A phonograph turntable, initially rotating at 0.75 rev/s, slows down and stops in 30 s. The magnitude of its average angular acceleration in rad/s² for this process is: A) 1.5 B) 1.5 $\pi$ C) $\pi$ /40 D) $\pi$ /20 E) 0.75

Question

Quizplus · Accepted Answer

The average angular acceleration is given by:$$\alpha = \frac{\omega_f - \omega_i}{t}$$where:$$\omega_f$$ is the final angular velocity (0 rad/s)$$\omega_i$$ is the initial angular velocity (0.75 rev/s)t is the time (30 s)Substituting these values into the equation, we get:$$\alpha = \frac{0 - 0.75 	imes 2\pi}{30} = -0.157 \ rad/s^2$$The magnitude of the average angular acceleration is therefore:$$|\alpha| = 0.157 \ rad/s^2$$which is choice D.

A Phonograph Turntable, Initially Rotating at 0 π\piπ C) π\piπ /40 D) π\piπ

A Phonograph Turntable, Initially Rotating at 0 $\pi$ C) $\pi$ /40
D) $\pi$