Question

In the design of a timing mechanism, the motion of pin P in the fixed circular slot is controlled by the guide A, which is being elevated by its lead screw. Guide A starts from rest with pin P at the lowest point in the circular slot, and accelerates upward at a constant rate until it reaches a speed of 60 mm/s at the halfway point of its vertical displacement. The guide then decelerates at a constant rate and comes to a stop with pin P at the uppermost point in the circular slot. Determine the n- and t-components of acceleration of pin P once the pin has traveled 30° around the slot from the starting position.

Answer 1

Answer: Got It!

Explanation: Guide A Starts From Rest With Pin P At The Lowest Point In The Circular Slot, And Accelerates Upward At A Constant Rate Until It Reaches A Speed Of 175 Mm/s At The ... In the design of a timing mechanism, the motion of pin P in the fixed circular slot is controlled by the guide A, which is being elevated by its lead screw.

Answer 2

The timing mechanism is used to enable optimal operation of an engine and prevent damage

The n-component of the acceleration is approximately 0.005 m/s²

The t-component of the acceleration is approximately 0.003 m/s²

The reason the above values are correct are as follows;

Question: The diagram in the question that appear required can be described as follows;

The radius of the slot, r = 30 cm

The angle of the start position from the vertical = 20°

The  known parameters are;

The speed of the guide A at the start = 0

The speed of the guide A at the halfway point = 60 mm/s = 0.06 m/s

Required:

The n- and t- components of acceleration of pin P once the pin has traveled 30° around the slot from the starting position

Solution:

The vertical distance from the start to the mid position, s = 30 × cos(20°) ≈ 28.2 cm = 0.282 m

v² = u² + 2·a·s

a = v²/(2·s)

∴ a = (0.06 m/s)²/(2 × 0.282 m) ≈ 0.0064 m/s²

At 30° around the slot, we have; s₁ = 30 × cos(20°) - 30 × cos(50°) ≈ 8.9

The vertical distance moved, s₁ ≈ 8.9 cm = 0.089 m

v₁² ≈ 2 × 0.0064 m/s² × 0.089 m

v₁ ≈ √(2 × 0.0064 m/s² × 0.089 m) ≈ 0.03375 m/s = 33.75 mm/s

The velocity of the pin, v = v₁/(cos(θ) = 33.75 mm/s/(cos(30°)) ≈ 38.97 mm/s

The velocity of the pin, at the required point, v ≈ 38.97 mm/s = 0.03897 m/s

The normal acceleration towards the center, $a_n = \dfrac{v^2}{r}$, therefore;

$a_n = \dfrac{0.03897^2}{0.3} \approx 0.0050625 \ m/s^2$

The n-component (normal) of the acceleration, aₙ ≈ 0.005 m/s²

The tangential acceleration, $a_t$ = aₙ × tan(θ)

∴ $a_t$ = 0.0050625 × tan(30°) ≈ 0.003 m/s²

The t-component (tangential) of the acceleration, $a_t$ ≈ 0.003 m/s²

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