The flywheel of a steam engine runs with a constant angular speed of 165 rev/min.
When steam is shut off�, the friction of the bearings and of the air brings the wheel to
rest in 2.30 hrs
(a) What is the constant angular acceleration of the wheel, in rev/min2 ?
(b) How many rotations will the wheel make before coming to rest?
(c) What is the tangential linear acceleration of a point 54.2 cm from the axis of
rotation when the flywheel is turning at 75.2 rev/min
(d) What is the magnitude of the total linear acceleration of the point in part (c)?
I will be happy to critique your work.
To solve these problems, we need to use the equations of rotational motion. Let's go through each question one by one.
(a) To find the constant angular acceleration, we can use the formula:
Angular acceleration (α) = (Final angular speed - Initial angular speed) / Time taken
Given:
Initial angular speed (ω1) = 165 rev/min
Final angular speed (ω2) = 0 rev/min
Time taken (t) = 2.30 hours = 2.30 * 60 = 138 minutes
Using the formula, we have:
α = (0 - 165) / 138
Simplifying the equation, we get:
α = -1.1957 rev/min^2
Therefore, the constant angular acceleration of the wheel is approximately -1.1957 rev/min^2.
(b) To find the number of rotations the wheel makes before coming to rest, we can use the formula:
Number of rotations = (Initial angular speed)^2 / (2 * angular acceleration)
Given:
Initial angular speed (ω1) = 165 rev/min
Angular acceleration (α) = -1.1957 rev/min^2
Using the formula, we have:
Number of rotations = (165)^2 / (2 * -1.1957)
Simplifying the equation, we get:
Number of rotations ≈ 1369.252
Therefore, the wheel will make approximately 1369.252 rotations before coming to rest.
(c) To find the tangential linear acceleration of a point 54.2 cm from the axis of rotation when the flywheel is turning at 75.2 rev/min, we can use the formula:
Linear acceleration (a) = Radius * Angular acceleration
Given:
Radius (r) = 54.2 cm
Angular acceleration (α) = -1.1957 rev/min^2
First, we need to convert the radius to meters:
Radius (r) = 54.2 cm = 0.542 meters
Using the formula, we have:
a = 0.542 * -1.1957
Simplifying the equation, we get:
a ≈ -0.6485 m/s^2
Therefore, the tangential linear acceleration of the point 54.2 cm from the axis of rotation is approximately -0.6485 m/s^2.
(d) To find the magnitude of the total linear acceleration of the point in part (c), we need to consider both the tangential linear acceleration and the radial acceleration. The radial acceleration can be calculated using the formula:
Radial acceleration (ar) = (angular speed)^2 * Radius
Given:
Angular speed (ω) = 75.2 rev/min
Radius (r) = 54.2 cm = 0.542 meters
First, we need to convert the angular speed to radians per second:
Angular speed (ω) = 75.2 rev/min = 75.2 * 2π / 60 radians/second
Using the formula, we have:
ar = (75.2 * 2π / 60)^2 * 0.542
Simplifying the equation, we get:
ar ≈ 4.3409 m/s^2
The total linear acceleration (at) is then the vector sum of the tangential linear acceleration (a) and the radial acceleration (ar). Since they are in opposite directions, the magnitude of the total linear acceleration is given by:
|at| = |a| - |ar|
|at| ≈ |-0.6485| - |4.3409|
|at| ≈ 5.9894 m/s^2
Therefore, the magnitude of the total linear acceleration of the point in part (c) is approximately 5.9894 m/s^2.