It is a rotational motion problem from Kleppner mechanics:

Two masses $m_a$ and $m_b$ are connected by a string of length $l$ and lie on a frictionless table. The system is twirled and released with $m_a$ instantaneously at rest and $m_b$ moving with instantaneous velocity $v_0$ at a right angle to the line of the centres as shown below. Find the subsequent motion of the system and tension in the string.

free body diagram

After I tried for two days I can find tension by polar coordinates system! (Here I assumed there is zero radial velocity of the reduced mass, according to central force problem.)

But my confusion is: As the tension is pulling the mass $m_a$ in direction along the line of the string but it has zero velocity initially; so the string should be collapsed after a certain amount of time but if I think about it the problem is invalid! At this point my mind blows up!

I am really confused. Where I am getting it wrong? 🤕🤕

  • $\begingroup$ I cannot think of any reason of why the masses will fall towards each other. They will maintain a constant seperation of L. $\endgroup$ Sep 24 '20 at 13:06
  • $\begingroup$ @ManvendraSomvanshi why they will maintain constant separation L? As mass Ma has instantaneous acceleration along the string but no tangential velocity! So the string should collapse after some times! May be I am wrong but I can't think beyond this till now! $\endgroup$ Sep 24 '20 at 13:19
  • $\begingroup$ The answer given by gandalf61 is pretty clear. I think his explanation should clarify your doubt. But if not, you should try writing conservation of angular momentum and momentum for the system. $\endgroup$ Sep 24 '20 at 13:24
  • $\begingroup$ Please do not post formulae and variable names like $M_a$ as plain text, but use MathJax instead. $\endgroup$
    – ACuriousMind
    Sep 24 '20 at 14:56

By conservation of momentum the centre of mass of the system must move to the right with constant speed

$\displaystyle \frac {m_b}{m_a+m_b}v_0 = \frac {m_b}{M}v_0$

where $M=m_a+m_b$.

If you work in the reference frame in which the COM is stationary (note that this is an inertial reference frame) then you will find that $m_a$, which is at a distance $\frac {m_b} M l$ from the COM, initially moves to left with speed $\frac {m_b} M v_0$, so has angular speed $\frac {v_0} l$ rad/s anticlockwise about the COM.

Similarly $m_b$, which is at distance $\frac {m_a} M l$ from the COM, initially moves to right with speed $\frac {m_a} M v_0$, so it also has angular speed $\frac {v_0} l$ rad/s anticlockwise about the COM.

So both masses revolve about the COM with the same constant angular speed $\frac {v_0} l$ rad/s. Once you know this you can calculate the tension in the string - and also convince yourself that the string does not collapse.


First, consider the initial motion, and try to imagine the system as a rigid body


The center of mass must have a horizontal velocity equal to

$$ v_{\rm COM} = \frac{a}{\ell} v_0 $$

The velocity is maintained through-out the motion since no external forces act here.

In addition, the motion is decomposed to a rotation about the center of mass with a rotational speed

$$ \omega = \frac{v_0}{\ell} $$

This means the motion of mass $m_a$ tracks a circle around the center of mass with radius $a$ and the mass $m_b$ tracks a circle around the center of mass with radius $b$.

The tension is such as to force both of these motions

$$ T = m_a \omega^2 a = m_b \omega^2 b $$

and remember that the center of mass is defined by $a = \frac{m_b}{m_a+m_b} \ell$ and $b = \frac{m_a}{m_a+m_b} \ell$.


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