In a Do‑Nothing Machine, two sliders are confined to perpendicular channels and are connected by a rod. The rod can be rotated, causing the sliders to move back and forth in their respective channels. Suppose that the channels line up with the x‑ and y ‑directions and that the slider in the x ‑channel has velocity vx at some instant. Let theta be the instantaneous angle of the connecting rod with respect to the x ‑direction. The length of the rod between the two sliders is L. Find an expression for the instantaneous velocity vy of the slider in the y‑channel. Express your answer in terms of vx, L, and theta.

Question

Amaya

Physics

29 April, 02:21

In a Do‑Nothing Machine, two sliders are confined to perpendicular channels and are connected by a rod. The rod can be rotated, causing the sliders to move back and forth in their respective channels. Suppose that the channels line up with the x‑ and y ‑directions and that the slider in the x ‑channel has velocity vx at some instant. Let theta be the instantaneous angle of the connecting rod with respect to the x ‑direction. The length of the rod between the two sliders is L. Find an expression for the instantaneous velocity vy of the slider in the y‑channel. Express your answer in terms of vx, L, and theta.

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Answers (1)

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Royce Conner · Answer 1

Vy = - Vx * cotg (θ)

Explanation:

From the Pythagoras theorem:

X^2 + Y^2 = L^2

The first derivative respect of time is:

2*X*Vx + 2*Y*Vy = 0

X*Vx = - Y*Vy

Vy = - Vx * X/Y

From trigonometry:

X = L * cos (θ)

Y = L * sin (θ)

X/Y = (L * cos (θ)) / (L * sin (θ))

X/Y = cotg (θ)

Vy = - Vx * cotg (θ)

The speed does not depend on the length of the arm L.