What Is the Angular Acceleration of a Steam Engine Flywheel?

[sana masih]

The flywheel of a steam engine runs with a constant angular speed of a 149 rev/min. When the steam is cut off, the friction of the bearings and of the air brings the wheel to rest in 1.9 hours.
a) assuming it is constant, what is the angular acceleration of the wheel?

b) how many rotations will the wheel make before coming to rest?

c) what is the tangential linear acceleration of the particle 49 cm from axis of rotation when the flywheel when it is turning at 74 rev/min?

d) what is the magnitude of the total linear acceleration of the particle in the Part C?

Can someone Please help me on this question, and walk me through the steps.

 

[hage567]

Do you know your rotational kinematic equations? For (a) you are trying to find the angular acceleration. Which equation has the three things you know (that would be initial and final angular speeds, time), so you can solve for angular acceleration?

Be careful of your units, remember to convert them if necessary so they are consistent with each other.

 

[ogb p]

a) The angular acceleration of the wheel can be calculated using the formula: angular acceleration = (final angular velocity - initial angular velocity) / time. In this case, the final angular velocity is 0 rev/min since the wheel comes to rest, the initial angular velocity is 149 rev/min, and the time is given as 1.9 hours. To convert hours to minutes, we multiply by 60, so the time becomes 1.9 hours x 60 minutes/hour = 114 minutes. Substituting these values into the formula, we get: angular acceleration = (0 rev/min - 149 rev/min) / 114 minutes = -1.31 rev/min^2.

b) The number of rotations the wheel makes before coming to rest can be calculated by dividing the initial angular velocity by the angular acceleration. In this case, the initial angular velocity is 149 rev/min and the angular acceleration is -1.31 rev/min^2. So, the number of rotations can be calculated as 149 rev/min / (-1.31 rev/min^2) = -113.74 rotations. However, since we cannot have a negative number of rotations, we can take the absolute value of this answer, which gives us 113.74 rotations. This means that the wheel will make approximately 113.74 rotations before coming to rest.

c) The tangential linear acceleration of a particle is given by the formula: tangential linear acceleration = radius x angular acceleration. In this case, the radius is given as 49 cm and the angular acceleration is given as -1.31 rev/min^2 (as calculated in part a). To convert rev/min^2 to cm/s^2, we multiply by (2π/60)^2, which gives us -0.0034 cm/s^2. So, the tangential linear acceleration can be calculated as 49 cm x (-0.0034 cm/s^2) = -0.1666 cm/s^2.

d) The total linear acceleration of a particle is given by the formula: total linear acceleration = √(radial linear acceleration^2 + tangential linear acceleration^2). In this case, the radial linear acceleration is equal to 0 since the particle is located on the axis of rotation. So, the total linear acceleration can be calculated as √(0^2 + (-0.1666 cm/s^2)^2) =