A rod of length L has non-uniform linear mass
density given by ρ(x) = a + b( x/L )² , where a
and b are constants and 0 ≤ x ≤ L. The value
of x for the centre of mass of the rod is at :

Solution

Physics sir · 5 steps

Density changes along the rod, so integrate: total mass on the bottom, total moment x dm on top.

1. Set up. A small piece of length dx at position x has mass dm = ρ(x) dx. The centre of mass is the mass-weighted average of x over the whole rod.

   x_(cm) = (∫₀^L x ρ(x) dx)/(∫₀^L ρ(x) dx)

2. Total mass. ∫₀^L (a + (b x²)/L²)dx = aL + bL/3 = L(3a+b)/3.
3. Total moment. ∫₀^L (ax + (b x³)/L²)dx = aL²/2 + bL²/4 = L²(2a+b)/4.
4. Divide. x_(cm) = (L²(2a+b)/4)/(L(3a+b)/3) = 3/4((2a+b)/(3a+b))L. Check: with b = 0 the rod is uniform and this gives 3/4 × 2/3L = L/2, as it should.
5. So. The centre of mass is at 3/4((2a+b)/(3a+b))L, which is option (c).

The idea to keep. For a rod with varying density, x_(cm) = ∫ x dm / ∫ dm. Then put b = 0 (uniform rod): any correct answer must give L/2.

Answer from the NTA final answer key. Solution by PrepWiser.