A horizontal 810-N merry-go-round of radius 1.40 m is started from rest by a constant horizontal force of 55 N applied tangentially to the merry-go-round. Find the kinetic energy of the merry-go-round after 2.0 s. (Assume it is a solid cylinder. Also assume the force is applied at the outside edge.)

52.218 J

Explanation:

Mass of the merry go round   m = weight / g

                                              = 810 / 9.8

                                              = 82.65 kg

radius   r = 1.40   m

torque applied   τ = r * F

                          = 1.40 * 55

                         = 77   N - m

Moment of inertia (of cylinder) I = (1/2) * m * r²

                                                                   = 0.5 * 82.65 * 1.96

                                                  = 80.997   kg-m²

angular acceleration     α = τ / I

                                      = 77 / 119.43

                                      = 0.6447   rad/s²

According to first equation in angular motion     final angular velocity      ω   =   ω0 + α * t

ω   = 0 + 0.6447 * 2.0

    =   1.2894   rad/s

Hence k.e. after 2 seconds    K   =   (1/2) * I * ω²

                                                                 = 0.5 * 80.997 * 1.2894

                                                                =   52.218    J

explanation:

could i give you a exaplo that could you?

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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