Collars A and B slide along the fixed right-angle rods and are connected by a cord of length L. Determine the acceleration of collar B as a function of y if collar A is given a constant upward velocity vA.
A
−L2v2A(L2−y2)3/2
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B
−Lv2A(L2−y2)3/2
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C
−L2v2A(L−y)3
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D
−L2vA(L2−y2)3/2
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Solution
The correct option is A−L2v2A(L2−y2)3/2 At the given instant, the velocities and distances of the collars have been shown in figure.
From the right triangle, x2+y2=L2 ⇒x=(L2−y2)1/2
Differentiating both sides w.r.t time: 2xdxdt+2ydydt=0...(i) ⇒dxdt=−ydydtx...(ii)
Now using ˙x=dxdt,˙y=dydt, eqn. (ii) can be written as, ⇒˙x=−y˙yx
Again differentiating eqn. (i) w.r.t t gives, (dxdt)2+xd2xdt2+(dydt)2+yd2ydt2=0 ⇒˙x2+x¨x+˙y2+y¨y=0...(iii)
where ˙y=dydt=vA= constant. It represents the rate of change of distance y, for collar A ∴¨y=d2ydt2=0
Hence from Eq (iii), acceleration of collar B: d2xdt2=−˙x2+˙y2x=−(−y˙yx)2+˙y2x ⇒d2xdt2=−⎛⎜
⎜
⎜
⎜
⎜⎝y2×(dydt)2x2⎞⎟
⎟
⎟
⎟
⎟⎠+(dydt)2x
Putting the values of dydt=vA,x2+y2=L2 gives: d2xdt2=−vA2(x2+y2)x3 ∴d2xdt2=−L2v2A(L2−y2)3/2