Problem Statement
A uniform rod of mass m and length l rotates about one end in a horizontal plane with angular velocity ω. A particle of mass m₀ is placed at the midpoint and held by a pin. Find the force on the pin.
Given
Rod mass m, length l, ω. Particle mass m₀ at l/2. Horizontal rotation.
Concepts & Formulas
Centripetal force needed by particle: F_particle = m₀ω²(l/2). Centripetal force needed by rod element at r: dF = λω²r dr, total for rod = mω²l/2. Pin force = total centripetal needed.
Step-by-Step Solution
Step 1: For the particle at l/2: F_p = m₀ω²(l/2).
Step 2: For the rod from 0 to l: F_rod = ∫₀ˡ (m/l)ω²r dr = mω²l/2.
Step 3: The pin at midpoint must supply centripetal for both particle AND the outer half of the rod.
Step 4: Outer rod (l/2 to l): F_outer = (m/l)∫_{l/2}^{l} ω²r dr = (m/l)ω²(3l²/8) = 3mω²l/8.
Step 5: F_pin = F_particle + F_outer_rod = m₀ω²l/2 + 3mω²l/8.
Worked Calculation
F_pin = ω²l(m₀/2 + 3m/8).
Boxed Answer
F_pin = ω²l(4m₀ + 3m)/8
Physical Interpretation
The pin must bear the centripetal load of the particle plus the outer half of the rod, less what the inner half can self-support through its own attachment at the pivot.
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