Problem — Minimum speed at the top of a loop-the-loop

Problem

Show that the minimum speed a (point-like) ball must have at the lowest point of a circular loop-the-loop of radius $R$, in order to complete the loop with no friction, is $v = \sqrt{5gR}$. Start from the condition at the highest point.

Solution

Condition at the highest point. At the highest point of the loop the ball leaves the track if gravity alone is not enough to supply the centripetal acceleration. The limiting condition (contact about to vanish, normal force $N = 0$) is that the weight alone provides the entire centripetal force: $m\,g = \frac{m\,v_\text{alto}^2}{R} \quad\Rightarrow\quad v_\text{alto}^2 = g\,R$ This is the smallest possible speed at the top: below it the ball falls away from the track.

Conservation of energy between the lowest point (height $0$) and the highest point (height $2R$), with no friction: $\frac{1}{2}\,m\,v_\text{basso}^2 = \frac{1}{2}\,m\,v_\text{alto}^2 + m\,g\,(2R)$ Cancel $m$ and substitute $v_\text{alto}^2 = gR$: $v_\text{basso}^2 = v_\text{alto}^2 + 4gR = gR + 4gR = 5gR$ $\ev{v = \sqrt{5gR}}$

Links

Topics: Work and energy Concepts: Conservation of mechanical energy Skills: Conservation of energy Objects: Loop-the-loop