A park merry-go-round consists of a 255 kg circular wooden platform 4.10 m in diameter.?

Question

Four children running alongside push tangentially along the platform's circumference until, starting from rest, the merry-go-round reaches a steady speed of one complete revolution every 2.8 s.

(a) If each child exerts a force of 26 N, how far does each child run?
m
(b) What is the angular acceleration of the merry-go-round?
rad/s2
(c) How much work does each child do?
J
(d) What is the kinetic energy of the merry-go-round?
J

Anakin, you got the last part, D, right. The other stuff is wrong. Thanks though.

Answer 1

Look, I have an idea, if it's not correct, please let me know, but, I will do this :

Yes, I was thinking about it on the afternoon, and I have this idea : Let's apply torques :

Tangential force = 26 N

Total tangential force = 26*4 = 104 Newtons

So, here we apply torques :

F*R = I*a

104*(4.1/2) = (255*(4.1/2)^2) / 2*a

a = angular acceleration

radius = 2.05 meters

Angular acceleration = 0.4 rad / s^2

Correcting it :

Now we have the angular acceleration, we have the initial angular velocity and the final angular velocity : 2pi / 2,8

initial angular velocity = 0 rad / s

So now, we can find the angle that the merry went around :

wf^2 = wo^2 + 2a(angle)

(2pi/2.8)^2 = 2*0.4*angle

angle = 6.29 radians

Then we can find how far every child runs :

angle*radius = 6.29*2.05 = 12.9 meters >>> distance

Moment of inertia : I = M*R^2 / 2 = 536 kg*m^2

work done by each child = force*distance = 26*12.9 = 335.4 Joules

We have the final velocity of the merry round = 2pi / 2.8 rad/s

kinetic energy = rotation kinetic energy = I.w^2 / 2

536*(2pi / 2.8)^2 / 2 = 1350 Joules

Hope that helped