Question

# A solid sphere is rolling down an inclined plane without slipping. If the inclined plane has inclination $$\theta$$ with the horizontal, then the coefficient of friction $$\mu$$ between the sphere and the inclined plane should be

A
μ2/7cotθ
B
μ2/7tanθ
C
μ2/7cosθ
D
μ4/7sinθ

Solution

## The correct option is B $$\mu \ge 2/7\tan { \theta }$$f is the frictional force is upward direction along inclined plane.$$\alpha$$ is the angular acceleration of the sphere.Moment of inertia of the sphere about centre of mass is $$I=\cfrac{2}{5}mR^2$$.N is the normal reactional force on the sphere &N $$=mgcos\theta$$$$f=\mu N=\mu mgcos\theta$$Torque about centre of mass sphere is$$I\alpha = f.R$$For downward motion let the acceleration along inclined plane $$\alpha$$ So, $$mgsin\theta -f=ma$$for pure rotation $$\alpha =\cfrac{a}{R}$$So, $$mgsin\theta -f=m\alpha R$$$$\implies mgsin\theta = f+\cfrac{5}{2}f$$$$\implies mgsin\theta =\cfrac{7}{2}f$$$$\implies \cfrac{7}{2}\mu mg cos\theta = mgsin\theta$$$$\implies \mu = \cfrac{2}{7}tan\theta$$This is the minimum coefficient of friction So, it should be $$\mu \geq \cfrac{2}{7}tan\theta$$Physics

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