Question

A playground merry-go-round is a solid disk of radius 2 m and mass 500 kg. If four children with masses of 27 kg each are standing on the edge (holding on), how hard would an AP Physics student have to push tangent to the edge to accelerate the whole thing at 3.1 rad/s^2

Answer 1

F = 2220 N

Step-by-step explanation:

• There exists a fixed relationship, that resembles Newton's 2nd Law (applied to rigid bodies) between an external net torque (τ), the rotational inertia (I) of the body, and the angular acceleration (α) due to the torque, as follows:

`\tau = I * \alpha (1)`

• By definition, the magnitude of the torque is given by the following expression:
• τ = F* r* sinθ (2)
• where F is the applied force, r is the distance to the center of rotation, and θ is the angle between the force and the position vector from the center of rotation to the line of action of the force.
• In this case, as the force is tangential to the edge, both vectors are perpendicular each other, so (2) becomes simply as follows:
• τ = F*r (3)
• Replacing in (1), we have:
• F*r = I*α (4)
• I is the total rotational inertia, that is composed by two terms, one is the rotational inertia of the disk, Idisk, which is given by the following expression:

`I_(disk) = m*(r^(2) )/(2) (5)`

• The other component of I, is due to the four children.
• Assuming that we can treat them as point masses, the rotational inertia due to any of them is as follows:

`I_(child) = m_(child) * r^(2) (6)`

• So, the rotational inertia will be (5) plus (6) times 4 (due to there are 4 children at the same distance r from the center of rotation), as follows:

`I_(t) = I_(disk) + 4* I_(child) (7)`

• Replacing by the givens, we get:

`I_(t) = (500kg*(2.00m)^(2))/(2) + (4* 27 kg* (2.00m)^(2)) = 1432 kg*m2 (8)`

• Replacing by the values of I, r and α in (4), and solving for F, we get:

`F = (I*\alpha )/(r) = (1432kg*m2*3.1 rad/sec2)/(2.00m) = 2220 N (9)`