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## The slotted link is pinned at O, and as a result of the constant angular velocity u # = 3 rad>s it drives the peg P for a short distance

Question

The slotted link is pinned at O, and as a result of the constant angular velocity u # = 3 rad>s it drives the peg P for a short distance along the spiral guide r = (0.4 u) m, where u is in radians. Determine the velocity and acceleration of the particle at the

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Physics
4 months
2021-09-02T09:03:50+00:00
2021-09-02T09:03:50+00:00 1 Answers
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## Answers ( )

Answer:

A) vᵣ = 1.2 m/s

B) vₜ = 0.4π m/s = 1.257 m/s

C) aᵣ = – 1.2π m/s² = – 3.771 m/s²

D) aₜ = 7.2 m/s²

Explanation:

Given,

θ’ = 3 rad/s

r = 0.4 θ

Note that

θ’ = (dθ/dt) = 3 rad/s

θ” = (d²θ/dt²) = (d/dt) (3) = 0 rad/s²

r’ = (dr/dt) = (d/dt) (0.4θ) = 0.4 (dθ/dt) = 0.4 × θ’ = 0.4 × 3 = 1.2 m/s

r” = (d²r/dt²) = 0.4 (d²θ/dt²) = 0 m/s²

A) Radial component of the velocity of P at the instant θ=π/3rad.

From the kinematics of a particle in a plane,

vᵣ = r’ = 1.2 m/s

B) Transverse component of the velocity and acceleration of P at the instant θ=π/3rad.

vₜ = rθ’ = (0.4 θ) (3) = 1.2 θ = 1.2 (π/3) = 0.4π m/s = 1.257 m/s

C) Radial component of the acceleration of P at the instant θ=π/3rad.

aᵣ = r” – r(θ’²) = 0 – (0.4θ)(3²) = – 3.6θ = – 3.6 (π/3) = – 1.2π m/s² = – 3.771 m/s²

D) Transverse component of the acceleration of P at the instant θ=π/3rad

aₜ = rθ” + 2r’θ’ = (0.4θ)(0) + (2)(1.2)(3) = 0 + 7.2 = 7.2 m/s²

Hope this Helps!!!