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OLEGan [10]
3 years ago
5

Sylvia and Jadon now want to work a problem. Imagine a puck of mass 0.5 kg moving as in the simulation. Suppose that the tension

in the string is 1.0 N, and that the radius of its circular path is 0.7 m. What will Jadon and Sylvia find for the tangential speed of the puck? m/s
Physics
1 answer:
Pavlova-9 [17]3 years ago
5 0

Answer:

1.2 m/s

Explanation:

The puck in this problem is moving with uniform circular motion, so the net force acting on it (the tension in the string) must be equal to the centripetal force.

So we can write:

T=m\frac{v^2}{r}

where:

T is the tension in the string

m is the mass of the puck

v is its tangential speed

r is the radius of the circular path

For the puck in this problem, we have:

m = 0.5 kg

T = 1.0 N

r = 0.7 m

Substituting and solving for v, we find the tangential speed:

v=\sqrt{\frac{Tr}{m}}=\sqrt{\frac{(1.0)(0.7)}{0.5}}=1.2 m/s

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So, RCF, or relative centrifugal force should be equal to RCF = <img src="https://tex.z-dn.net/?f=11%2C18%2Ar%2A%28RPM%2F1000%29
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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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