The distance of centre of mass from end A of a one dimensional rod (AB) having mass density ρ=ρ₀(1-x²/L²) kg / m and length L (in meter) is (3 L)/α m. The value of α is ___________. (where x is the distance from end A)

Solution

Physics sir · 4 steps

The density changes along the rod, so cut it into thin slices and weight each one by its mass.

1. Set up slices. Take a thin slice of width dx at distance x from A. Its mass is dm = ρ dx = ρ₀(1-x²/L²)dx, and the centre of mass is the mass-weighted average of x.

   x_(cm) = (∫₀^L x ρ₀(1-x²/L²)dx)/(∫₀^L ρ₀(1-x²/L²)dx)

2. Total mass. The denominator is ρ₀(L - L³/3L²) = ρ₀(L - L/3) = (2ρ₀ L)/3.
3. First moment. The numerator is ρ₀(L²/2 - L⁴/4L²) = ρ₀(L²/2 - L²/4) = (ρ₀ L²)/4.
4. So. x_(cm) = (ρ₀ L²/4)/(2ρ₀ L/3) = 3L/8. Comparing with 3L/α gives α = 8.

The common slip: 12. Taking the total mass as ρ₀ L instead of (2ρ₀ L)/3 gives x_(cm) = L/4 = 3L/12, so α = 12.

The idea to keep. For a rod with varying density, x_(cm) = ∫ x dm / ∫ dm. Always find the true total mass by integrating; never use ρ₀ L unless the density is uniform.

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