At what minimum speed a roller coaster be traveling so that passengers upside down at the top of the circle (Fig 5-48) do not fall out? Assume a radius of curvature of 8.6 m.

Apply Newton’s second law to find the equation for normal force. The normal force is proportional to the velocity. Therefore, when the speed is high, the normal force is positive and vice versa. The passenger will leave the contact when the normal force is zero, which is the condition of free fall.

The radius of curvature of roller coaster is \(r = 8.6\;m\).

The figure below represents the free body diagram:

In the free body diagram, consider positive direction is downward. According to the Newton’s second law is:

\(\begin{aligned}\sum F &= ma\\{F_N} + mg &= m\left( {\frac{{{v^2}}}{r}} \right)\\{F_N} &= m\left( {\frac{{{v^2}}}{r} - g} \right)\end{aligned}\)

Here, \({F_N}\) is the normal force, \(m\) is the mass of passanger, \(v\) is the velocity and \(r\) is the radius of curvature.

When the normal force is zero, then the speed will be minimum. The minimum speed can be calculated as:

\(\begin{aligned}\frac{{v_{\min }^2}}{r} - g &= 0\\{v_{\min }} &= \sqrt {rg} \\ &= \sqrt {\left( {9.8\;{\rm{m/}}{{\rm{s}}^2}} \right)\left( {8.6\;{\rm{m}}} \right)} \\ &= 9.2\;{\rm{m/s}}\end{aligned}\)

Thus, the minimum velocity of roller coaster is \(9.2\;{\rm{m/s}}\).