Answer:
a) α = ¹²/₇ L/g
b) v = √(²⁷/₁₄ gL)
Explanation:
The net torque on the rod is equal to its moment of inertia about its pivot times the angular acceleration. The torque is equal to the force of gravity times the distance between the center and the pivot. Energy is conserved, so as the rod falls, the potential energy is converted to rotational kinetic energy.
a) Draw a free body diagram. There are two forces on the rod, weight force mg pulling down at the center of the rod (which is ¼ L from the pivot), and reaction force R at pivot P. Sum of moments about the pivot P:
∑τ = Iα
mg (¼ L) = Iα
The moment of inertia of a rod about its center is I = ¹/₁₂ mL². Using parallel axis theorem, the moment of inertia offset from the center by ¼ L is:
I = ¹/₁₂ mL² + m (¼ L)²
I = ¹/₁₂ mL² + ¹/₁₆ mL²
I = ⁴/₄₈ mL² + ³/₄₈ mL²
I = ⁷/₄₈ mL²
Substituting:
mg (¼ L) = ⁷/₄₈ mL² α
g = ⁷/₁₂ L α
α = ¹²/₇ L/g
b) The initial potential energy equals the final kinetic rotational energy.
PE = RE
mgh = ½ Iω²
The height of the rod's center relative to the final position is h = ¼ L.
mg (¼ L) = ½ (⁷/₄₈ mL²) ω²
¼ mgL = ½ m (⁷/₄₈ ω²L²)
½ gL = ⁷/₄₈ ω²L²
g = ⁷/₂₄ ω²L
ω² = ²⁴/₇ g/L
ω = √(²⁴/₇ g/L)
The tangential velocity (v) of the bottom of the rod is equal to the angular velocity (ω) times the distance between the bottom of the rod and the pivot (¾ L).
v = ω × ¾ L
v = ¾ L √(²⁴/₇ g/L)
v = √(⁹/₁₆ × ²⁴/₇ gL)
v = √(²⁷/₁₄ gL)
