A uniform solid sphere of mass m and radius r | vedclass

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(B) For a body rolling without slipping on an inclined plane,the condition for no slipping is given by $\mu \geq \frac{k^2}{r^2 + k^2} 	an 	heta$.Fo

Vedclass · Accepted Answer

$\frac{2}{7}$

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(B) For a body rolling without slipping on an inclined plane,the condition for no slipping is given by $\mu \geq \frac{k^2}{r^2 + k^2} 	an 	heta$.For a uniform solid sphere,the radius of gyration $k$ is given by $k^2 = \frac{2}{5}r^2$.Substituting this into the formula,we get $\mu \geq \frac{\frac{2}{5}r^2}{r^2 + \frac{2}{5}r^2} 	an 	heta$.$\mu \geq \frac{\frac{2}{5}}{\frac{7}{5}} 	an 	heta = \frac{2}{7} 	an 	heta$.Given the angle of inclination $	heta = 45^{\circ}$,we have $	an 45^{\circ} = 1$.Therefore,$\mu \geq \frac{2}{7} 	imes 1 = \frac{2}{7}$.The minimum frictional coefficient is $\frac{2}{7}$.

$A$ uniform solid sphere of mass $m$ and radius $r$ rolls without slipping down an inclined plane,inclined at an angle $45^{\circ}$ to the horizontal. Find the magnitude of the minimum frictional coefficient at which slipping is absent.

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