Question

Unlike a roller coaster, the seats in a Ferris wheel swivel so that the rider is always seated upright. An 80-ft-diameter Ferris wheel rotates once every 24 s.What is the apparent weight of a 40 kg passenger at the lowest point of the circle?What is the apparent weight of a 40 kg passenger at the highest point of the circle?

Answer 1

Answer:

a

   $F_A =425.42 \ N$

b

  $F_A_H = 358.58 \ N$

Explanation:

From the question we are told that

    The diameter of the Ferris wheel is  $d = 80 \ ft = \frac{80}{3.281} = 24.383$

    The  period of the Ferris wheel is  $T = 24 \ s$

     The  mass of the passenger is  $m_g = 40 \ kg$

The  apparent weight of the passenger at the lowest point is mathematically represented as

           $F_A_L = F_c + W$

Where  $F_c$ is the centripetal force on the passenger,  which is mathematically represented as

         $F_c =m * r * w^2$

Where $w$ is the angular velocity which is mathematically represented as

         $w = \frac{2* \pi }{T}$

substituting values

         $w = \frac{2* 3.142 }{24}$

         $w = 0.2618 \ rad/s$

and  r  is the radius which is evaluated as $r = \frac{d}{2}$

   substituting values

         $r = \frac{24.383}{2}$

         $r = 12.19 \ ft$

So

          $F_c = 40 * 12.19* (0.2618)^2$

          $F_c = 33.42 \ N$

W is the weight which is mathematically represented as

           $W = 40 * 9.8$

           $W = 392 \ N$

So

         $F_A = 33.42 + 392$

         $F_A =425.42 \ N$

The  apparent weight of the passenger at the highest point is mathematically represented as

          $F_A_H = W- F_c$

substituting values

         $F_A_H = 392 - 33.42$

         $F_A_H = 358.58 \ N$