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:

Explanation:

Answer:

a

    The apparent weight is  $N = 424.5 N$

b

    The apparent weight is  $N = 358.5 N$

Explanation:

From the question we are told that

    The diameter of the Ferris wheel is d  =  80ft

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

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

     

The radius of the Ferris wheel is evaluated as  $r = \frac{d}{2}$ substituting values

          $r = \frac{80}{2}$

         $r = 40 \ ft$

converting to meters

           $r = 40 * 0.3048 = 12.20 \ m$

The angular velocity of the Ferris wheel is mathematically represented as  

                $w = \frac{2 \pi}{T}$

substituting values  

                $w = \frac{2* 3.142 }{24}$

                $w = \frac{2* 3.142 }{24}$

                $w = 0.2618 \ rad/s$

   At the lowest the point the  apparent weight of the passenger is  equal to the normal force on the chair which is  mathematically represented as

             $N = m (g + a_c)$

Where  $a_c$ is the centripetal acceleration which is mathematically represented  as  

             $a = w^2 R$

So  

          $N = m (g + w^2 R)$

substituting values  

              $N = 40 (9.8 + (0.2618)^2 * 12.2)$

              $N = 424.5 N$

   At the highest the point the  apparent weight of the passenger is  equal to the normal force on the chair which is  mathematically represented as

         $N = m (g - a_c)$

          $N = m (g - w^2 R)$

          $N = 40 (9.8 - (0.2618)^2 * 12.2)$

          $N = 358.5 N$