Question

1. determine the angular acceleration of a gear that speeds up from 25 rpm to 45 rpm in 3 seconds.

0.35 $\frac{rad}{s^{2}}$
0.70 $\frac{rad}{s^{2}}$
14 $\frac{rad}{s^{2}}$
-0.70 $\frac{rad}{s^{2}}$

Explanation:

Step1: Convert RPM to rad/s

First, convert the initial and final angular - velocities from RPM to rad/s. The conversion factor is $1\ RPM=\frac{2\pi}{60}\ rad/s$.
The initial angular velocity $\omega_0 = 25\ RPM=25\times\frac{2\pi}{60}\ rad/s=\frac{5\pi}{6}\ rad/s$.
The final angular velocity $\omega = 45\ RPM = 45\times\frac{2\pi}{60}\ rad/s=\frac{3\pi}{2}\ rad/s$.

Step2: Use the angular - acceleration formula

The formula for angular acceleration $\alpha=\frac{\omega-\omega_0}{t}$, where $t = 3\ s$.
$\alpha=\frac{\frac{3\pi}{2}-\frac{5\pi}{6}}{3}$.
First, find a common denominator for the numerator: $\frac{3\pi}{2}-\frac{5\pi}{6}=\frac{9\pi - 5\pi}{6}=\frac{4\pi}{6}=\frac{2\pi}{3}$.
Then, $\alpha=\frac{\frac{2\pi}{3}}{3}=\frac{2\pi}{9}\approx0.70\ rad/s^{2}$.

Answer:

$0.70\frac{rad}{s^{2}}$