The flywheel of an old steam engine is a solid homogeneous metal disk of mass M = 131 kgand radius R = 80 cm.The engine rotates the wheel at 580 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 = 120 N.If the coefficient of kinetic friction between the pad and the flywheel is ìk = 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 answer these questions, we need to use the principles of rotational dynamics. Let's break down the problem into smaller steps.
First, let's find the moment of inertia of the flywheel. The moment of inertia for a solid disk rotating about its center is given by the formula:
I = (1/2) * M * R^2
where M is the mass of the disk and R is its radius. Plugging in the given values, we get:
I = (1/2) * 131 kg * (0.8 m)^2 = 41.92 kg * m^2
Next, let's calculate the angular velocity of the flywheel when the brake is applied. The angular velocity is given by the formula:
ω = (2π * n) / 60
where ω is the angular velocity in radians per second and n is the rotational speed in revolutions per minute. Plugging in the given values, we get:
ω = (2π * 580) / 60 = 60.94 rad/s
Now, let's find the torque τ exerted by the brake pad using the equation:
τ = F * R
where F is the force applied by the brake pad and R is the radius of the flywheel. Plugging in the given values, we get:
τ = 120 N * 0.8 m = 96 N * m
Since the torque is equal to the moment of inertia multiplied by the angular acceleration, we can calculate the angular acceleration α using the equation:
τ = I * α
Rearranging the equation, we get:
α = τ / I
Plugging in the values, we get:
α = 96 N * m / 41.92 kg * m^2 ≈ 2.29 rad/s^2
Next, we can use the equation for angular acceleration to find the time it takes for the flywheel to come to rest. The equation is:
ω = ω0 + αt
where ω0 is the initial angular velocity, α is the angular acceleration, and t is the time. The initial angular velocity ω0 is 60.94 rad/s, and we want to find the time it takes for ω to become zero.
0 = 60.94 rad/s + 2.29 rad/s^2 * t
Solving for t, we get:
t = -60.94 rad/s / (2.29 rad/s^2)
t ≈ -26.62 s
Since we cannot have a negative time, we discard this negative value. Therefore, it takes approximately 26.62 seconds for the flywheel to come to rest.
Finally, to calculate the work done by the torque during this time, we can use the equation:
Work = τ * θ
where θ is the angle through which the torque is applied. In this case, since the flywheel comes to rest, the angle θ is equal to 2π radians (or one full revolution). Plugging in the values, we get:
Work = 96 N * m * 2π ≈ 603.19 J
Therefore, the work done by the torque during this time is approximately 603.19 Joules.
To summarize:
- The flywheel makes approximately 26.62 revolutions before coming to rest.
- It takes approximately 26.62 seconds for the flywheel to come to rest.
- The work done by the torque during this time is approximately 603.19 Joules.