The machine as shown has 2 rods of length 1, | vedclass

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(B) Let $l = 1\, m$ be the length of each rod. Let $y$ be the height of the weight from the floor and $x$ be the horizontal distance between the fixed

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decreasing speed

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(B) Let $l = 1\, m$ be the length of each rod. Let $y$ be the height of the weight from the floor and $x$ be the horizontal distance between the fixed pivot and the roller.From the geometry of the isosceles triangle formed by the two rods,we have the relation: $(x/2)^2 + y^2 = l^2$.Substituting $l = 1$,we get $x^2/4 + y^2 = 1$,which simplifies to $x^2 + 4y^2 = 4$.Differentiating both sides with respect to time $t$:$2x(dx/dt) + 8y(dy/dt) = 0$.Let $v_r = dx/dt$ be the constant speed of the roller and $v_w = dy/dt$ be the speed of the weight.Then $2x v_r + 8y v_w = 0$,which gives $v_w = -(x v_r) / (4y)$.Since the weight is moving up,we consider the magnitude: $v_w = (x v_r) / (4y)$.Substituting $x = \sqrt{4 - 4y^2} = 2\sqrt{1 - y^2}$,we get $v_w = (2\sqrt{1 - y^2} \cdot v_r) / (4y) = v_r \cdot \frac{\sqrt{1 - y^2}}{2y}$.As the roller moves towards the right,the distance $x$ decreases,which means the height $y$ increases.As $y$ increases,the term $\frac{\sqrt{1 - y^2}}{2y}$ decreases.Therefore,the speed of the weight $v_w$ decreases as it moves up.

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