Question

The flywheel of a steam engine runs with a constant angular velocity of 180 rev/min. When steam is shut off, the friction of the bearings and of the air stops the wheel in 1.1 h. (a) What is the constant angular acceleration, in revolutions per minute-squared, of the wheel during the slowdown? (b) How many revolutions does the wheel make before stopping? (c) At the instant the flywheel is turning at 90.0 rev/min, what is the tangential component of the linear acceleration of a flywheel particle that is 29 cm from the axis of rotation? (d) What is the magnitude of the net linear acceleration of the particle in (c)?

Answer 1

Solution:

Part a:

w0 = 180 rev/min

w = w0 + ?t

Therefore

0 = 180 + ?*1.1*60

Therefore

? = - 2.727 rev/min^2

-ve means retardation

So ? = 2.727 rev/min^2

Part B:

= angular displacement

Using the equation

= W_i t + (0.5) t²

= 180 (1.1) + (0.5) (- 2.727 ) (1.1)²

= 196.35

number of rotations = /6.28 = 196.35/6.28 = 31.2659

Part c:

Linear Tangential Acceleration= r a

a_tan= 90.0* 2.727 = 245.43