A teenager pushes tangentially on a small hand-driven merry-go-around and is abl

Question

A teenager pushes tangentially on a small hand-driven merry-go-around and is able to accelerate it from rest to a frequency of 15 rpm in 10.0 s. Assume the merry-go-round is a uniform disk of radius 2.5 m and has a mass of 760 kg, and two children (each with a mass of 25 kg) sit opposite each other on the edge. Upon reaching 15rpm, the teenager stops pushing. Both children walk radially inward to the center of the merry-go-round. Determine the final rotational velocity of the merry-go-round system and the net work performed by the children. also calculate the angular displacement. The

Explanation / Answer

I * delta w = T * delta t.
(T for torque, w for angular speed)

15rpm is the change in angular speed. Convert it to radians.
The I is the rotational inertia. For a disk it is MR^2/2. Plug in the numbers for that. You will need to ADD to this the rotational inertia of the children. Assume each is a point mass, for which the rotational inertia is MR^2. (R is their distance from the axis.)
The delta t is of course 10 seconds.

Plug all of those into the rotational impulse equation, and you can calculate the torque (the only unknown left).

(You can use a similar equation, T = I * alpha, but then you have to calculate the angular acceleration before proceeding, so the rotational impulse way is simpler.)

Once you have the torque, you know that

T = F * R (the torque is the force times the radius at which it is applied. Since it is applied at the edge, R is the radius of the merry-go-round.)

plug in the torque and the radius and solve for the force.

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