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PilotLPTM [1.2K]
2 years ago
10

So, RCF, or relative centrifugal force should be equal to RCF =

%5E2" id="TexFormula1" title="11,18*r*(RPM/1000)^2" alt="11,18*r*(RPM/1000)^2" align="absmiddle" class="latex-formula">
The radius is measured in centimeters and RPM is revolutions per minute.

What I don't know is, if I plug all the data in the formula, will the result equal to acceleration in meters per second squared, or it will be measured in g's so that if I multiply the g's by 9.8 meters per second square, it will give me the actual acceleration.
Physics
1 answer:
zysi [14]2 years ago
6 0

After plugging all the data into the equation, the result of the relative centrifugal force (RCF)  is measured in terms of g.

<h3>What is relative centrifugal force?</h3>

The relative centrifugal force (RCF) or the g force is the radial force generated by the spinning rotor as expressed relative to the earth's gravitational force.

RCF = ac/g

where;

  • ac is centripetal acceleration
  • g is acceleration due to gravity

RCF = \frac{\omega ^2 r}{g} = 1.118\times 10^{-5} \ (RPM)^2 r = 11.18r\ (RPM/1000)^2

where;

  • r is radius in cm

<h3>For example, </h3>

Find the maximum RCF of the JS-4.2 rotor can be obtained from its maximum speed (4200 rpm) and its rmax (250 mm);

RCF = 11.18 \times 25\ cm \times (\frac{4200 \ RPM}{1000} )^2 = 4,930.3 \times g

Thus, after plugging all the data into the equation, the result is measured in terms of g.

Learn more about relative centrifugal force here: brainly.com/question/26887699

#SPJ1

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A ferry coming into port is sailing at 12 m/s. It takes 2.5km to come to rest in the port. Calculate the deceleration of the fer
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Answer:    -0.0288 m/s^2

Explanation:

Let's suppose that the ferry decelerates at a constant rate A (deceleration is an acceleration in the opposite direction to the original motion of an object)

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a(t) = -A

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To get the velocity equation of the ferry, we need to integrate with respect to the time, t, we will get:

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then p0 = 0m

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Now we can replace this in the position equation because we know that the ferry needs 2.5 km or 2500 meters to come to rest.

p(  (12m/s)/A) = 2500m =  (-A/2)*( (12m/s)/A)^2 + (12m/s)*((12m/s)/A)

2500m = (-72 m^2/s^2)/A + (144m^2/s^2)/A

2500m = (72 m^2/s^2)/A

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A = (72 m^2/s^2)/2500m = 0.0288 m/s^2

and the acceleration of the ferry was -A, then the acceleration of the ferry is:

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