A solid cylinder and a solid sphere of same m | vedclass

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(B) For an object of mass $M$,radius $R$,and moment of inertia $I = kMR^2$ rolling down an inclined plane of angle $	heta$,the force of friction $f$

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(B) For an object of mass $M$,radius $R$,and moment of inertia $I = kMR^2$ rolling down an inclined plane of angle $	heta$,the force of friction $f$ is given by the formula:$f = \frac{Mg \sin	heta}{1 + \frac{MR^2}{I}} = \frac{Mg \sin	heta}{1 + \frac{1}{k}}$For a solid cylinder,$I = \frac{1}{2}MR^2$,so $k = \frac{1}{2}$. Thus,$f_{cylinder} = \frac{Mg \sin	heta}{1 + 2} = \frac{1}{3}Mg \sin	heta$.For a solid sphere,$I = \frac{2}{5}MR^2$,so $k = \frac{2}{5}$. Thus,$f_{sphere} = \frac{Mg \sin	heta}{1 + \frac{5}{2}} = \frac{Mg \sin	heta}{3.5} = \frac{2}{7}Mg \sin	heta$.Comparing the two,$\frac{1}{3} \approx 0.333$ and $\frac{2}{7} \approx 0.285$.Since $\frac{1}{3} > \frac{2}{7}$,the force of friction is more for the cylinder.

$A$ solid cylinder and a solid sphere of same mass and radius roll without slipping on a rough inclined plane. The force of friction is ..........

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