup$
    – anna v
    Jun 7 at 9:29
  • $\begingroup$ @annav By free here I mean that it can leave the path, it's not guided(Like a ball in a closed circular tube). $\endgroup$ Jun 7 at 9:45
  • $\begingroup$ So you mean that a radial force will act to keep it on the circle inwardly, but not outwardly? And you want the minimum velocity for it to not leave the path at the top? $\endgroup$
    – sqek
    Jun 7 at 10:59
  • $\begingroup$ @annav how come it ? specify more like for example if at $\theta=180$ and the normal reaction $N=0$ how will the body not become a projectile instantly ? $\endgroup$ Jun 7 at 11:47
  • $\begingroup$ @sqek no there is no radial force we want to launch the ball with a initial launch velocity to make it complete a revolution without returning or falling off $\endgroup$ Jun 7 at 11:51

2 Answers 2


From what I think you mean from

Like a ball in a closed circular tube

the radial or normal force from the tube, $N$ in the diagram above, can only be positive. If $N$ is negative, the ball will fall away from the wall of the tube.

Radially, the ball needs a centripetal acceleration of $\frac{V^2}{R}$. So using $f=ma$ in the radial direction, $N-mg\cos\theta=m\frac{V^2}{R}$.

Assuming no friction, conservation of energy (kinetic + gravitational potential) gives


So $mV^2=mV_0^2+2mgR(\cos\theta-1)$

So $N=\frac{1}{R}mV_0^2+3mg\cos\theta-2mg$

$N$ is at its minimum when $\theta=180^o$, when $\cos\theta=-1$. The limiting speed is when $N=0$ at this instant - $N$ can't be negative because the tube can't pull the ball outward, so if $N<0$ the ball falls away. So $N_{min}=\frac{1}{R}mV_{0,min}^2-5mg\ge0$ (one of the $mg$ is from gravity pulling the ball down, the rest is from the change in velocity from $V_0$ as the ball has risen $2R$)

So $V_{0,min}=\sqrt{5gR}$


To do a loop over the top the velocity at the top ($v_{top}$) has to be such that the centripetal acceleration is at least $g$ (otherwise it will fall down). This is:

$$a= v^2 / r$$


$$g = {v_{top}}^2 / r$$ $${v_{top}}^2 = gr$$

In turning the half circle, the ball will gain potential energy $2mgr$. That must translate to a loss of kinetic energy, so:

$$\frac 1 2 m{v_{0}}^2 - \frac 1 2 m{v_{top}}^2 = 2mgr$$ $${v_{0}}^2 - {v_{top}}^2 = 4gr$$ but ${v_{top}}^2 = gr$, so $${v_0}^2 - gr = 4gr$$ $${v_0} = \sqrt{5gr}$$


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.