Question

A Ferris wheel with a 15-m radius makes three complete revolutions in one minute about its horizontal axis. What is the magnitude and direction of a passengers acceleration at the lowest point during the ride?

Answer 1

Answer:

$1.48 m/s^2$, upward

Explanation:

The passenger on the wheel experiences a centripetal acceleration, which is the one that keeps him in a circular motion.

The direction of this acceleration is always towards the centre of the circular trajectory: so when the passenger is at the lowest point of the ride, the acceleration is upward.

Concerning the magnitude, it is given by

$a=\omega^2 r$

where

$\omega$ is the angular velocity

r = 15 m is the radius

We need to find the angular velocity; we know that the wheel completes 3 revolutions in one minute. Each revolution corresponds to an angle of $2\pi$ rad, so the total angular displacement is

$\theta = 3 \cdot 2\pi = 6\pi$ rad

And the time is

$t = 1 min = 60 s$

So the angular velocity is

$\omega = \frac{6\pi}{60 s}=0.314 rad/s$

And substituting into the equation of the acceleration,

$a=(0.314)^2(15)=1.48 m/s^2$