A sphere of mass M and radius R is released from the top of an inclined plane of inclination θ . The minimum coefficient of friction between the plane and the sphere so that it rolls down the plane without sliding is given by

Question

A sphere of mass M and radius R is released from the top of an inclined plane of inclination θ . The minimum coefficient of friction between the plane and the sphere so that it rolls down the plane without sliding is given by (A) μ =tan θ  (B) μ =(2/3)tan θ  (C) μ =(2/5)tan θ  (D) μ =(2/7)tan θ

Tardigrade Admin · Accepted Answer

μ ≥ (tan θ /1 + (R2)K2) ⇒ μ min=(tan θ /1 + (R2)K2)

Q. A sphere of mass $M$ and radius $R$ is released from the top of an inclined plane of inclination $θ .$ The minimum coefficient of friction between the plane and the sphere so that it rolls down the plane without sliding is given by

$μ = t an θ$

$μ = \frac{2}{3} t an θ$

$μ = \frac{2}{5} t an θ$

$μ = \frac{2}{7} t an θ$

$μ \geq \frac{t an θ}{1 + \frac{R ^{2}}{K ^{2}}}$
$\Rightarrow μ_{min} = \frac{t an θ}{1 + \frac{R ^{2}}{K ^{2}}}$

Q. A sphere of mass M and radius R is released from the top of an inclined plane of inclination θ. The minimum coefficient of friction between the plane and the sphere so that it rolls down the plane without sliding is given by

μ=tanθ

μ=32​tanθ

μ=52​tanθ

μ=72​tanθ