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?

Explanation:

Step1: Recall the centripetal acceleration formula

The centripetal acceleration formula is \(a_{c}=\frac{v^{2}}{r}\), where \(a_{c}\) is the centripetal acceleration, \(v\) is the speed, and \(r\) is the radius of curvature.

Step2: Substitute the given values into the formula

We are given that \(a_{c} = 1.50g\) and \(r=15.0\ m\), and \(g = 9.8\ m/s^{2}\). So \(a_{c}=1.50\times9.8\ m/s^{2}=14.7\ m/s^{2}\).
From \(a_{c}=\frac{v^{2}}{r}\), we can solve for \(v\). Rearranging the formula gives \(v=\sqrt{a_{c}r}\).
Substitute \(a_{c}=14.7\ m/s^{2}\) and \(r = 15.0\ m\) into the formula: \(v=\sqrt{14.7\times15.0}\).

Step3: Calculate the value of \(v\)

\(14.7\times15.0=220.5\). Then \(v=\sqrt{220.5}\approx14.85\ m/s\).

Answer:

\(14.9\ m/s\) (rounded to three significant figures)