A rod of length 3 m is placed along the positive x-axis with one end at the origin. Its linear mass density varies as lambda(x) = 2 + x kg/m. Find the position of the centre of mass of the rod from the origin.

7/3 m
12/7 m
10/7 m
9/7 m

Correct answer: 12/7 m

Solution

x_cm = (integral₀³ x*(2+x) dx) / (integral₀³ (2+x) dx). Numerator = integral₀³ (2x + x²) dx = [x² + x³/3]₀³ = 9 + 9 = 18. Denominator = [2x + x²/2]₀³ = 6 + 9/2 = 12/2 + 9/2 = 21/2. x_cm = 18 / (21/2) = 36/21 = 12/7 m.

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