A merry-go-round in the park has a radius of 1.8 m and a rotational inertia of 900kgm^2. A child pushes the merry-go-round with a constant force of 80 N applied at the edges and parallel to the edge. A frictional torque of 12 Nm acts at the axle of the merry-go-round.

a. what is the net torque acting on the merry-go-round about its axle?
b. what is the rotational acceleration of the merry-go-round?

To answer these questions, we need to understand the concept of torque and how it relates to rotational motion. Torque is a measure of the tendency of a force to rotate an object about a given axis. It is calculated by multiplying the force applied to an object by the distance from the axis of rotation.

a. To find the net torque acting on the merry-go-round, we need to consider both the torque applied by the child and the frictional torque. The torque applied by the child is equal to the force applied multiplied by the radius of the merry-go-round.

Given:
Force applied by the child (F) = 80 N
Radius of the merry-go-round (r) = 1.8 m
Frictional torque (τ_friction) = 12 Nm

The torque applied by the child (τ_child) = F * r = 80 N * 1.8 m = 144 Nm

The net torque (τ_net) can be calculated by subtracting the frictional torque from the torque applied by the child:

τ_net = τ_child - τ_friction = 144 Nm - 12 Nm = 132 Nm

Therefore, the net torque acting on the merry-go-round about its axle is 132 Nm.

b. The rotational acceleration of the merry-go-round can be calculated using the formula:

τ_net = I * α

Where τ_net is the net torque, I is the rotational inertia, and α is the rotational acceleration.

Given:
Net torque (τ_net) = 132 Nm
Rotational inertia (I) = 900 kgm^2

By rearranging the formula, we can solve for the rotational acceleration:

α = τ_net / I = 132 Nm / 900 kgm^2

Using the given values:

α = 0.147 m/s^2

Therefore, the rotational acceleration of the merry-go-round is 0.147 m/s^2.