The flywheel of an old steam engine is a solid homogeneous metal disk of mass M = 120 kg and radius R = 80 cm. The engine rotates the wheel at 500 rpm. In an emergency, to bring the engine to a stop, the flywheel is disengaged from the engine and a brake pad is applied at the edge to provide a radially inward force F = 100 N. If the coefficient of kinetic friction between the pad and the flywheel is uk = 0.2, how many revolutions does the flywheel make before coming to rest? How long does it take for the flywheel to come to rest? Calculate the work done by the torque during this time.
To solve this problem, we can use the principles of rotational motion and the work-energy theorem.
First, let's find the initial angular velocity of the flywheel.
Given:
- Mass of the flywheel (M) = 120 kg
- Radius of the flywheel (R) = 80 cm = 0.8 m
- Rotational speed (ω) = 500 rpm
We can convert the rotational speed to radians per second:
Angular velocity (ω) = (500 rpm) * (2π rad/1 min) * (1 min/60 s) = 52.36 rad/s
Using the moment of inertia formula for a solid disk:
Moment of inertia (I) = (1/2) * M * R^2 = (1/2) * 120 kg * (0.8 m)^2 = 38.4 kg·m^2
The torque (τ) acting on the flywheel is given by:
Torque = Force * Lever arm
Lever arm (r) = radius of the flywheel = 0.8 m
Torque (τ) = Force (F) * Lever arm (r) = 100 N * 0.8 m = 80 N·m
Now, let's find the angular acceleration (α):
Torque (τ) = Moment of inertia (I) * Angular acceleration (α)
α = τ / I = 80 N·m / 38.4 kg·m^2 = 2.08 rad/s^2
Using the kinematic equation:
Final angular velocity (ωf)^2 = Initial angular velocity (ωi)^2 + 2 * α * θ
Since the flywheel comes to rest when ωf = 0 rad/s, we have:
0 = (52.36 rad/s)^2 + 2 * 2.08 rad/s^2 * θ
θ = - (52.36 rad/s)^2 / (2 * 2.08 rad/s^2) = - 662.78 rad
We take the negative value of θ because the flywheel rotates in the opposite direction when it comes to rest.
The number of revolutions the flywheel makes before coming to rest can be calculated using the formula:
Number of revolutions = θ / (2π rad/rev) = -662.78 rad / (2π rad/rev) ≈ -105.5 rev
Note that we take the negative value because the flywheel rotates in the opposite direction.
To find the time it takes for the flywheel to come to rest, we can use the equation:
ω = ωi + α * t
0 = 52.36 rad/s + (-2.08 rad/s^2) * t
t = - 52.36 rad/s / (-2.08 rad/s^2) ≈ 25.19 s
The time it takes for the flywheel to come to rest is approximately 25.19 seconds.
To calculate the work done by the torque during this time, we can use the work-energy theorem:
Work = Change in kinetic energy
Since the initial kinetic energy is zero (the flywheel starts from rest), the work done by the torque is equal to the final kinetic energy.
The final kinetic energy (Kf) can be calculated using the rotational kinetic energy formula:
Kf = (1/2) * I * ωf^2
Plugging in the values, we get:
Kf = (1/2) * 38.4 kg·m^2 * (0 rad/s)^2 = 0 J
Therefore, the work done by the torque during this time is 0 Joules.