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!!!
