Question

You rev your car's engine and watch the tachometer climbsteadily from 1200rpm to 5500rpm in 2.7s.

A) What is the angular acceleration of the engine?

B) What is the tangential acceleration of a point on the edgeof the engine's 3.5-cm-diameter crankshaft?

C) How many revolutions does the engine make during thistime?

Answer 1

initial angular speed w = 1200 rpm

                                  = 1200 * 2π rad / 60 s

                                  = 125.66 rad / s

final angular speed w ' = 5500 rpm

                                 = 5500 * 2π rad / 60 s

                                 = 575.95 rad / s

time t = 2.7 s

So, angular accleration α = ( w ' - w ) / t

                                     = 166.777 rad / s ^ 2

(B). the tangential acceleration of a point on the edge of theengine's 3.5-cm-diameter crankshaft is a = r α

where r = 3.5 cm / 2 = 1.75 cm = 0.0175 m

So, a = 2.918 m / s ^ 2

(C). Angular displacement θ = wt + ( 1/ 2) α t ^2

where t = 2.7 s

substitue values we get θ value

So, No.of revolutions N = θ / 2π