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Question: Two small rings \( O \) and \( O' \) are put on two vertical stationary rods \( AB \) and \( A'B' \)...

Two small rings OO and OO' are put on two vertical stationary rods ABAB and ABA'B' , respectively. One end of an inextensible thread is tied at the point AA' . The thread passes through the ring OO' and its other end is tied to the ring OO . Assuming that ring OO moves downwards at a constant velocity v1{v_1} , the velocity v2{v_2} of the ring OO when AOO=α\angle AOO' = \alpha is

(A) v1[2sin2(α/2)cosα]\left( A \right){\text{ }}{{\text{v}}_1}\left[ {\dfrac{{2{{\sin }^2}(\alpha /2)}}{{\cos \alpha }}} \right]
(B) v1[2cos2(α/2)sinα]\left( B \right){\text{ }}{{\text{v}}_1}\left[ {\dfrac{{2{{\cos }^2}(\alpha /2)}}{{\sin \alpha }}} \right]
(C) v1[3cos2(α/2)sinα]\left( C \right){\text{ }}{{\text{v}}_1}\left[ {\dfrac{{3{{\cos }^2}(\alpha /2)}}{{\sin \alpha }}} \right]
(D) None of these\left( D \right){\text{ None of these}}

Explanation

Solution

Since, in this question, we have an inextensible thread. Also since the reference frame is fixed to ring OO . So in this system, the velocity of the ring will be equal to v1/cosα{v_1}/\cos \alpha and it will be directed upwards. So from this, we will be able to find the velocity of the ring OO which will be relative to the straight line AAAA' .

Formula used:
The trigonometric identity used is,
1cosα=2sin2(α/2)1 - \cos \alpha = 2{\sin ^2}(\alpha /2) .

Complete step by step solution:
So for solving this question, we will go to the reference frame which is fixed to ring OO' in the system, and from this, we will get the velocity of the ring will be equal to v1/cosα{v_1}/\cos \alpha . Since we know that the velocity of the ring OO which will be relative to the straight line AAAA' is stationary with respect to the ground.
So from this, the equation made will be,
v1=v1cosα+v2cosα\Rightarrow {v_1} = {v_1}\cos \alpha + {v_2}\cos \alpha
Solving for the value of v2{v_2} , we will get the above equation as
v2=v1(1cosα)cosα\Rightarrow {v_2} = \dfrac{{{v_1}\left( {1 - \cos \alpha } \right)}}{{\cos \alpha }}
And as we know that 1cosα=2sin2(α/2)1 - \cos \alpha = 2{\sin ^2}(\alpha /2)
So on substituting the values in the above equation, we will get the equation as
v2=v1(2sin2(α/2))cosα\Rightarrow {v_2} = \dfrac{{{v_1}\left( {2{{\sin }^2}(\alpha /2)} \right)}}{{\cos \alpha }} .
Hence, the option (A)\left( A \right) is correct.

Note:
For solving this type of question the most important thing is a figure, so a body diagram should always be made while solving this type of question as it makes the question very clear and also it makes precise which helps to find the required values quickly.