Alternate Forms of Moment Equation (example continued)

Strategy (continued)

Next apply Euler’s Second Law in the following form using point A for P.

Σ M_P = (r_PC x m a_P)_z + I^P_zz α

Euler’s Second Law- Arbitrary Point P

Note: a_A = 0 i + 0 j

Σ M_A = I^A_zz α Use the parallel axis theorem I^A_zz = I^C_zz + m(L/2)² = 1/3 mL²

M_A = ( L/2 cosθ i + L/2 sinθ j) x (-W) j = - (LW/2) cosθ k = [1/3 mL²] α k

So α = - (3/2) (g/L) cos θ Result: α = - (3/2) (g/L) cos θ k

α = dω/dt = (dω/dθ) (dθ/dt) = ω dω/dθ = - (3/2) (g/L) cos θ

Integration gives ½ ω² = - (3/2) (g/L) sin θ + C

Use the initial condition when θ = 90^o ω = 0. C = (3/2) (g/L)

The result is ω² = 3g/L[1 - sin θ ] So ω = - √ 3g/L[1 – sin θ] k rad/sec

Strategy: To find the pin reactions at A apply Euler’s First Law.

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