Question

a solid cylinder is rolling without slipping on a plane having inclination theta and the coefficient of static friction.....​

Answer 1

Given:

A solid cylinder is rolling without slipping on a plane having inclination $\theta$ and coefficient of friction is $\mu$.

To find:

Relation between $\theta \:and\: \mu$.

Calculation:

Considering Translational Motion:

$a = g \sin( \theta) - \mu g \cos( \theta) \: \: \: ........(1)$

Considering Rotational Motion:

$\therefore \: \tau = I \times \alpha$

$= > \mu mg \cos( \theta) \times r = \dfrac{m {r}^{2} }{2} \times \dfrac{a}{r}$

$= > a = 2 \mu g \cos( \theta) \: \: \: \: ......(2)$

Comparing eq. (1) and (2):

$\therefore \: a = g \sin( \theta) - \mu g \cos( \theta)$

$= > \: 2 \mu g \cos( \theta) = g \sin( \theta) - \mu g \cos( \theta)$

$= > \: 3\mu g \cos( \theta) = g \sin( \theta)$

$= > \: 3\mu \cos( \theta) = \sin( \theta)$

$= > \tan( \theta) = 3 \mu$

So, we can say that for rolling motion:

$\therefore \: \mu \geqslant \dfrac{ \tan( \theta) }{3}$

$= > \: 3\mu \geqslant \tan( \theta)$

$= > \: \tan( \theta) \leqslant 3 \mu$

So, final answer is:

$\boxed{ \red{ \sf{ \large{ \: \tan( \theta) \leqslant 3 \mu}}}}$

Answer 2

$\huge\underline\frak{\fbox{Answer️}}$

⟹ tanθ ≤ 3μs

$\huge\rm\underline\purple{Explanation:-}$

Torque about point O To = fR

where f=μs•mgcos •θ

Using To = Iα..

=====> fR= 1/MR²α. (for rolling a=Rα)

For Cylinder not to slip,

Ma ≥ Mg sinθ − f,

Or 2f ≥ Mg sinθ − f ,

Or 3f ≥ Mg sinθ,

Or 3μs Mg cosθ ≥ Mg sinθ.

====>. tanθ ≤ 3μs.

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