Question

A woman rides a carnival Ferris wheel at radius 21 m, completing 4.5 turns about its horizontal axis every minute. What are (a) the period of the motion, and the magnitude of her centripetal acceleration at (b) the highest point and (c) the lowest point?

Answer 1

(a) The woman completes 4.5 turns every minute.

The period is the amount of time it take to make 1 revolution.

1 min = 60 sec.

Therefore,

The period, T = 60 / 4.5 = 13.33 sec (Answer)

Now here's where you have to think.

We know by algebra that the rotations per second is 1/13.33 = 0.075 rotations / sec

Circumference of the wheel, s = 2*pi*r

= 2*3.141*21 = 131.9 m

Therefore, the linear speed of the woman, v = s*0.075 = 131.9*0.075 = 9.89 m/s

centripetal acceleration = v^2 / r

= 9.89^2 / 21 = 4.66 m/s^2 (Answer)

(b) Think about this centripetal force pushes out (thats why you can swing a bucket over your head and not get wet)

So at the top of the wheel, gravity is pushing down against the centripetal acceleration (which is pushing out or in this case up).

At the bottom it is pushing with it.

So,

Centripetal acceleration at the top = 4.66 - 9.8 = -5.14 m/s^2 (Answer)

(c) Like-wise,

Centripetal acceleration at the lowest point = 4.66 + 9.8 = 14.46 m/s^2 (Answer)