Chapter 6, Problem 3T

(a)

To determine

To find: The angular speed of the rotor.

Answer to Problem 3T

The angular speed of the rotor is $240\pi \mathrm{rad}/\min$ or $753.98\mathrm{rad}/\min$.

Explanation of Solution

Given:

Length of the rotor blades of a helicopter are $16\mathrm{ft}$.

Rotation per minute is $120$.

Formulas used:

Angular speed is $\omega =\frac{\theta }{t}$, where $\theta$ is the angle change in a rotation and $t$ is time taken for angular rotation.

Calculation:

For one complete rotation a circle is covered.

Angle for one rotation is $2\pi$.

In $1\min$ the fan rotates $120$ times.

Angle change will be,

$\begin{array}{c}\theta =120\times 2\pi \\ =240\pi \end{array}$

In $\text{one}\min$, the angle $\theta$ changes by $240\pi \mathrm{rad}$.

Formula for angular speed is $\omega =\frac{\theta }{t}$.

Substitute $240\pi \mathrm{rad}$ for $\theta$ and $1\min$ for $t$,

$\begin{array}{c}\omega =\frac{240\pi }{1}\\ =240\pi \\ \approx 753.98\end{array}$

Thus the angular speed of the rotor is $240\pi \mathrm{rad}/\min$ or $753.98\mathrm{rad}/\min$.

(b)

To determine

To find: The linear speed of a point on the tip of a blade.

Answer to Problem 3T

The linear speed of the a point on the tip of a blade is $12063.7\mathrm{ft}/\min$ or $137\mathrm{mi}/h$.

Explanation of Solution

Given:

Length of the rotor blades of a helicopter are $16\mathrm{ft}$.

Rotation per minute is $120$.

Formulas used:

Linear speed is $v=\frac{s}{t}=\frac{r\theta }{t}=r\left(\frac{\theta }{t}\right)=r\omega$

Here $\theta$ is the angle change in a rotation, $s$ is the distance the point travels in time $t$ and $\omega$ is the angular speed.

Conversion of $\mathrm{ft}/\min$ to $\mathrm{mi}/\mathrm{h}$,

$\begin{array}{c}1\mathrm{ft}/\min =\frac{1}{88}\mathrm{mi}/\mathrm{h}\\ =0.0113636\mathrm{mi}/h\end{array}$

Calculation:

For $1$ complete rotation a circle is covered.

Angle for $1$ rotation is $2\pi$.

In $1\min$, the fan rotates $120$ times.

Angle change will be,

$\begin{array}{c}\theta =120\times 2\pi \\ =240\pi \end{array}$

In $1\min$ the angle $\theta$ changes by $240\pi \mathrm{rad}$.

The distance a point travels in time $t$ is $s=r\theta$.

Substitute $16\mathrm{ft}$ for $r$ and $240\pi \mathrm{rad}$ for $\theta$.

$\begin{array}{c}s=16\times 240\pi \\ =3840\pi \\ \approx 12063.7\end{array}$

The distance travelled by blades in $1\min$ for $120\mathrm{rpm}$ is $12063.7\mathrm{ft}$

Formula for linear speed is $v=\frac{s}{t}$,

Substitute $12063.7\mathrm{ft}$ for $s$ and $1\min$ for $t$,

$\begin{array}{c}v=\frac{12063.7}{1}\\ =12063.7\end{array}$

The linear speed of the a point on the tip of a blade is $12063.7\mathrm{ft}/\min$.

Convert $\mathrm{ft}/min$ to $\mathrm{mi}/h$,

$\begin{array}{c}12063.7\times 0.0113636=137.0875\\ \approx 137\mathrm{mi}/h\end{array}$

Thus the linear speed of the a point on the tip of a blade is $12063.7\mathrm{ft}/\min$ or $137\mathrm{mi}/h$.