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The flywheel of a steam engine runs with a constant angular velocity of 130 rev/

ID: 1910102 • Letter: T

Question

The flywheel of a steam engine runs with a constant angular velocity of 130 rev/min. When steam is shut off, the friction of the bearings and of the air stops the wheel in 2.4 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 65.0 rev/min, what is the tangential component of the linear acceleration of a flywheel particle that is 27 cm from the axis of rotation? (d) What is the magnitude of the net linear acceleration of the particle in (c)? Answer to a)-.90277 re/min^2 b)9360.08064 rev Need C and D - thanks

Explanation / Answer

a) =/t =0-130/(2.4h)(60min/h) =-0.90 rev/min2

b)t=2.4*60=144min

the number of revolutions is

=ot+(1/2)t2

=130*144+(1/2)(-0.9)(144)2 =9.39*103 rev

c)the tangential acceleration is

at =r =(-0.9*2/60)*0.37 =-0.035m/s2

d)the angular speed of flywheel

=65*(2/60)=6.8rad/s

the radial acceleration

ar =r2r =6.82*0.37=17.1m/s2

a=(ar2+at2) =(17.12+(-0.035)2) 17.1m/s2

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