The hoop is cast on the rough surface such that it has an angular velocity w=4rad/s and an angular acceleration a=5rad/s^2. also, its center has a velocity v0=5m/s and a deceleration a0=2m/s^2. determine the acceleration of point a at this instant

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 = ?

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°

answer:

a

explanation: