Motion of two bodies at both ends of a string

Question

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.

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? 🤕🤕

Answer 1

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.

Answer 2

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 $b = \frac{m_a}{m_a+m_b} \ell$.