A Ferris wheel is a vertical, circular amusement ride with radius 10 m. Riders s

Question

A Ferris wheel is a vertical, circular amusement ride with radius 10 m. Riders sit on seats that swivel to remain horizontal. The Ferris wheel rotates at a constant rate, going around once in 8 s. Consider a rider whose mass is 52 kg.

,  , 0> N

,  , 0> N

,  , 0> kgm/s/s

,  , 0> N

,  , 0> N

A Ferris wheel is a vertical, circular amusement ride with radius 10 m. Riders sit on seats that swivel to remain horizontal. The Ferris wheel rotates at a constant rate, going around once in 8 s. Consider a rider whose mass is 52 kg. At the bottom of the ride, what is the rate of change of the rider's momentum? vector f capby seat = N vector f capgrav = N At the top of the ride, what is the vector force exerted by the seat on the rider? (dp^^- > )/(dt) = kg½m/s/s At the top of the ride, what is the vector gravitational force exerted by the Earth on the rider? vector f capby seat = N Next consider the situation at the top of the ride. At the top of the ride, what is the rate of change of the rider's momentum? vector f capgrav = N At the bottom of the ride, what is the vector force exerted by the seat on the rider? (dp^^- > )/(dt) = kg½m/s/s At the bottom of the ride, what is the vector gravitational force exerted by the Earth on the rider?

Explanation / Answer

          

     wheel 's radious = R = 10 m

       since . this wheel rotates 1 oscillation in 8 sec

       thus, time period of the wheel = T = 8 sec

                so, angular velocity = W = 2 ? / T

                                                       = 2(3.14) / 8   rad/s

                        

                                                       = 0.785 rad/s

      since , wheel is in vertical position , the rider moves in the horizontial direction

                  thus, linear velocity of the wheel is only in the Horizontail direction

                                    v   = R W = 8 (0.785)    = 6.28 m/s

   thus, parallel component of mometum is only acting

           vertical component of momentum is 0

            thus, momentum along the horizontila direction is

                        m v   = 52 (6.28)

                                   = 326.56 kg/m /sec

Thus, rate of change of momentum = dp /dt = 326.56 / 8

                                                                       = 40.82 kgm/s/s

vertical component is     = 0   kg m/s/s

gravitationla force exreted by the earth on the rider is = M g

                                                                                  = 52 (9.8)

                                                                                  = 509.6 N

   

what is the vector force exerted by the seat on the rider   is also have the same magnitude

but have opp. direction

                                      =   force = Mg = (52)(9.8)

                                                               = - 509.6 N

      Plz. post ramaining seperately ..,                            

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