91Ó°ÊÓ

i) Frequency of propeller,$f=2000\text{ r±ð±¹/³¾¾±²Ô}$

ii) Radius,$r=1.5\text{″¾}$

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$\mathbf{\text{0.150″¾}}$.

The angular velocity in terms of frequency is given as-

$\mathbf{Ó\neg }\mathbf{=}\mathbf{2}\mathbf{Ï€´Ú}$

The centripetal acceleration is given as-

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

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

For angular velocity of the propeller,

$\begin{array}{c}\mathrm{Ó\neg }=2\mathrm{Ï€}f\\ =\left(2000\frac{\text{rev}}{\text{min}}\right)\left(\frac{2\mathrm{Ï€}\text{rad}}{1\text{ r±ð±¹}}\right)\left(\frac{1\text{″¾in}}{60\text{ s}}\right)\\ =209\text{ r²¹»å}/\text{s}\end{array}$

Now,

$\begin{array}{c}a=\frac{v^2}{r}\\ ={\mathrm{Ó\neg }}^{2}r\end{array}$

For point A, radial distance is $r=1.5\text{″¾}$, therefore,

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

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

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

$a_p=6552.15\text{″¾}/{\text{s}}^2$

Hence, difference in their acceleration is,

$\begin{array}{c}\mathrm{Δ}a=a_A−a_p\\ =65521.5\text{ }\frac{\text{m}}{{\text{s}}^{\text{2}}}−6552.15\text{ }\frac{\text{m}}{{\text{s}}^{\text{2}}}\\ =5.9×{10}^4\text{″¾}/{\text{s}}^2\end{array}$

Hence, the difference in the magnitudes of the centripetal acceleration of point$A$and of a point at radial distance$0.150\text{″¾}$is$5.9×{10}^4\text{″¾}/{\text{s}}^2$.

Slope of the plot a versus radial distance is,

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

Hence, the slope of versus radial distance along the blade is $4.39×{10}^4\text{″¾}/{\text{s}}^2$.