MHB Particles rotating about each other with uniform angular speed

[Fantini]

Here's the problem:

Two particles of mass $m$ and $M$ undergo uniform circular motion about each other at a separation $R$ under the influence of an attractive force $F$. The angular velocity is $\omega$ radians per second. Show that

$$R = \frac{F}{\omega^2} \left( \frac{1}{m} + \frac{1}{M} \right).$$

I don't understand what is meant by undergo circular motion about each other.

 

[I like Serena]

Here's the problem:

Two particles of mass $m$ and $M$ undergo uniform circular motion about each other at a separation $R$ under the influence of an attractive force $F$. The angular velocity is $\omega$ radians per second. Show that

$$R = \frac{F}{\omega^2} \left( \frac{1}{m} + \frac{1}{M} \right).$$

I don't understand what is meant by undergo circular motion about each other.

Hi Fantini,

For instance 2 bodies attracted by gravity do that.
In that case they both make a circular movement around the common center of gravity.

 

[Fantini]

Thank you, ILS. This confirms my initial sketch of the situation was correct. A colleague managed to solve it during a brainstorm today. Here's the solution.

For each body we can equate the resultant force to mass times the acceleration it feels, therefore we have the set of equations

$$F = \frac{m \omega^2 R}{2}, \text{ and } F = \frac{M \omega^2 R}{2}.$$

Isolating $R$ in both gives

$$R = \frac{2F}{m \omega^2} \text{ and } R = \frac{2F}{M \omega^2}.$$

Adding them results in

$$2R = \frac{2F}{\omega^2} \left( \frac{1}{m} + \frac{1}{M} \right)$$

and we have

$$R = \frac{F}{\omega^2} \left( \frac{1}{m} + \frac{1}{M} \right).$$

 

[I like Serena]

Thank you, ILS. This confirms my initial sketch of the situation was correct. A colleague managed to solve it during a brainstorm today. Here's the solution.

For each body we can equate the resultant force to mass times the acceleration it feels, therefore we have the set of equations

$$F = \frac{m \omega^2 R}{2}, \text{ and } F = \frac{M \omega^2 R}{2}.$$

I believe... that is not correct.
It seems your colleague is assuming that the common point of rotation is halfway the 2 masses, but as I see it, there is nothing that suggests that.
It appears to be accidental that the right answer came out.

As I see it, mass $m$ has a distance of $r_1$ to the common point of rotation.
And mass $M$ has a distance of $r_2$ to the common point of rotation.

I believe that the proper set of equations is:
\begin{cases}
F=m\omega^2 r_1 \\
F=M\omega^2 r_2 \\
R=r_1+r_2
\end{cases}

 

[Fantini]

I see. So if mass $m$ is closer to the center of rotation than mass $M$ then it should have a smaller speed because it covers less distance, while mass $M$ has greater speed because it covers greater distance.

Either way, they are always at an angle of $\pi$ with each other such that $R$ is constant, correct?

 

[I like Serena]

I see. So if mass $m$ is closer to the center of rotation than mass $M$ then it should have a smaller speed because it covers less distance, while mass $M$ has greater speed because it covers greater distance.

Either way, they are always at an angle of $pi$ with each other such that $R$ is constant, correct?

Exactly.

The force $F$ is the centripetal force that dictates the relation between the $\omega$'s and the respective radiuses.
The separation $R$ can only be constant if the $\omega$'s are the same, meaning the angle is a constant $\pi$.