A non-uniform cylinder of mass m,length l,and | vedclass

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(B) The cylinder is in a horizontal position. The buoyant force $F_b$ acts at the geometric center of the cylinder,while the weight $mg$ acts at the c

Vedclass · Accepted Answer

$\frac{\pi \rho g l^2 r^2}{4I}$

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(B) The cylinder is in a horizontal position. The buoyant force $F_b$ acts at the geometric center of the cylinder,while the weight $mg$ acts at the center of mass $(CM)$.The volume of the cylinder is $V = \pi r^2 l$.The buoyant force is $F_b = V ho g = \pi r^2 l ho g$.The distance between the geometric center and the center of mass is $d = l/4$.The torque $	au$ about the center of mass is due to the buoyant force:$	au = F_b \cdot d = (\pi r^2 l ho g) \cdot \frac{l}{4} = \frac{\pi ho g l^2 r^2}{4}$.Using the rotational analog of Newton's second law,$	au = I \alpha$,where $I$ is the moment of inertia about the center of mass:$\frac{\pi ho g l^2 r^2}{4} = I \alpha$$\alpha = \frac{\pi ho g l^2 r^2}{4I}$.

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