Question

A Ferris wheel has a radius of 10 m, and the bottom of the wheel passes 1 m above the ground. If the Ferris wheel makes one complete revolution every 18 s, find an equation that gives the height above the ground of a person on the Ferris wheel as a function of time

Answer 1

Answer:$1+10(1-\cos (\frac{\pi t}{9}))$

Step-by-step explanation:

Given

radius of wheel $r=10 m$

Time period of Wheel $T=18 s$

and $T\cdot \omega =2\pi$ , where $\omega =angular velocity of wheel$

$\omega =\frac{2\pi }{18}$

Let at any angle $\theta$with vertical position of a point is given by

$x=r\sin \theta$

$y=y_0+r(1-\cos \theta )$

and $\theta =\omega \times t$

for velocity differentiate x and y to get

$v_x=r\cos \theta =r\cos (\omega t)$

$v_y=0+r(\sin \theta )=r\sin (\omeag t)$

Height at any time t is given by

$h=1+10(1-\cos \theta )=1+10(1-\cos (\frac{\pi t}{9}))$