Answer:
a) 0.31 rad/sec b) 101 J c) 6.73 W
Explanation:
a) If the merry-go-round starts from rest, we can apply the definition of angular acceleration, to get the value of the angular speed, as follows:
ωf = γ *t (1)
We don´t know the value of the angular acceleration, but we know that there was an external force applied tangentially to the edge of the body, which produced a torque.
Newton´s 2nd Law has also a rotational equivalent, so we can say the following:
τ = I * γ (2)
Applying the definition of torque, we can write the following equation:
τ = F*r (as both are perpendicular) = 18.0 N * 2.4 m = 43.2 N.m
Replacing in (2) and solving for the angular acceleration γ:
γ = 0.02 rad/sec2
Now, replacing this value in (1) we can get the value of the angular speed, as follows:
ωf = 0.31 rad/sec.
b) Applying the work-energy theorem, we know that the change in the kinetic energy, is equal to the work done by the child (neglecting any friction).
As the merry-go-round only rotates, the kinetic energy is only rotational.
As it started from rest, the change is just equal to the final value, as follows:
∆K = ½ I * ωf2 = ½ * 2100 kg.m2 * (0.31)2 rad/sec2 = 101 J
W= 101 J
c) By definition, the average power is the work done in a given time, so we can write the following equation:
P = W/t = 101 J / 15.0 sec = 6.73 W