The flywheel of a steam engine runs with a constant angular velocity of 160 rev/

Question

The flywheel of a steam engine runs with a constant angular velocity of 160 rev/min. When steam is shut off, the friction of the bearings and of the air stops the wheel in 2.8 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 80.0 rev/min, what is the tangential component of the linear acceleration of a flywheel particle that is 48 cm from the axis of rotation? (d) What is the magnitude of the net linear acceleration of the particle in (c)?

Explanation / Answer

a) wf = wi + alpha x t

     0 = 160 + alpha x (2.8 x 60)

alpha = - 0.952 rev / min^2

b) rev. = wi x t + alpha x t^2 /2

         = 160 x 2.8 x 60 + (- 0.952 ) x (2.8 x 60)^2 /2

         = 13445.376 rev

c) tang, acc. = alpha. r   =[ 0.952 x 2 x pi / (60 x 60)] x 0.48 = 7.97 x 10-4 m/s2

d) centripiatla acc. = w^2 .r   = (80 x 2 x pi / 60)^2 x 0.48 = 33.69 m/s2

net = sqrt ( tan.^2 + centri. ^2 ) = 33.70 m/s2

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