A 190-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.700 rev/s in 2.00 s? (State the magnitude of the force.)
To solve this problem, we can use the rotational equivalent of Newton's second law, which states that the torque applied to an object is equal to the moment of inertia times the angular acceleration. The torque can be calculated by multiplying the force applied at the edge of the merry-go-round by the radius of the disk.
The moment of inertia for a solid disk rotating about its central axis is given by the formula I = (1/2)*m*r^2, where I is the moment of inertia, m is the mass of the disk, and r is the radius of the disk.
Given:
Mass of the disk (m) = 190 kg
Radius of the disk (r) = 1.50 m
Initial angular speed (ω₁) = 0 rad/s
Final angular speed (ω₂) = 0.700 rev/s = 0.700 * (2π rad/rev) = 1.40π rad/s
Time taken for acceleration (t) = 2.00 s
First, we need to find the change in angular speed (Δω):
Δω = ω₂ - ω₁
= (1.40π rad/s) - (0 rad/s)
= 1.40π rad/s
Next, we can calculate the angular acceleration (α) using the formula:
α = Δω / t
α = (1.40π rad/s) / (2.00 s)
= 0.70π rad/s²
Now, we can calculate the moment of inertia (I) using the formula:
I = (1/2) * m * r²
I = (1/2) * (190 kg) * (1.50 m)²
= 213.75 kg•m²
Finally, we can find the force (F) applied to the rope using the formula for torque (τ) and rearranging it:
τ = I * α
τ = F * r
F * r = I * α
F = (I * α) / r
F = (213.75 kg•m² * 0.70π rad/s²) / 1.50 m
= 660.57 N
Therefore, a constant force of 660.57 N must be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.700 rev/s in 2.00 s.