Modern roller coasters have vertical loops like the one shown in Figure. The radius of curvature is smaller at the top than on the sides so that the downward centripetal acceleration at the top will be greater than the acceleration due to gravity, keeping the passengers pressed firmly into their seats. What is the speed of the roller coaster at the top of the loop if the radius of curvature there is \({\bf{15}}.{\bf{0}}{\rm{ }}{\bf{m}}\) and the downward acceleration of the car is \({\bf{1}}.{\bf{50g}}\)?.

The rate of change of transverse velocity is known as centripetal acceleration.

Radius of curvature \({\rm{r = 15 m}}\)

Acceleration of the car \({{\rm{a}}_{\rm{c}}}{\rm{ = 1}}{\rm{.5 g}}\)

Speed of the roller coaster car at the top of the loop \({\rm{v = ?}}\)

\({{\rm{v}}^{\rm{2}}}{\rm{ = r}} \times {{\rm{a}}_{\rm{c}}}\)

Here,\({\rm{v}}\)is the velocity of the roller coaster car,\({\rm{r}}\)is the radius of curvature and\({{\rm{a}}_{\rm{c}}}\)is the centripetal acceleration of the roller coaster car.

\(\begin{aligned}{l}{\rm{v = }}\sqrt {{\rm{r}} \times {{\rm{a}}_{\rm{c}}}} \\{\rm{v = }}\sqrt {{\rm{15}} \times {\rm{1}}{\rm{.5}} \times {\rm{9}}{\rm{.8}}} \\{\rm{v = 14}}{\rm{.85 m/s}}\end{aligned}\)

The speed of the roller coaster at the top of the loop is 14.85 m/s.