The ring $M_{1}$ and block $M_{2}$ are held in the position shown in $F i g . 6.176$ . Now the system is released. If $M_{1} > M_{2}$ . find $V_{1} / V_{2}$ when the ring $m_{1}$ slides down along the smooth fixed vertical rod by the distance h.

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Solution

Method 1 :

Let ring has moved down a distance y.
$y^{2} + l^{2} = s^{2}$ .......(i)
Here is the length of string which is constant. Differentiating (i) w.r.t. time, we get

$2 y \frac{d y}{d t} + 0 = 2 s \frac{d s}{d t}$ or $2 y \frac{d y}{d t} = \frac{s}{y} \frac{d s}{d t}$ ......(ii)
Here $\frac{d y}{d t} = v_{1}$ and $\frac{d s}{d t} = v_{2}$
For $y = s, s = \sqrt{h^{2} + l^{2}}$
Therefore, Eq. (ii) takes the form
$v_{1} = \frac{\sqrt{h^{2} + l^{2}}}{h} v_{2}$ or $\frac{v_{1}}{v_{2}} = \frac{\sqrt{h^{2} + l^{2}}}{h}$

Method 2 :

Let in time t the ring falls to point A which is a distance h below the ring's initial position.
At this position, the string makes an angle $θ$ with the rod in small time interval dt, the ring down to A' and the block rises by ${d y}_{2}$ . Let us draw a perpendicular AP form A to AB.
For very small displacement' $A B \approx P B$
As the length of the string is constant
$∴$ $A^{'} P = {d y}_{2}$
But $A^{'} P = {d y}_{1} cos θ$
$∴$ ${d y}_{1} cos θ = {d y}_{2}$
Dividing both sides by dt, we get $(\frac{{d y}_{1}}{d t}) cos θ = (\frac{{d y}_{2}}{d t})$ .
Here $\frac{{d y}_{1}}{d t} = v_{1}$ , the rate of change of position of ring with time
and $\frac{{d y}_{2}}{d t} = v_{2}$ , the rate of change position of block with time.
Thus, we have $V_{1} cos θ = V_{2}$
or $\frac{v_{1}}{v_{2}} = \frac{1}{cos θ} = \frac{1}{\frac{h}{\sqrt{h^{2} + l^{2}}}} = \frac{\sqrt{h^{2} + l^{2}}}{h}$

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