How Does Accelerating a Car Affect Its Engine's Angular Metrics?

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SUMMARY

The discussion focuses on the angular metrics of a car's engine during acceleration, specifically analyzing the tachometer's readings from 1200 rpm to 5500 rpm over 2.5 seconds. The angular acceleration is calculated to be 180.12 rad/s², while the tangential acceleration at the edge of the 3.8 cm diameter crankshaft is determined to be 3.42 m/s². The engine completes approximately 139.58 revolutions during this time, and the torque generated by the engine is 1.62 mN. Finally, the kinetic energy of the crankshaft at maximum rpm is 1492.78 J, with an angular momentum calculated accordingly.

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As you press down on the accelerator of your car you see the tachometer climb steadily from 1200 rpm to 5500 rpm in 2.5 s.
(a) What is the angular acceleration of the car?
(b) What is the tangential acceleration of a point on the edge of the engine’s 3.8 cm diameter crankshaft?
(c) How many revolutions does the engine make during this time?
(d) If the moment of inertia of the crankshaft is 0.009 kg.m2, what is the torque generated by the engine?
(e) What is the kinetic energy of the crankshaft at its maximum rpm? What is its angular momentum at that time?


I like to make sure I'm grasping this:

angular velocity intial and final (wi and wf)
a. wi=125.66 wf=575.96
wf=wi+at
575.96=125.66-a2.5
ang=180.12 rads-2

b.ang=a/r
ang*r=a
180.12*.019=a
a=3.42 ms-2

c.theta=wit+(1/2)at^2
theta=314.14+562.88
theta=877.02 rad
rev=theta/2pi
rev=877.02/2pi
rev=139.58

d. torque=I*ang
torque=.009*180.12
torque=1.62 mN

e. K=(1/2)Iw^2
K=(1/2)(.009)(575.98(^2
K=1492.78 J
 
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