The angular speed of an automobile engine is increased from 1200 rev/min to 3000 rev/min in 12 s. (a) What is its angular acceleration, assuming it to be uniform? (b) How many revolutions does the engine make during this time?

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Answer 1 · 2018-04-26T11:55:07

The angular acceleration is #=15.71rads^-2# and the number of revolutions is #=419.9#

The initial angular speed is #omega_0=1200*2pi/60=40pirads^-1#

The final angular speed is #omega_1=3000*2pi/60=100pirads^-1#

The time is #t=12s#

Apply the equation of motion

#omega_1=omega_0+alphat#

The angular acceleration is

#alpha=(omega_1-omega_0)/(12)=(100pi-40pi)/12#

#=15.71rads^-2#

Apply the equation

#omega_1^2=omega_0^2+2alphatheta#

The angle travelled is

#theta=(omega_1^2-omega_0^2)/(2alpha)=((100pi)^2-(40pi)^2)/(2*15.71)#

#=2638.6rad#

#=419.9 " revolutions"#