What is the minimum coefficient of friction f | vedclass

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Question

(A) For a body of mass $m$ and radius $r$ rolling down an inclined plane of angle $	heta$,the acceleration $a$ is given by $a = \frac{mg \sin 	heta}

Vedclass · Accepted Answer

$\frac{2}{7} 	an 	heta$

Vedclass · Answer

(A) For a body of mass $m$ and radius $r$ rolling down an inclined plane of angle $	heta$,the acceleration $a$ is given by $a = \frac{mg \sin 	heta}{m + \frac{I}{r^2}}$.The frictional force $f$ required for pure rolling is $f = ma_{cm} \cdot \frac{I}{mr^2} = \frac{mg \sin 	heta}{1 + \frac{mr^2}{I}}$.For a solid sphere,the moment of inertia $I = \frac{2}{5} mr^2$.Substituting $I$ into the friction equation: $f = \frac{mg \sin 	heta}{1 + \frac{mr^2}{(2/5)mr^2}} = \frac{mg \sin 	heta}{1 + 2.5} = \frac{mg \sin 	heta}{3.5} = \frac{2}{7} mg \sin 	heta$.Since $f \le \mu N$ and $N = mg \cos 	heta$,we have $\mu \ge \frac{f}{N} = \frac{(2/7) mg \sin 	heta}{mg \cos 	heta} = \frac{2}{7} 	an 	heta$.Thus,the minimum coefficient of friction is $\mu_{min} = \frac{2}{7} 	an 	heta$.

What is the minimum coefficient of friction for a solid sphere to roll without slipping on an inclined plane of inclination $\theta$?

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