A thin rod lies along x from 0 to L = 1.6 m. Its linear density is λ(x) = λ₀(1 + ax/L), with λ₀ = 0.32 kg m⁻¹ and dimensionless a = 1.1. (a) Integrate to find its mass and centre-of-mass coordinate. (b) Where should a single vertical support be placed to balance the horizontal rod in a uniform gravitational field? Explain why L/2 is not correct.

Question

Accepted Answer

Mass 0.7936 kg; centre x_cm = 0.8946 m from x = 0. Place the support at x_cm, to the right of L/2. A short element dx has mass dm = λ(x)dx. Because density varies, replacing the rod by a uniform rod would incorrectly give equal masses to equal lengths. Integrate M = ∫₀ᴸ λ₀(1 + ax/L) dx = λ₀L(1 + a/2) = 0.7936 kg. The integral has units of mass because linear density is multiplied by length. The first moment is ∫₀ᴸ x dm = λ₀L²(1/2 + a/3). Divide this by M to obtain x_cm = L(1/2 + a/3)/(1 + a/2). Substitution gives x_cm = 0.8946 m. A single support at this position lies under the line of action of the total weight, so the gravitational torque about it vanishes. The centre is beyond L/2 because the right-hand side is denser, yet it remains within the rod. Setting a = 0 recovers L/2, a useful check on the nonuniform-density calculation.

Centre of mass of a nonuniform rod · set 6

What a complete response should do

How did this one feel?