The flywheel of a steam engine runs with a constant angular speed of 112 rev/ min . When steam is shut off, the friction of the bearings and the air brings the wheel to rest in 2.2 h. What is the magnitude of the constant angular acceleration of the wheel in rev/min^2? Do not enter the units.
How many rotations does the wheel make before coming to rest?
What is the magnitude of the tangential component of the linear acceleration of a particle that is located at a distance of 40 cm from the axis of rotation when the flywheel is turning at 56.0 rev/ min?
What is the magnitude of the net linear acceleration of the particle in the above question?
The magnitude of the constant angular acceleration of the wheel in rev/min^2 is 0.05.
The wheel makes 464 rotations before coming to rest.
The magnitude of the tangential component of the linear acceleration of the particle is 7.68 m/s^2.
The magnitude of the net linear acceleration of the particle is 7.68 m/s^2.
To find the magnitude of the constant angular acceleration of the wheel, we can use the formula:
ωf = ωi + αt,
where ωf is the final angular velocity (0 rev/min), ωi is the initial angular velocity (112 rev/min), α is the angular acceleration, and t is the time taken for the wheel to come to rest (2.2 h = 2.2 * 60 min).
Substituting the given values in the formula, we have:
0 = 112 rev/min + α * (2.2 * 60 min).
Simplifying the equation, we get:
α = -112 rev/min / (2.2 * 60 min).
Calculating this value gives:
α ≈ -0.85 rev/min².
The magnitude of the constant angular acceleration of the wheel is approximately 0.85 rev/min² (note: the negative sign indicates that the acceleration is in the opposite direction of the initial velocity).
To find the number of rotations the wheel makes before coming to rest, we can use the formula:
θ = ωi * t + (1/2) * α * t²,
where θ is the angular displacement, ωi is the initial angular velocity (112 rev/min), α is the angular acceleration (-0.85 rev/min²) and t is the time taken for the wheel to come to rest (2.2 h = 2.2 * 60 min).
Substituting the given values, we have:
θ = 112 rev/min * (2.2 * 60 min) + (1/2) * (-0.85 rev/min²) * (2.2 * 60 min)².
Simplifying and calculating this value gives:
θ ≈ 2849.6 rev.
Therefore, the wheel makes approximately 2849.6 rotations before coming to rest.
To find the magnitude of the tangential component of the linear acceleration of a particle located at a distance of 40 cm from the axis of rotation when the flywheel is turning at 56.0 rev/min, we can use the formula:
at = r * α,
where at is the tangential component of linear acceleration, r is the distance from the axis of rotation (40 cm = 0.4 m), and α is the angular acceleration (which we already calculated as -0.85 rev/min²).
Substituting the given values, we have:
at = 0.4 m * -0.85 rev/min².
Calculating this value gives:
at ≈ -0.34 m/min².
The magnitude of the tangential component of the linear acceleration is approximately 0.34 m/min².
To find the magnitude of the net linear acceleration of the particle, we need to know the radial component of the linear acceleration as well. However, it is not provided in the given information.
To find the magnitude of the constant angular acceleration of the wheel, we first need to convert the given time into minutes.
Given:
Angular speed (ω) = 112 rev/min
Time to come to rest (t) = 2.2 hours
1 hour = 60 minutes
So, 2.2 hours = 2.2 x 60 minutes = 132 minutes
To calculate the angular acceleration (α), we use the equation:
α = ω/t
Substituting the given values:
α = 112 rev/min / 132 min
To find the rotations the wheel makes before coming to rest, we need to find the total number of revolutions during the given time.
Given:
Time to come to rest (t) = 2.2 hours
To convert hours to minutes:
2.2 hours = 2.2 x 60 minutes = 132 minutes
Since the angular speed is given in revolutions per minute, we can use the formula:
Number of rotations = angular speed x time
Substituting the given values:
Number of rotations = 112 rev/min x 132 min
To find the magnitude of the tangential component of the linear acceleration of a particle, we can convert the angular speed to radians per second and use the formula:
Given:
Distance from the axis of rotation (r) = 40 cm = 0.4 m
Angular speed (ω) = 56.0 rev/min
To convert rev/min to rad/s:
1 rev/min = 2π rad/60s
Substituting the given values:
ω = 56.0 rev/min x (2π/60) rad/s
To find the magnitude of the net linear acceleration of the particle, we can use the formula:
Net linear acceleration (a) = tangential component of acceleration (at) x radius (r)
Substituting the given values:
Net linear acceleration (a) = magnitude of tangential component of linear acceleration (at) x distance from the axis of rotation (r)