The flywheel of an old steam engine is a solid homogeneous metal disk of mass M = 114 kgand radius R = 80 cm.The engine rotates the wheel at 540 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 = 140 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 find the number of revolutions the flywheel makes before coming to rest, we need to calculate the angular acceleration and then use it to find the time taken for the flywheel to come to rest.
Step 1: Convert the rotational speed from rpm to rad/s.
Given: Rotational speed (ω) = 540 rpm
1 revolution = 2π radians
1 minute = 60 seconds
Convert rpm to rad/s:
ω = (540 rpm) × (2π rad/1 min) × (1 min/60 sec) = 56.55 rad/s
Step 2: Calculate the moment of inertia of the flywheel.
Given: Mass of the flywheel (M) = 114 kg
Radius of the flywheel (R) = 80 cm = 0.8 m
Moment of inertia (I) for a solid disk rotating about its center is given by:
I = (1/2)MR^2
Substitute the given values into the equation:
I = (1/2)(114 kg)(0.8 m)^2 = 36.48 kg·m²
Step 3: Calculate the torque acting on the flywheel.
Given: Force (F) = 140 N
Coefficient of kinetic friction (μk) = 0.2
Torque (τ) = Fr
Substitute the values into the equation:
τ = (140 N)(0.8 m) = 112 N·m
Step 4: Use Newton's second law for rotational motion to find the angular acceleration (α).
τ = Iα
α = τ/I
Substitute the values into the equation:
α = (112 N·m)/(36.48 kg·m²) = 3.07 rad/s²
Step 5: Use the equation for angular acceleration to find the time taken for the flywheel to come to rest.
ω = ω0 + αt
Where ω0 is the initial angular velocity, which is 56.55 rad/s.
Rearrange the equation to solve for time (t):
t = (ω - ω0)/α
t = (0 - 56.55 rad/s)/(3.07 rad/s²) = -18.44 s
Note: The negative sign indicates that we are dealing with deceleration, which is expected since the flywheel is coming to rest.
Step 6: Calculate the number of revolutions made by the flywheel before coming to rest.
The number of revolutions (N) is given by:
N = (ω0 - ω)/2π
N = (56.55 rad/s - 0 rad/s)/(2π rad/1 revolution) = 9 revolutions (approximately)
Step 7: Calculate the work done by torque during this time.
The work done (W) is given by:
W = τθ, where θ is the angle rotated.
Since the flywheel makes 9 revolutions before coming to rest, the angle rotated (θ) is given by:
θ = N × 2π = 9 revolutions × 2π rad/1 revolution = 18π rad
Substitute the values into the equation:
W = (112 N·m)(18π rad) = 20179.08 N·m (or J)
Therefore, the flywheel makes approximately 9 revolutions before coming to rest, takes about 18.44 seconds to come to rest, and the work done by the torque during this time is 20179.08 N·m (or J).
To calculate the number of revolutions the flywheel makes before coming to rest, we need to determine the angular deceleration caused by the applied force and the moment of inertia of the flywheel.
1. First, let's find the moment of inertia (I) of the flywheel using the formula for a solid disk:
I = (1/2) * M * R^2
= (1/2) * 114 kg * (0.8 m)^2
= 36.48 kg m^2
2. Next, we need to find the torque (τ) acting on the flywheel due to the applied force (F). The torque can be calculated using the formula:
τ = F * R
= 140 N * 0.8 m
= 112 Nm
3. Now, we can calculate the angular deceleration (α) using the torque (τ) and the moment of inertia (I) using the equation:
τ = I * α
α = τ / I
α = 112 Nm / 36.48 kg m^2
≈ 3.07 rad/s^2
4. To calculate the number of revolutions, we need to convert the given angular velocity in RPM to radians per second (rad/s). We can use the conversion factor:
ω (rad/s) = ω (RPM) * (2π / 60)
ω = 540 RPM * (2π / 60)
ω ≈ 56.55 rad/s
5. Using the formula for angular deceleration (α) and final angular velocity (ωf) with an initial angular velocity (ωi) of 56.55 rad/s, we can calculate the time (t) it takes for the flywheel to come to rest:
ωf = ωi + α * t
0 = 56.55 rad/s + (-3.07 rad/s^2) * t
t ≈ 18.42 s
6. Finally, to calculate the number of revolutions the flywheel makes before coming to rest, we can use the formula:
Number of revolutions = (ωi * t) / (2π)
Number of revolutions = (56.55 rad/s * 18.42 s) / (2π)
Number of revolutions ≈ 165.85 revolutions
Therefore, the flywheel makes approximately 165.85 revolutions before coming to rest.
Regarding the work done by the torque during this time, the work-energy principle states that the work done is equal to the change in kinetic energy:
Work = ΔKE
Since the flywheel comes to rest, the change in kinetic energy is equal to the initial kinetic energy:
Work = KE_initial
The initial kinetic energy of the flywheel can be calculated using the formula:
KE_initial = (1/2) * I * ωi^2
KE_initial = (1/2) * 36.48 kg m^2 * (56.55 rad/s)^2
KE_initial ≈ 114,047.65 J
Therefore, the work done by the torque during the time it takes for the flywheel to come to rest is approximately 114,047.65 Joules.