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Two particles of mass m and M Undergo uniform circular motion about each other at a seperation R under the influence of an attractive force F . The angular velocity is w (Omega) radians per second. Show that R=(F/w)(1/m+ 1/M).
To show that R = (F/w)(1/m + 1/M), we need to apply Newton's second law of motion to the particles undergoing circular motion.
The centripetal force required for an object to move in a circle of radius R with angular velocity w is given by:
F = mRw^2
where F is the centripetal force, m is the mass of one particle, R is the separation between the particles, and w is the angular velocity.
Now, let's solve for R:
F = mRw^2 (equation 1)
Divide both sides of equation 1 by mw^2:
F/(mw^2) = R
Now, let's multiply and divide the right side of the equation by (1/m + 1/M):
R = (F/(mw^2)) * (1/m + 1/M)
R = (F/w) * (1/m + 1/M)
Therefore, we have shown that R = (F/w)(1/m + 1/M), as required.
To show that R = (F/w)(1/m + 1/M), we can start with the basic principles of circular motion and use Newton's law of gravitation.
Circular motion can be described in terms of centripetal force, which is the force acting towards the center of the circle. In this case, the attractive force F between the two particles provides the centripetal force.
The centripetal force can also be written in terms of the angular velocity and the distance R between the particles. The centripetal force is given by F = mw²R, where w is the angular velocity.
Now let's consider the gravitational force between the two particles. According to Newton's law of gravitation, the gravitational force F is given by F = G(m)(M) / R², where G is the gravitational constant.
Comparing the two expressions for the force F, we have:
mw²R = G(m)(M) / R².
To eliminate the unknown angular velocity w, we can rearrange the equation to isolate R:
R³ = (G(m)(M))/w².
Now, divide both sides of the equation by MW:
(R³)/(M)(m) = (G)/(w²).
Finally, take the reciprocal of both sides:
1/R = (F/w)(1/m + 1/M).
Hence, we have shown that R = (F/w)(1/m + 1/M).