Question

II An electric fan goes from rest to$1800\text{rpm}$ in $4.0\text{s}$. What is its angular acceleration?

Short answer

The angular acceleration of the fan is $47.12\text{rad}/{\text{s}}^2$.

Step-by-step solution

Step 1. Given information

The electric fan goes from rest to$1800\text{rpm}$ in $4.0\text{s}$.

Step 2. Explanation

The angular acceleration of the fan is,

$\alpha =\frac{{\omega }_2-{\omega }_1}{t}$

Here, $\alpha$is the angular acceleration of the fan, ${\omega }_2$is the final angular velocity of the fan, ${\omega }_1$is the initial velocity of the fan, and $t$ is the period of time.

At initial position, the fan is at rest. Thus ${\omega }_1=0\text{rad}/\text{s}$.

Substitute $0\text{rad}/\text{s}$for ${\omega }_1,1800\text{rpm}$for ${\omega }_2$and $4.0\text{s}$for $t$in the above equation to find $\alpha$.

$\alpha =\frac{\left(1800\text{rpm}\times \frac{1\text{min}}{60\text{s}}\times \frac{2\text{mmd}}{1\text{rv}}\right)-0\text{rad}/\text{s}}{4.0\text{s}}=47.12\text{rad}/{\text{s}}^2$

Thus, the angular acceleration of the fan is $47.12\text{rad}/{\text{s}}^2$.