A merry-go-round in the park has a radius of 1.8 m and a rotational inertia of 900 kgm^2. A child pushes the merry-go-round with a constant force of 80 N applied at the edge 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 mery-go-round about its axle?
b. What is the rotational acceleration of the merry-go-round?
c. At this rate, what will the rotational velocity of the merry-go-round be after 15s if it starts from rest?
d. What is the rotational acceleration of the merry-go-round if the child stops pushing after 15 s.? How long will it take for the merry-go-round to stop turning?
To answer these questions, we'll need to use the equations of rotational motion and the concept of torque. Here's how we can solve each part of the problem:
a. The net torque acting on the merry-go-round can be found by subtracting the frictional torque from the applied torque. Since torque is the product of force and radius, we have:
Applied torque = Force x Radius = 80 N x 1.8 m = 144 Nm
Net torque = Applied torque - Frictional torque = 144 Nm - 12 Nm = 132 Nm
b. The rotational acceleration can be calculated using the equation τ = Iα, where τ is the net torque, I is the rotational inertia, and α is the rotational acceleration. Rearranging the equation, we have α = τ / I. Substituting the given values, we get:
α = 132 Nm / 900 kgm^2 = 0.1467 rad/s^2
c. To determine the rotational velocity after 15s, we can use the equation ω = ω0 + αt, where ω0 is the initial rotational velocity, α is the rotational acceleration, and t is the time. Since the merry-go-round starts from rest, ω0 is 0. Substituting the values:
ω = 0 + (0.1467 rad/s^2) x (15 s) = 2.2 rad/s
d. If the child stops pushing after 15s, the net torque becomes zero. When the net torque is zero, the rotational acceleration also becomes zero. Therefore, the merry-go-round will not experience any further rotational acceleration. To find the time it takes for the merry-go-round to stop turning, we can use the equation ω = ω0 + αt and set ω = 0. Substituting the values:
0 = 0 + (0.1467 rad/s^2) x t
t = 0 / 0.1467 rad/s^2 = 0s
Since the acceleration is zero, the merry-go-round will stop turning instantly after the child stops pushing.