Problem — Loop-the-loop

Problem

Loop-the-loop. A ball starts from rest at the top of a frictionless track of height $h$ and travels around a vertical loop of radius $R$, as shown in the figure. Derive the speed at the highest point of the loop and the minimum condition on $h$ for the ball not to leave the track.

Solution

Speed at the highest point. The track is frictionless: mechanical energy is conserved. The ball starts from rest at height $h$; the highest point of the loop is at height $2R$. Conservation of energy between the start and the top of the loop: $m\,g\,h = \frac{1}{2}\,m\,v^2 + m\,g\,(2R)$ Cancel $m$ and solve for $v$: $v^2 = 2g\,(h - 2R) \quad\Rightarrow\quad \ev{v = \sqrt{2g(h-2R)}}$

Minimum condition on $h$. At the highest point the ball stays in contact with the track as long as the required centripetal force is at least equal to the weight. The limiting case (normal force $N=0$) is $m\,g = \frac{m\,v^2}{R} \quad\Rightarrow\quad v^2 \geq g\,R$ Substitute $v^2 = 2g(h-2R)$: $2g\,(h - 2R) \geq g\,R \quad\Rightarrow\quad 2h - 4R \geq R \quad\Rightarrow\quad h \geq \frac{5R}{2}$ $\ev{h_\text{min} = \frac{5R}{2}}$

Links

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