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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 -

1. (1/4) ((a + b)/(2a + 3b)) L
2. (3/2) ((2a + b)/(3a + b)) L
3. (3/2) ((a + b)/(2a + b)) L
4. (3/4) ((2a + b)/(3a + b)) L

Correct answer: (3/4) ((2a + b)/(3a + b)) L

Solution

With rho=a+b(x/L)^2: total mass = L(3a+b)/3 and integral of x*rho = L^2(2a+b)/4. So x_cm = [L^2(2a+b)/4]/[L(3a+b)/3] = (3/4)*((2a+b)/(3a+b))*L.

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