Question

Two small rings $O$ and $O^{\prime }$ are put on two vertical stationary rods $AB$ and $A^{\prime }{B}^{\prime }$, respectively. One end of an inextensible thread is tied at a point $A^{\prime }$. The thread passes through ring $O^{\prime }$ and its other end is tied to ring $O$. Assuming that ring $O^{\prime }$ moves downwards at a constant velocity $v_1$, then velocity $v_2$ of the ring
$O$, when $\angle AO{O}^{\prime }$= $\alpha$, is

• A. $v_1\left[\frac{2{\sin }^2\alpha /2}{\cos \alpha }\right]$
• B. $v_1\left[\frac{2{\cos }^2\alpha /2}{\sin \alpha }\right]$
• C. $v_1\left[\frac{3{\cos }^2\alpha /2}{\sin \alpha }\right]$
• D. None of these

Answer 1

The correct option is A $v_1\left[\frac{2{\sin }^2\alpha /2}{\cos \alpha }\right]$

Thread length will remain constant.Therefore,
A'O'+O'O = Constant
Therefore,
$\frac{d\left(A\text{'}O\text{'}\right)}{dt}=-\frac{d(O\text{'}O)}{dt}$
$\Rightarrow v_2\cos \alpha -(-v_1\cos \alpha )=v_1\Rightarrow v_2\cos \alpha +v_1\cos \alpha =v_1\Rightarrow v_2=v_1\left(\frac{2{\sin }^2(\alpha /2)}{\cos \alpha }\right]$