Question

A Ferris wheel is a vertical, circular amusement ride with radius 9.5 m. Riders sit on seats that swivel to remain horizontal. The Ferris wheel rotates at a constant rate, going around once in 7.6 s. Consider a rider whose mass is 57 kg.
At the bottom of the ride, what is the rate of change of the rider's momentum?

Answer 1

Answer:

928.34 N

Explanation:

Radius=r=9.5 m

Time period of revolution=T=7.6 s

Mass of rider=m=57 kg

We have to find the rate of change of the rider's momentum at the bottom of the ride

According to Newton's second law of motion

$F_{net}=ma_c$

$N-mg=mr\omega^2$

$N=mg+mr(\frac{2\pi}{T})^2$

Where $\pi=3.14,g=9.8 m/s^2$

Using the formula

$N=57\times 9.8+57\times 9.5(\frac{2\times 3.14}{7.6})^2$

$N=928.34 N$

Hence, the rate of change of the rider's momentum=928.34 N