Question 64 (1 point)
1 answer:
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Given Information:
Angular velocity = ω = 4 rad/s
Angular acceleration = α = 5 rad/s²
Center deceleration = a₀ = 2 m/s
Required Information:
Acceleration of point A at this instant = ?
Answer:
Acceleration of point A at this instant = 5.94 m/s²
Explanation:
Refer to the attached diagram of the question,
The acceleration of point A is given by
a = a₀ + rα - rω²
Where r is the radial distance between the center and point A, a₀ is the deceleration of center, α is the angular acceleration and ω is the angular velocity.
a = -2i + 0.3j*5k - 0.3j*4²
a = -2i + 1.5(j*k) - 0.3j*16
a = -2i + 1.5(-i) - 4.8j
a = -2i - 1.5i - 4.8j
a = -3.5i - 4.8j
The magnitude of acceleration vector is
a = √(-3.5)² + (-4.8)²
a = √35.29
a = 5.94 m/s²
Therefore, the acceleration of point A is 5.94 m/s²
The angle is given by
θ = tan⁻¹(y/x)
θ = tan⁻¹(-4.8/-3.5)
θ = 53.9°
