Each of the space shuttle's main engines is fed liquid hydrogen bya high-pressure pump. Turbine blades inside the pump rotateat 617 rev/s. A point on one of the blades traces out acircle with a radius of 0.020m as the blade rotates. a) What is themagnitude of the centripetal acceleration that the blade mustsustain at this point? b) Express this acceleration as amultiple of g = 9.80m/s2 .
answer: 3.0*105m/s2

Respuesta :

Answer:

a) [tex] a= \frac{4 \pi^2 (0.02 m)^2}{0.00162 s}=9.74 \frac{m}{s^2}[/tex]

b) [tex] k = \frac{9.8}{9.74}=1.006[/tex]

Explanation:

Part a

For this case we can begin finding the period like this:

[tex] T= \frac{1}{w} =\frac{1}{617 rad/s}=0.00162 s[/tex]

Then we know that the centripetal acceleration is given by:

[tex] a= \frac{v^2}{r}[/tex]

And the velocity is given by:

[tex] v=\frac{2\pi r}{T}[/tex]

If we replace this into the acceleration we got:

[tex] a = \frac{(\frac{2\pi r}{T})^2}{r}= \frac{4 \pi^2 r}{T^2}[/tex]

And we can replace the values and we got:

[tex] a= \frac{4 \pi^2 (0.02 m)^2}{0.00162 s}=9.74 \frac{m}{s^2}[/tex]

Part b

For this case we want to find a value of k such that:

[tex] a= k 9.8[/tex]

Where a = 9.74, so then we can solve for k like this:

[tex] k = \frac{9.8}{9.74}=1.006[/tex]

a) The magnitude of the centripetal acceleration be "300581.71 m/s²".

b) The acceleration be "30671.5 g"

Centripetal Acceleration

According to the question,

Radius, r = 0.020m

g = 9.8 m/s²

[tex]\omega[/tex] = 617 rev/s

By converting it into "rads/sec",

  = 617 × [tex]\frac{2 \pi}{1 \ rev}[/tex]

  = 3876.73 rads/sec

We know the formula,

(a)

Centripetal acceleration, [tex]a_c[/tex] = [tex]\frac{v^2}{r}[/tex]

                                               = r[tex]\omega^2[/tex]

By substituting the values, we get

                                               = 0.020 × 3876.73

                                               = 300581.71 m/s²

(b)

As a multiple of "g", the acceleration be:

[tex]a_c[/tex] = 30671.5 g

Thus the answer above is appropriate.

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