A merry-go-round at a park is subject to a constant torque with a magnitude of 605 N · m as a parent pushes the ride.

Question

(a) How much work is performed by the torque as the merry-go-round rotates through 2.40 revolutions? 

= ? J

(b) If it takes the merry-go-round 4.00 s to go through the 2.40 revolutions, what is the power transferred by the parent to the ride? 

= ? W

Answer 1

Without loss of generality we can assume that the parent's force

is tangential to the merry-go-round, i.e., perpendicular to its radius.

Let

F (unknown) be the magnitude of the parent's force,

r (unknown) be the distance of the point of pushing from the center of rotation,

z = 2.4 be the number of iterations,

t = 4 s be the time it take to go through z iterations.

We are given

r×F = 605 Nm

(a)

In general, when a force is in the same direction as movement (as is the case here),

then work

W = s×F

where s is the distance over which the force is applied.

In our case

s = z×2×π×r

So

W = z×2×π×r×F

Substituting actual numbers

W = 2.4×2×π×605 = 9123 J

(b)

In general, power

P = W/t

Substituting actual numbers

P = 9123/4 = 2281 W

(c)

Bonus question:

Who is the parent?

Answer:

The parent is the Superman, because ordinary mortals can generate only about 100 W.