A sphere rolls without slipping on an inclined plane of inclination - Then the minimum value of coefficient of friction so that the sphere rolls with out slipping is

The correct option is A μ_min = tan θ/ 1 + R^2/K^2 As , moment of inertia of solid shere = nbsp; 2/5 M nbsp; R^2 nbsp; ⇒ μ_min = nbsp;tan θ/1 + 5/2 = ...

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Byju's Answer

Standard XII

Physics

Friction General Understanding

A sphere roll...

Question

A sphere rolls without slipping on an inclined plane of inclination $\theta$ . Then the minimum value of coefficient of friction so that the sphere rolls with out slipping is

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Solution

The correct option is A

$μ_{m i n}$ = $\frac{t a n θ}{1 + \frac{R^{2}}{K^{2}}}$ As , moment of inertia of solid shere = $\frac{2}{5}$ M $R^{2}$

$\Rightarrow$ $μ_{m i n}$ = $\frac{t a n θ}{1 + \frac{5}{2}}$ = $\frac{2}{7}$ tan $θ$

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A sphere rolls without slipping on an inclined plane of inclination . Then the minimum value of coefficient of friction so that the sphere rolls with out slipping is

The correct option is A μmin = tanθ1+R2K2 As , moment of inertia of solid shere = 25 M R2 ⇒ μmin = tanθ1+52 = 27 tan θ

A sphere rolls without slipping on an inclined plane of inclination $\theta$ . Then the minimum value of coefficient of friction so that the sphere rolls with out slipping is

The correct option is A

$μ_{m i n}$ = $\frac{t a n θ}{1 + \frac{R^{2}}{K^{2}}}$ As , moment of inertia of solid shere = $\frac{2}{5}$ M $R^{2}$

$\Rightarrow$ $μ_{m i n}$ = $\frac{t a n θ}{1 + \frac{5}{2}}$ = $\frac{2}{7}$ tan $θ$