A playground merry-go-round has radius 2.10m and moment of inertia 2500kg*m^2 ab

Question

A playground merry-go-round has radius 2.10m and moment of inertia 2500kg*m^2 about a vertical axle through its center, and it turns with negligible friction.

A)A child applies an 18.0 N force tangentially to the edge of the merry-go-round for 16.0s . If the merry-go-round is initially at rest, what is its angular speed after this 16.0s interval?

in rad/s
B)How much work did the child do on the merry-go-round?


C)What is the average power supplied by the child?

I understand the jist of the question but am stuck on where to start

Explanation / Answer

Radius of the wheel r = 2.1 m Moment of inertia I = 2500 Kg m2 Tangential force applied F = 18 N Time interval t = 16 s Initial angular speed 1 = 0 Final angular speed 2 = ? ------------------------------------------------------------------------------------------------------------------------------- Let be the angular acceleration. Torque applied = I                        Fr = I Angular acceleration = Fr/I                                      = (18 N * 2.1 m)/(2500 Kg m2)                                      = 0.015 rad/s^2 (a)From rotational kinematic relation              Final angular speed 2 = 1 + t                                                   = 0 + (0.015 rad/s^2 * 16 s)                                                   = 0.24 rad/s ------------------------------------------------------------------------------------------------------------------------- (b) Work done W = (1/2)I2^2 - (1/2)I1^2                        = 0.5*( 2500 Kg m^2)(0.24 rad/s)^2 - 0                        =  72 J ------------------------------------------------------------------------------------------------------------------------- (c) Average power supplied by the child P = W/t = 72 J/16 s                                                                          = 4.5 watt                           Radius of the wheel r = 2.1 m Moment of inertia I = 2500 Kg m2 Tangential force applied F = 18 N Time interval t = 16 s Initial angular speed 1 = 0 Final angular speed 2 = ? ------------------------------------------------------------------------------------------------------------------------------- Let be the angular acceleration. Torque applied = I                        Fr = I Angular acceleration = Fr/I                                      = (18 N * 2.1 m)/(2500 Kg m2)                                      = 0.015 rad/s^2 (a)From rotational kinematic relation              Final angular speed 2 = 1 + t                                                   = 0 + (0.015 rad/s^2 * 16 s)                                                   = 0.24 rad/s ------------------------------------------------------------------------------------------------------------------------- (b) Work done W = (1/2)I2^2 - (1/2)I1^2                        = 0.5*( 2500 Kg m^2)(0.24 rad/s)^2 - 0                        =  72 J ------------------------------------------------------------------------------------------------------------------------- (c) Average power supplied by the child P = W/t = 72 J/16 s                                                                          = 4.5 watt                          

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