The machine as shown has 2 rods of length1 m connected by a pivot at… | JEE Main 2017 (9 Apr, Shift 1)

Q40JEE Main 2017 · 9 APR · SHIFT 1Difficult

Torque, Couple & Equilibrium of Rigid Bodies

The machine as shown has 2 rods of length1 m connected by a pivot at the top. The end of one rod is connected to the floor by a stationary pivot and the end of the other rod has a roller that rolls along the floor in a slot. As the roller goes back and forth, a 2 kg weight moves up and down. If the roller is moving towards right at a constant speed, the weight moves up with a:

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Answer: (b)

Rods of length 1 m form an isosceles triangle: height $h$ , base $x$ , so $h^2 + \frac{x^2}{4} = 1$ .
Differentiate: $2h\frac{dh}{dt} + \frac{x}{2}\frac{dx}{dt} = 0$ . The roller moves right at speed $v$ , so $\frac{dx}{dt} = -v$ and $\frac{dh}{dt} = \frac{x v}{4h}$ .
As the roller moves right, $x$ falls and $h$ rises, so $\frac{x}{h}$ falls and the upward speed decreases.
At $h = 0.4$ m: $x = 2\sqrt{0.84} \approx 1.83$ m, speed $\approx 1.15v$ , not $\frac{3}{4}v$ .
So the answer is (b) decreasing speed.

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