Question

Figure $\mathbf{\text{10-58}}$shows a propeller blade that rotates at $\mathbf{\text{2000\hspace{0.17em}rev/min}}$about a perpendicular axis at point B. Point A is at the outer tip of the blade, at radial distance $\mathbf{\text{1.50\hspace{0.17em}m}}$. (a) What is the difference in the magnitudes a of the centripetal acceleration of point A and of a point at radial distance $\mathbf{\text{0.150\hspace{0.17em}m}}$? (b) Find the slope of a plot of a versus radial distance along the blade.

Short answer

a) The difference in the magnitudes of the centripetal acceleration is$5.92\times {10}^4\text{\hspace{0.17em}m}/{\text{s}}^2$ .

b) The slope of versus radial distance along the blade is $4.39\times {10}^4\text{\hspace{0.17em}m}/{\text{s}}^2$.

Step-by-step solution

Step 1: Given

i) Frequency of propeller,$f=2000\text{\hspace{0.17em}rev/min}$

ii) Radius,$r=1.5\text{\hspace{0.17em}m}$

the concept

From the given frequency of propeller, find the angular velocity of the propeller. Then, by using the relation between angular velocity and linear velocity into a formula of centripetal acceleration, find the difference in the magnitudes of the centripetal acceleration of point and of a point at radial distance$\mathbf{\text{0.150\hspace{0.17em}m}}$.

The angular velocity in terms of frequency is given as-

$\mathbf{\omega }\mathbf{=}\mathbf{2}\mathbf{\pi f}$

The centripetal acceleration is given as-

$\begin{array}{c}\mathbf{a}\mathbf{=}{\mathbf{v}}^2\mathbf{/}\mathbf{r}\\ \mathbf{=}{\mathbf{\omega }}^2\mathbf{r}\end{array}$

where, v is velocity, r is radius, a is acceleration, f is frequency and $\mathbf{\omega }$ is angular frequency.

(a) Determining the difference in the magnitudesof the centripetal acceleration

For angular velocity of the propeller,

$\begin{array}{c}\omega =2\pi f\\ =\left(2000\frac{\text{rev}}{\text{min}}\right)\left(\frac{2\pi \text{rad}}{1\text{\hspace{0.17em}rev}}\right)\left(\frac{1\text{\hspace{0.17em}min}}{60\text{\hspace{0.17em}s}}\right)\\ =209\text{\hspace{0.17em}rad}/\text{s}\end{array}$

Now,

$\begin{array}{c}a=\frac{v^2}{r}\\ ={\omega }^{2}r\end{array}$

For point A, radial distance is $r=1.5\text{\hspace{0.17em}m}$, therefore,

$\begin{array}{c}{a}_A={\omega }^{2}r\\ ={\left(209\frac{\text{rad}}{\text{s}}\right)}^2\times (1.5\text{m})\\ =65521.5\text{m}/{\text{s}}^2\end{array}$

And, for pointp, radial distance is$r=0.150\text{m},$therefore,

$\begin{array}{c}{a}_p={\omega }^{2}r\\ ={\left(209\frac{\text{rad}}{\text{s}}\right)}^2\times (0.150\text{m})\end{array}$

$a_p=6552.15\text{\hspace{0.17em}m}/{\text{s}}^2$

Hence, difference in their acceleration is,

$\begin{array}{c}\Delta a=a_A-a_p\\ =65521.5\frac{\text{m}}{{\text{s}}^{\text{2}}}-6552.15\frac{\text{m}}{{\text{s}}^{\text{2}}}\\ =5.9\times {10}^4\text{\hspace{0.17em}m}/{\text{s}}^2\end{array}$

Hence, the difference in the magnitudes of the centripetal acceleration of point$A$and of a point at radial distance$0.150\text{\hspace{0.17em}m}$is$5.9\times {10}^4\text{\hspace{0.17em}m}/{\text{s}}^2$.

(b) Determining the slope of  versus radial distance along the blade 

Slope of the plot a versus radial distance is,

$\begin{array}{c}\frac{a}{r}={\omega }^2\\ ={\left(209\frac{\text{rad}}{\text{s}}\right)}^2\\ =4.37\times {10}^4\frac{\text{m}}{{\text{s}}^{\text{2}}}\end{array}$

Hence, the slope of versus radial distance along the blade is $4.39\times {10}^4\text{\hspace{0.17em}m}/{\text{s}}^2$.