A 346 kg merry-go-round in the shape of a
horizontal disk with a radius of 1.7 m is set in
motion by wrapping a rope about the rim of
the disk and pulling on the rope.
How large a torque would have to be exerted to bring the merry-go-round from rest
to an angular speed of 4.7 rad/s in 3.4 s?
angular acceleration = a = 4.7/3.4 = 1.38 radians/s^2
I = (1/2)mr^2 = (1/2)346(1.7^2) = 500kg m^2
Torque = I a = 500 (1.38) = 690 Nm
To calculate the torque required to bring the merry-go-round to the given angular speed, we need to use the equation:
Torque = Moment of Inertia × Angular Acceleration
First, let's find the moment of inertia of the merry-go-round. Since it is in the shape of a disk, the moment of inertia can be calculated using the formula:
Moment of Inertia (I) = (1/2) × Mass × Radius^2
Given:
Mass (m) = 346 kg
Radius (r) = 1.7 m
Substituting the values into the formula:
I = (1/2) × 346 kg × (1.7 m)^2
Solving this equation, we find:
I ≈ 522.43 kg·m²
Next, let's find the angular acceleration. We can use the formula:
Angular Acceleration (α) = (Final Angular Speed - Initial Angular Speed) / Time
Given:
Final angular speed (ω_f) = 4.7 rad/s
Initial angular speed (ω_i) = 0 rad/s
Time (t) = 3.4 s
Substituting the values into the formula:
α = (4.7 rad/s - 0 rad/s) / 3.4 s
Solving this equation, we find:
α ≈ 1.382 rad/s²
Finally, let's calculate the torque using the formula provided earlier:
Torque = Moment of Inertia × Angular Acceleration
Substituting the values:
Torque = 522.43 kg·m² × 1.382 rad/s²
Solving this equation, we find:
Torque ≈ 721.31 N·m
Therefore, the torque required to bring the merry-go-round from rest to an angular speed of 4.7 rad/s in 3.4 s is approximately 721.31 N·m.