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Question

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(L2y2)3/2
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B
Lv2A(L2y2)3/2
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C
L2v2A(Ly)3
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D
L2vA(L2y2)3/2
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Solution

The correct option is A L2v2A(L2y2)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=(L2y2)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(L2y2)3/2

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