# When does a free body moving on a smooth circular path make a complete revolution?

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? • 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, Jun 7 at 9:29
• @annav By free here I mean that it can leave the path, it's not guided(Like a ball in a closed circular tube). Jun 7 at 9:45
• 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?
– sqek
Jun 7 at 10:59
• @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 ? Jun 7 at 11:47
• @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 Jun 7 at 11:51

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

$$\frac{1}{2}mV^2-mgR\cos\theta=$$constant$$=\frac{1}{2}mV_0^2-mgR$$

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:

$$a= v^2 / r$$

So:

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

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 $${v_0}^2 - gr = 4gr$$ $${v_0} = \sqrt{5gr}$$