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An automobile engine slows down from 4500 rpm to 1200 rpm in 2.5 seconds. (A) Calculate its angular acceleration. (Give your answer in radians per second squared written as rad/s^2)

User Mikael
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1 Answer

3 votes

Answer:

-138.23
rad/s^2

Step-by-step explanation:

The initial angular velocity = 4500 rpm

The final angular velocity = 1200 rpm

1 revolution = 2π radians

1 minute = 60 seconds

So,


Initial\ velocity=\frac {4500* 2\pi}{60}\ radians/sec=471.2389\ radians/sec


Final\ velocity=\frac {1200* 2\pi}{60}\ radians/sec=125.6637\ radians/sec

Time = 2.5 seconds

Angular acceleration is:


\alpha=\frac {\omega_f-\omega_i}{t}


\alpha=\frac {125.6637-471.2389}{2.5}\ rad/s^2

Angular acceleration =- 138.23
rad/s^2

User Jordan Soltman
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