Question

1. a park ride has people stand around a 6.0 - m radius circle with their backs to a wall. the ride then spins them with a 2.1 - s period of revolution. what is the centripetal acceleration of the riders? (hint: 2 step problem) 18.0\\(\frac{m}{s^{2}}\\) 322.3\\(\frac{m}{s^{2}}\\) 17.1\\(\frac{m}{s^{2}}\\) 53.7\\(\frac{m}{s^{2}}\\) clear all

Explanation:

Step1: Calculate the angular velocity

The formula for angular velocity \(\omega\) is \(\omega=\frac{2\pi}{T}\), where \(T\) is the period of revolution.
Given \(T = 2.1\ s\), then \(\omega=\frac{2\pi}{2.1}\ rad/s\approx 2.99\ rad/s\)

Step2: Calculate the centripetal acceleration

The formula for centripetal acceleration \(a_c\) is \(a_c=\omega^{2}r\), where \(r\) is the radius of the circular path.
Given \(r = 6.0\ m\) and \(\omega\approx 2.99\ rad/s\), then \(a_c=(2.99)^{2}\times6.0\)
\(a_c\approx 53.7\ m/s^{2}\)

Answer:

\(53.7\frac{m}{s^{2}}\) (the fourth option)