Explanation
In equilibrium, when sliding stops
$$N=mg\cos\theta…(1)$$
And $$f=mg\sin\theta=\mu N…(1)$$
Dividing $$(2)$$ by$$(1)\\ \Rightarrow \mu=\tan\theta=0.75=\cfrac{3}{4}\\ \Rightarrow \theta=37°$$
For downward direction
Force equation in perpendicular direction is given as
$$ N=mg\cos \theta $$
$$ {{f}^{'}}=\mu mg\cos \theta $$
Force equation parallel to incline is
$$ f+{{f}^{'}}=mg\sin \theta $$
$$ f+\mu mg\cos \theta =mg\sin \theta $$
$$ f=mg\sin \theta -\mu mg\cos \theta ..............(1) $$
For upward direction
$$ {{f}^{'}}=\mu N $$
$$ f'=\mu mg\cos \theta $$
$$ 2f=mg\sin \theta +f' $$
$$ 2f=mg\sin \theta +\mu mg\cos \theta ............(2) $$
From equation (1),equation (2) becomes
$$ 2mg\sin \theta -2\mu mg\cos \theta =mg\sin \theta +\mu mg\cos \theta $$
$$ mg\sin \theta =3\mu mg\cos \theta $$
$$ \tan \theta =3\mu $$
$$ \mu =\frac{\tan \theta }{3} $$
$${ T }_{ max }=\cfrac { m{ v }_{ max }^{ 2 } }{ r } +mg$$
where $${ T }_{ max }$$ is maximum tension
$$m$$ is mass
$$x$$ is radius
and $$g$$ is acceleration due to gravity
$${ T }_{ max }=m{ { w }_{ max }^{ 2 } }r+mg$$
where $$w=\cfrac { v }{ r } $$ is angular velocity
$$\cfrac { { T }_{ max } }{ m } =m{ { w }_{ max }^{ 2 } }r+g$$
$$\cfrac { 30 }{ 0.5 } ={ { w }_{ max }^{ 2 } }2+10\\ \Rightarrow { w }_{ max }^{ 2 }=25\\ \Rightarrow { w }_{ max }=5 rad/s$$
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