The uniform slender rod has a mass m.

if it is released from rest when theta = 0 degree, determine the magnitude of the reactive force exerted on it by pin b when theta = 90 degree.

3/2mg

Explanation:

Express the moment equation about point B

MB = (M K)B

-mg cosθ (L/6) = m[α(L/6)](L/6) – (1/12mL^2 )α

α = 3g/2L cosθ

express the force equation along n and t axes.

Ft = m (aG)t

mg cosθ – Bt = m [(3g/2L cos) (L/6)]

Bt = ¾ mg cosθ

Fn = m (aG)n

Bn -mgsinθ = m[ω^2 (L/6)]

Bn =1/6 mω^2 L + mgsinθ

Calculate the angular velocity of the rod

ω = √(3g/L sinθ)

when θ = 90°, calculate the values of Bt and Bn

Bt =3/4 mg cos90°

= 0

Bn =1/6m (3g/L)(L) + mg sin (9o°)

= 3/2mg

Hence, the reactive force at A is,

FA = √(02 +(3/2mg)^2

= 3/2 mg

The magnitude of the reactive force exerted on it by pin B when θ = 90° is 3/2mg

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