The flywheel of a steam engine runs with a constant angular velocity of 140 rev/

Question

The flywheel of a steam engine runs with a constant angular velocity of 140 rev/min. When steam is shut off, the friction of the bearings and of the air stops the wheel in 1.8 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 70.0 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) using wf = wi + alpha*t

0 = 140 + alpha*(1.8x60)

alpha = 1.30 rev/min^2

b) using , wf^2 - wi^2 = 2 x alha x theta

0^2 - 140^2 = 2 x 1.30 x rev

rev = 7560 rev.

c) alpha = 1.30 rev / min = 1.30 x 2pi / 60x60 = 2.27 x 10^-3 rad/s^2

a_t = alpha*r = 2.27 x 10^-3 x 0.48 = 1.09 x 10^-3 m/s^2

d) w = 70 x 2pi /60 =7.33 rad/s^2

radial acc. = w^2 r =7.33^2 x 0.48 =25.79 m/s^2

a = sqrt(ac^2 + at^2) =25.79 m/s^@

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