Question

1.) An electric motor can accelerate a Ferris wheel of moment of inertia I = 19500 kg·m2 from rest to 10.1 rev/min in 12.0 s. When the motor is turned off, friction causes the wheel to slow down from 10.1 to 8.1 rev/min in 10.0 s. (a) Determine the torque generated by the motor to bring the wheel to 10.1 rev/min. (b) Determine the power that would be needed to maintain this rotational speed. W

2.)A disk-shaped merry-go-round of radius 2.63 m and mass 155 kg rotates freely with an angular speed of 0.526 rev/s. A 59.4 kg person running tangential to the rim of the merry-go-round at 2.80 m/s jumps onto its rim and holds on. Before jumping on the merry-go-round, the person was moving in the same direction as the merry-go-round's rim. (a) Does the kinetic energy of the system increase, decrease, or stay the same when the person jumps on the merry-go-round? decrease stay the same increase (b) Calculate the initial and final kinetic energies for this system. Ki = kJ Kf = kJ

Answer 1

1.

An electric motor can accelerate a Ferris wheel of moment of inertia I = 19500 kg·m2 from rest to 10.1 rev/min in 12.0 s. When the motor is turned off, friction causes the wheel to slow down from 10.1 to 8.1 rev/min in 10.0 s. (a) Determine the torque generated by the motor to bring the wheel to 10.1 rev/min. (b) Determine the power that would be needed to maintain this rotational speed. W

I = moment of inertia = 19500 kgm²

w_o = initial angular velocity = 0 rad/s

w_f = final angular velocity = 10.1 rev/min = (10.1) (2/60) = 1.06 rad/s

t = time = 12 s

= torque = ?

Using the equation

t = I (w_f - w_o )

(12) = (19500) (1.06 - 0 )

= 1722.5 Nm

b)

P = power needed = ?

I = moment of inertia = 19500 kgm²

w_o = initial angular velocity = 1.06 rad/s

w_f = final angular velocity = 8.1 rev/min = (10.1) (2/60) = 0.85 rad/s

Power is given as

P = - (0.5) I (w²_f - w²_o)/t

P = - (0.5) (19500) ((0.85)² - (1.06)²)/10

P = 391.1 Watt