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(D) Given: Radius of sphere $= R$, mass of sphere $= m$, and angle of inclination $	heta = 45^{\circ}$.For a solid sphere rolling without slipping on

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$\frac{1}{7}$

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(D) Given: Radius of sphere $= R$, mass of sphere $= m$, and angle of inclination $	heta = 45^{\circ}$.For a solid sphere rolling without slipping on an inclined plane, the linear acceleration is given by:$a = \frac{g \sin 	heta}{1 + \frac{I}{m R^2}} = \frac{g \sin 	heta}{1 + \frac{2/5 m R^2}{m R^2}} = \frac{5}{7} g \sin 	heta$Applying Newton's second law for linear motion along the incline:$m g \sin 	heta - f = m a$$f = m g \sin 	heta - m \left( \frac{5}{7} g \sin 	heta ight) = \frac{2}{7} m g \sin 	heta$The normal force is $N = m g \cos 	heta$.The coefficient of static friction $\mu_s$ must satisfy:$\mu_s \geq \frac{f}{N} = \frac{\frac{2}{7} m g \sin 	heta}{m g \cos 	heta} = \frac{2}{7} 	an 	heta$Since $	heta = 45^{\circ}$, $	an 45^{\circ} = 1$, so $\mu_s \geq \frac{2}{7} \approx 0.2857$.Comparing the options:$(A)$ $\frac{3}{7} \approx 0.428 > 0.2857$$(B)$ $\frac{1}{2} = 0.5 > 0.2857$$(C)$ $\frac{5}{8} = 0.625 > 0.2857$$(D)$ $\frac{1}{7} \approx 0.1428 Since $\frac{1}{7}$ is less than $\frac{2}{7}$, the sphere will slip. Thus, option $(D)$ is the incorrect value for rolling without slipping.

$A$ uniform sphere of radius $R$ and mass $m$ is placed on an inclined plane which makes an angle $45^{\circ}$ to the horizontal. For which of the following values of the coefficient of friction does the sphere roll without slipping? Select the incorrect option.

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