Aplayground merry-go-round has radius 2.40 m and moment of inertia 2100 kgm2 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 15.0 s. if the merry-go-round is initially at rest, what is its angular speed after this 15.0-s interval? (b) how much work did the child do on the merry-go-round? (c) what is the average power supplied by the child?

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

reserve is the answer

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