A teenager pushes tangentially on a small hand-driven merry-go-round 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. Calculate the torque required to produce the acceleration, neglecting frictional torque. What force is required at the edge

a)  τ = 421.9 N m,  b)   F = 168.76 N

Explanation:

For this exercise we use Newton's second law for rotational motion

          τ = I α

Let's find the angular acceleration with kinematics

           w = w₀ + α t

as part of rest its initial angular velocity is zero wo = 0

           α= w / t

let's reduce the magnitudes to the SI system

            w = 15 rpm (2π rad / 1 rev) (1 min / 60s) = 1.57 rad / s

let's calculate

            α = 1.57 /10

            α = 0.157 rad / s²

Now let's look for the moment of inertia, which is the sum of the moment of inertia of the disk plus the moment of the children

disk moment                   I₁ = ½ M r²

moment of each child      I₂ = m r²

             I = I₁ + 2 I₂

             I = ½ M r² + 2 m r²

we substitute

             τ = (½ M r2 + 2 m r2) alpha

             τ = r² (½ M + 2 m) α

             τ = 2.5² (760/2 + 2 25) 0.157

             τ = 421.9 Nm

What force is applied

          τ = F r

          F = τ / r

          F = 421.9 / 2.5

          F = 168.76 N

it has to be b, about 97 percent.