Question

Breanna is standing beside a merry-go-round pushing 14° from the tangential direction and is able to accelerate the ride and her friends from rest to a frequency of 13 rpm in 9 seconds. Assume the merry-go-round is a uniform disc of radius 2.3 m and has a mass of 750 kg. Rachel (56 kg) and Tayler (57 kg) sit opposite each other on the edge of the ride.

What force did Breanna push with? (N) (include units with answer)

How much work did Breanna do? (J) (include units with answer)

Answer 1

Consider that the tangential component of the applied force as F.
Torque due to the applied force, = R * F ...(1)
Where R is the radius of the merry-go-round.

Moment of inertia of the merry-go-round = 1/2 * Md * R²
Where Md is the mass of the disk
Moment of inertia of Tayler and Rachel = M_T * R² + M_R * R²
Where M_T is the mass of Tayler and MR is the mass of Rachel
Total moment of inertia, I = 1/2 * Md * R² + M_T * R² + M_R * R²
= [1/2 Md + M_T + M_R] * R²
= [0.5 * 750 + 57 + 56] * 2.3²
= 2581.52

Initial angular velocity, i = 0
Final angular velocity, f = 13 rpm
= 13 * (2/60) rad/s = 1.36 rad/s
Time taken, t = 9 s
Using the formula, = [f - i]/t,
= [1.36 - 0] / 9
= 0.15 rad/s²

Torque, = I *
Substituting equation (1),
F * R = I *
F = [I * ] / R
= [2581.52 * 0.15] / 2.3
= 169.77 N

Consider Fa as the applied force
Fa cos = F
Fa = F / cos
= 169.77 / cos(14)
= 174.97 N
= 175 N

b)
Angular displacement, = i * t + 1/2 * t²
= 0 + 0.5 * 0.15 * 9²
= 6.13 rad

Work done = *
= R * F *
= 2.3 * 169.77 * 6.13
= 2392 J