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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
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B
μ2/7tanθ
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C
μ2/7cosθ
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D
μ4/7sinθ
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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 
$$\mu \geq \cfrac{2}{7}tan\theta$$

957317_761668_ans_36d466e1f3ac4d7ebb590ea7ce29ea9d.png

Physics

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