How to show that if two masses($m_1$ and $m_2$) are connected($r_1<r_2$) with a string moving around a point A( circular motion), they move with same angular velocity? In this special case if $r_1=0.5 m$ and $r_2=0.9 m$, $m_1=m_2=0.15 kg$ => $ \omega = 6.9 rad/s$.

Picure below depicts the situation besides that you have two objects in same wire, so the other "ball" has moved to next wire nearer($r_1$) than $r_2$.

picture which depicts almost the situation

  • $\begingroup$ Can you draw a diagram? It isn't clear what you're asking. $\endgroup$ Nov 30, 2012 at 10:29
  • $\begingroup$ How did you determine $\omega$? I am missing information here. $\endgroup$
    – Bernhard
    Nov 30, 2012 at 10:30

2 Answers 2


If the string remains straight, its "hands" turn at the same angle $d\phi$ for time interval $dt$: $d\phi_1 = d\phi_2$. The corresponding angular velocities are then equal: $\omega_1=d\phi_1/dt=\omega_2=d\phi_2/dt$


Assuming the string is free to slide around the pivot point the two masses can't move at the same angular velocity and stay stable.

The tension in the string due to the motion of the mass $M_1$ is $F_1 = M_1r_1\omega_1^2$ and likewise for the mass $M_2$. If the string doesn't slide $F_1$ must be equal to $F_2$ so:

$$ M_1r_1\omega_1^2 = M_2r_2\omega_2^2 $$


$$ \omega_1^2 = \frac{M_2r_2}{M_1r_1} \omega_2^2 $$

so in general $\omega_1$ will not be equal to $\omega_2$.

This seems obvious enough that I feel I must be missing something. Is the string free to slide at the pivot point $A$? Also, you use the word string at one point and $wire$ at another. Is the "string" actually a piece of rigid wire?


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