Engr Help

Rolling (example continued)

a_C = a_P + α x r_PC ˗ ω² r_PC = [1 + r/(R ˗ r)] r ω² j + α k x r j - ω² r j

a_C = r/(R ˗ r)] r ω² j - r α i = - (2)(˗ 3) i + [ 2 / (10 ˗ 2)] (2) (4²) j = 6 i + 8 j in/sec² (result)

Apply the relative velocity equation between points A and P. Recall v_P = 0 i + 0 j

v_A = v_P + ω x r_PA = v_P + ωk x (ri + rj) = - rω i + rω j

v_A = - 8 i + 8 j in/sec (result)

Next apply the relative acceleration equation between points A and P.

a_A = a_P + α x r_PA ˗ ω² r_PA = [1 + r/(R ˗ r)] r ω² j + α k x (ri + rj) - ω² (ri + r j)

a_A = 40 j - r α i + r α j - rω² i ˗ r ω² j = 6 i + 8 j + 6 i - 6 j - 32 i ˗ 32 j

a_A = [ 6 ˗ 32 ] i + [ 40 ˗ 6 ˗ 32 ] j = ˗ 26 i + 2 j in/sec² (result)

Apply the relative velocity equation between points C and O.

v_C = v_O + ω x r_OC = 0 i + 0 j + ω_OCk x [˗ (R ˗ r)] j] = 8ω_OC i

But v_C = - 8 i = 8ω_OC i which gives ω_OC = ˗ 1 So ω_OC = ˗ 1 k rad/sec (result)

Next apply the relative acceleration equation between points C and O.

a_C = a_O + α_OC x r_OC ˗ ω_OC² r_OC = 0 i + 0 j + α_OC k x [ ˗ (R ˗ r)] j] - ω_OC² [ ˗ (R ˗ r)] j]

a_C = (R ˗ r)] α_OC ( ˗ i ) + ω_OC² (R – r) j = 8 α_OC i + 8 j = 6 i + 8 j

Therefore 8 α_OC = 6 α_OC = ¾ α_OC = ¾ k rad/sec² (result)