A merry-go-round has a mass of 1580 kg and a radius of 7.25m. How much net work is required to accelerate it from rest to a rotation rate of 1 revolution per 7.1s? Assume it is a solid cylinder.
The work required equals the final kinetic energy.
The Kinetic Energy is (1/2) I w^2
The moment of inertia (for a solid disc) is (1/2)M R^2.
w = (2 pi radians)/7.1 s = 0.885 rad/s
is the angular velocity
Put it all together.
32522J
To calculate the net work required to accelerate the merry-go-round from rest to a specific rotation rate, we need to determine the rotational kinetic energy at that speed first.
The formula for rotational kinetic energy of a solid cylinder is:
K = (1/2) * I * ω^2
Where:
K is the rotational kinetic energy,
I is the moment of inertia of the solid cylinder,
ω is the angular velocity (in radians per second) squared.
The moment of inertia for a solid cylinder rotating about its central axis is given by the formula:
I = (1/2) * m * r^2
Where:
m is the mass of the merry-go-round,
r is the radius of the merry-go-round.
Now, let's calculate the moment of inertia I:
I = (1/2) * m * r^2
= (1/2) * 1580 kg * (7.25m)^2
I ≈ 32,252.19 kg·m²
Next, let's calculate the angular velocity ω:
ω = 2π / T
Where:
T is the period, given as 7.1 seconds.
ω = 2π / 7.1 s
≈ 0.885 rad/s
Now, substitute the values of I and ω into the formula for rotational kinetic energy K:
K = (1/2) * I * ω^2
≈ (1/2) * 32,252.19 kg·m² * (0.885 rad/s)^2
≈ 12,904.06 J (joules)
The rotational kinetic energy represents the net work done on the merry-go-round. Therefore, the net work required to accelerate the merry-go-round from rest to a rotation rate of 1 revolution per 7.1 seconds is approximately 12,904.06 joules.