A non-uniform rod OA of linear mass density l | vedclass

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(C) The center of mass $X_{cm}$ is given by:$X_{cm} = \frac{\int x dm}{\int dm} = \frac{\int_{0}^{L} x (\lambda_0 x) dx}{\int_{0}^{L} (\lambda_0 x) dx

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$\frac{4g}{3L}$

Vedclass · Answer

(C) The center of mass $X_{cm}$ is given by:$X_{cm} = \frac{\int x dm}{\int dm} = \frac{\int_{0}^{L} x (\lambda_0 x) dx}{\int_{0}^{L} (\lambda_0 x) dx} = \frac{\lambda_0 [x^3/3]_0^L}{\lambda_0 [x^2/2]_0^L} = \frac{L^3/3}{L^2/2} = \frac{2L}{3}$.The total mass $M$ of the rod is:$M = \int_{0}^{L} \lambda_0 x dx = \frac{\lambda_0 L^2}{2}$.The moment of inertia $I$ about the hinge $O$ is:$I = \int x^2 dm = \int_{0}^{L} x^2 (\lambda_0 x) dx = \lambda_0 \int_{0}^{L} x^3 dx = \frac{\lambda_0 L^4}{4}$.Just after the string is cut,the torque $	au$ about the hinge $O$ is due to gravity acting at the center of mass:$	au = Mg 	imes X_{cm} = \left( \frac{\lambda_0 L^2}{2} ight) g \left( \frac{2L}{3} ight) = \frac{\lambda_0 g L^3}{3}$.Using the relation $	au = I \alpha$:$\frac{\lambda_0 g L^3}{3} = \left( \frac{\lambda_0 L^4}{4} ight) \alpha$.Solving for $\alpha$:$\alpha = \frac{\lambda_0 g L^3}{3} 	imes \frac{4}{\lambda_0 L^4} = \frac{4g}{3L}$.

$A$ non-uniform rod $OA$ of linear mass density $\lambda = \lambda_0 x$ (where $\lambda_0$ is a constant) is suspended from the ceiling with a hinge joint at $O$ and a light string as shown in the figure. Find the angular acceleration of the rod just after the string is cut.

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