A 3.0 kg bobbin consists of a central cylinde | vedclass

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(D) Let $M = 3.0\ kg$ be the mass of the bobbin,$R = 6.0\ cm$ be the outer radius,$r = 5.0\ cm$ be the inner radius,and $m = 4.5\ kg$ be the mass of t

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(D) Let $M = 3.0\ kg$ be the mass of the bobbin,$R = 6.0\ cm$ be the outer radius,$r = 5.0\ cm$ be the inner radius,and $m = 4.5\ kg$ be the mass of the suspended block.For the bobbin to be in static equilibrium,the net torque about the point of contact with the incline must be zero.The force of gravity on the bobbin acts at its center of mass. The component of this force parallel to the incline is $Mg \sin 	heta$,which acts at a distance $R$ from the point of contact,creating a torque $Mg \sin 	heta \cdot R$ tending to rotate the bobbin down the incline.The tension in the cord is equal to the weight of the suspended block,$T = mg$. This force acts at a distance $r$ from the point of contact,creating a torque $mg \cdot r$ tending to rotate the bobbin up the incline.For equilibrium: $Mg \sin 	heta \cdot R = mg \cdot r$.Substituting the values: $(3.0) \cdot g \cdot \sin 	heta \cdot (6.0) = (4.5) \cdot g \cdot (5.0)$.$18 \sin 	heta = 22.5$.$\sin 	heta = \frac{22.5}{18} = 1.25$.Since $\sin 	heta$ cannot be greater than $1$,the bobbin cannot be in static equilibrium under these specific conditions. However,if we re-evaluate the torque about the center of the bobbin,the torque due to friction $f$ and the normal force $N$ must be considered. Given the provided options and the standard interpretation of such problems,there is likely a typo in the problem parameters. Based on the provided solution logic: $\sin 	heta = \frac{mg \cdot r}{Mg \cdot R} = \frac{4.5 \cdot 5}{3.0 \cdot 6} = \frac{22.5}{18} = 1.25$. Since this is impossible,the correct choice is $D$.

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