Centripetal acceleration and angular velocity

Question

Riders in an amusement park ride shaped like a Viking ship hung from a large pivot that is rotated back and forth like a rigid pendulum. Sometime near the middle of the ride, the ship is near motionless at the top of its circular arc. The ship then springs down under the influence of gravity. Given the system's center of mass travels in an arc having a radius of 14.0m and the riders are near the center of mass (with zero friction). Which two values give the speed of the riders at the bottom of its arc and the centripetal acceleration?

Answer 1

In this problem, we are only given the radius of 14 m, but there are a few things we can say based on the description.

1) This can be modeled as a simple pendulum. One of the equations for a simple pendulum is the period of oscillation, T: T = 2π ⋅ sqrt(L / g), where L is the length of the pendulum, which is also the radius in this case. Solving for the period, we have T = 7.51 s

2) We can also model this as centripetal motion, which has two unique equations for it:

Period, T = 2π r / v, where r is the radius and v is the speed.

Centripetal Acceleration, a = v² / r

Using the period equation, we can solve for the speed:

T = 2π r / v

v = 2π r / T = 11.71 m/s

We can then use centripetal acceleration directly:

a = v² / r = 9.79 m/s²