The linear mass density of a thin rod AB of l | vedclass

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(D) The moment of inertia $I$ is given by $I = \int r^{2} dm = \int_{0}^{L} x^{2} \lambda(x) dx$.Substituting $\lambda(x) = \lambda_{0}(1 + \frac{x}{L

Vedclass · Accepted Answer

$0.388$

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(D) The moment of inertia $I$ is given by $I = \int r^{2} dm = \int_{0}^{L} x^{2} \lambda(x) dx$.Substituting $\lambda(x) = \lambda_{0}(1 + \frac{x}{L})$:$I = \int_{0}^{L} x^{2} \lambda_{0}(1 + \frac{x}{L}) dx = \lambda_{0} \int_{0}^{L} (x^{2} + \frac{x^{3}}{L}) dx$$I = \lambda_{0} [\frac{x^{3}}{3} + \frac{x^{4}}{4L}]_{0}^{L} = \lambda_{0} (\frac{L^{3}}{3} + \frac{L^{3}}{4}) = \frac{7}{12} \lambda_{0} L^{3} \quad ...(i)$Now,calculate the total mass $M$:$M = \int_{0}^{L} \lambda(x) dx = \int_{0}^{L} \lambda_{0}(1 + \frac{x}{L}) dx = \lambda_{0} [x + \frac{x^{2}}{2L}]_{0}^{L} = \lambda_{0} (L + \frac{L}{2}) = \frac{3}{2} \lambda_{0} L$Thus,$\lambda_{0} L = \frac{2}{3} M \quad ...(ii)$Substituting $(ii)$ into $(i)$:$I = \frac{7}{12} (\lambda_{0} L) L^{2} = \frac{7}{12} (\frac{2}{3} M) L^{2} = \frac{14}{36} M L^{2} = \frac{7}{18} M L^{2}$$I \approx 0.388 M L^{2}$.

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