Engr Help

Relative Velocity and Acceleration withTranslating and Rotating Frames

Example: (continued)

a_P|_F = a_Q|_F + a_P|_B+ α_B x r_QP - ω_B² r_QP+ 2 ω_B x v_P|_B

0 i + 0 j = a_Q i + a_P|_B i₁+ α_B k x r i₁ - ω_B² r i₁+ 2 ω_B k x v_P|_B i₁

Next calculate cross products and collect terms using i₁ and j₁ components

0 i + 0 j = 0 i₁ + 0 j₁ = a_Q i + a_P|_B i₁+ r α_B j₁ - ω_B² r i₁+ 2 ω_B v_P|_B j₁

Recall from coordinate transformation i = cos θ i₁ ˗ sin θ j₁ so

0 i₁ + 0 j₁ = a_Q [ cos θ i₁ ˗ sin θ j₁ ] + a_P|_B i₁+ r α_B j₁ - ω_B² r i₁+ 2 ω_B v_P|_B j₁

0 i₁ + 0 j₁ = [ a_Q cos θ + a_P|_B ˗ ω_B² r ] i₁+ [ ˗ a_Q sin θ + r α_B + 2 ω_B v_P|_B ] j₁

Next equate i₁ and j₁ components.

0 = a_Q cos θ + a_P|_B ˗ ω_B² r or 0 = ˗ 36 (4/5) + a_P|_B ˗ (3²) 5 a_P|_B = 369/5

0 = ˗ a_Q sin θ + r α_B + 2 ω_B v_P|_B 0 = 36 (3/5) + 5 α_B + 2(3)(˗ 20) α_B = 492/25

So a_P|_B = 369/5 i₁ = 369/5 [ i cos θ + j sin θ ] = 1476/25 i + 1107/25 j m/sec² (result)

and α_B_\F = 492/25 k rad/sec² (result)