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Two masses $M$ and $m$ are attached to a vertical axis by weightless threads of combined length $\ell$. They are set in rotational motion in a horizontal plane about this axis with constant angular velocity $\omega$. If the tensions in the threads are the same during motion, the distance of $M$ from the axis is

1. $\dfrac{m\ell}{M+m}$
2. $\dfrac{M\ell}{M+m}$
3. $\dfrac{M+m}{M}\ell$
4. $\dfrac{M+m}{m}\ell$

Correct answer: $\dfrac{m\ell}{M+m}$

Solution

For equal tension, each mass must satisfy $T=m\omega^2 r_m$ and $T=M\omega^2 r_M$, so $m r_m = M r_M$. Also, the total length gives $r_m + r_M = \ell$. Solving these two equations gives the distance of mass $M$ from the axis as $r_M=\dfrac{m\ell}{M+m}$.

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