A slender rod of mass M and length L is hinge | vedclass

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(C) Let $S$ be the area of the cross-section of the rod.The linear mass density $\lambda(x)$ varies linearly from $\rho_0$ at $x=0$ to $2\rho_0$ at $x

Vedclass · Accepted Answer

$\frac{7ML^2}{18}$

Vedclass · Answer

(C) Let $S$ be the area of the cross-section of the rod.The linear mass density $\lambda(x)$ varies linearly from $\rho_0$ at $x=0$ to $2\rho_0$ at $x=L$.Thus,$\lambda(x) = S \cdot \rho(x) = S \left( \rho_0 + \frac{2\rho_0 - \rho_0}{L} x \right) = S \left( \rho_0 + \frac{\rho_0}{L} x \right)$.The total mass $M$ is given by:$M = \int_{0}^{L} \lambda(x) dx = S \int_{0}^{L} \left( \rho_0 + \frac{\rho_0}{L} x \right) dx = S \rho_0 \left[ x + \frac{x^2}{2L} \right]_{0}^{L} = S \rho_0 \left( L + \frac{L}{2} \right) = \frac{3}{2} S \rho_0 L$.Therefore,$S \rho_0 = \frac{2M}{3L}$.The moment of inertia $I$ about the hinge is:$I = \int_{0}^{L} x^2 dm = \int_{0}^{L} x^2 \lambda(x) dx = S \int_{0}^{L} x^2 \left( \rho_0 + \frac{\rho_0}{L} x \right) dx$.$I = S \rho_0 \int_{0}^{L} \left( x^2 + \frac{x^3}{L} \right) dx = S \rho_0 \left[ \frac{x^3}{3} + \frac{x^4}{4L} \right]_{0}^{L} = S \rho_0 \left( \frac{L^3}{3} + \frac{L^3}{4} \right) = S \rho_0 \left( \frac{7L^3}{12} \right)$.Substituting $S \rho_0 = \frac{2M}{3L}$:$I = \left( \frac{2M}{3L} \right) \left( \frac{7L^3}{12} \right) = \frac{14ML^2}{36} = \frac{7ML^2}{18}$.

$A$ slender rod of mass $M$ and length $L$ is hinged at one end to swing freely in a vertical plane. Its density is non-uniform and varies linearly from the hinged end to the free end,doubling its value. The moment of inertia of the rod about the rotation axis passing through the hinge point is:

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