Question

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

• A. $\frac{Ml}{M+m}$
• B. $\frac{ml}{M+m}$
• C. $\frac{M+m}{M}l$
• D. $\frac{M+m}{m}l$

Answer 1

The correct option is B $\frac{ml}{M+m}$

Since rotation is in the horizontal plane,
Tension in the thread $=$ centripetal force

Let the distance of $M$ from the axis of rotation be $x$. If both the masses are revolving about the axis $y{y}^{\prime }$ and tension in both the threads are equal, then

$M{\omega }^{2}x=m{\omega }^2(l-x)$
$\Rightarrow Mx=m(l-x)$
$\Rightarrow x=\frac{ml}{M+m}$