Question

1. determine the angular acceleration of a tire that accelerates from 750 rpm to 350 rpm in 6 seconds.

13.96 $\frac{rad}{s^{2}}$
-6.98 $\frac{rad}{s^{2}}$
6.98 $\frac{rad}{s^{2}}$
3.49 $\frac{rad}{s^{2}}$

Explanation:

Step1: Convert RPM to rad/s

First, convert initial and final angular - speeds from RPM to rad/s. The conversion factor is $2\pi$ radians per revolution and 60 seconds per minute. The initial angular speed $\omega_0$: $\omega_0 = 750\times\frac{2\pi}{60}= 25\pi$ rad/s. The final angular speed $\omega$: $\omega = 350\times\frac{2\pi}{60}=\frac{35\pi}{3}$ rad/s.

Step2: Use angular - acceleration formula

The formula for angular acceleration $\alpha$ is $\alpha=\frac{\omega-\omega_0}{t}$, where $t = 6$ s. Substitute the values: $\alpha=\frac{\frac{35\pi}{3}-25\pi}{6}=\frac{\frac{35\pi - 75\pi}{3}}{6}=\frac{-\frac{40\pi}{3}}{6}=-\frac{20\pi}{9}\approx - 6.98$ rad/s².

Answer:

-6.98 $\frac{\text{rad}}{\text{s}^2}$