Question

modern roller coasters have vertical loops like the one shown in the figure below. 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, in m/s, of the roller coaster at the top of the loop if the radius of curvature there is 15.0 m and the downward acceleration of the car is 1.50 g? m/s

Explanation:

Step1: Recall centripetal acceleration formula

The formula for centripetal acceleration is \(a_{c}=\frac{v^{2}}{r}\).

Step2: Substitute known values

We know that \(a_{c} = 1.50g\) (where \(g = 9.8\ m/s^{2}\)) and \(r=15.0\ m\). Substituting into the formula: \(1.50\times9.8=\frac{v^{2}}{15.0}\).

Step3: Solve for \(v\)

First, calculate \(1.50\times9.8 = 14.7\). Then, from \(\frac{v^{2}}{15.0}=14.7\), we can get \(v^{2}=14.7\times15.0 = 220.5\). Taking the square - root of both sides, \(v=\sqrt{220.5}\approx14.8\ m/s\).

Answer:

\(14.8\ m/s\)