Mechanics + Magnetism = Mechatism 3

I upload full solution later but you can see this question solution and also This Note

This question is combination of these two question's.

Update For calculating maximum velocity :

Here is my method when friction is present between only cylinders. By using concept of Instantaneous centre of rotation [ ICR ]

Because This system is an equivalent Rigid Body. Since There is no relative motion between point of contact of cylinder.

Figure

The whole system is assumed to be purely rotated about "O"

Then

so velocity of any point P of this combined system in ICR ( O ) frame is given By

${ V }_{ p/o }\quad =\quad { r }_{ p/0 }\quad \times \quad { \omega }_{ 0 }$ .

${ V }_{ 1 }\quad =\quad (2R\cos { \theta } ){ \omega }_{ 0 }\\ \\ { V }_{ 2 }\quad =\quad (2R\sin { \theta } ){ \omega }_{ 0 }\\ \\ \Rightarrow \quad { \omega }_{ 0 }\quad =\quad \cfrac { \sqrt { { V }_{ 1 }^{ 2 }\quad +\quad { V }_{ 2 }^{ 2 } } }{ 2R }$ .

Now using energy conservation of system in ICR frame we get

$2gR(1-\cos { \theta } )\quad =\quad \frac { 1 }{ 2 } { I }_{ o }{ { \omega }_{ 0 } }^{ 2 }\quad \quad \quad \longrightarrow (1)\\ \\ { I }_{ o }\quad =\quad { mR }^{ 2 }\quad +{ \quad mR }^{ 2 }\quad +\quad { 4mR }^{ 2 }\cos ^{ 2 }{ \theta } \quad +{ \quad 4mR }^{ 2 }\sin ^{ 2 }{ \theta } \\ \\ { I }_{ o }\quad =\quad 6{ mR }^{ 2 }\\ \\ \\ \Rightarrow \quad { \omega }_{ 0 }\quad =\quad \sqrt { \cfrac { 2g }{ 3R } (1-\cos { \theta } ) } \quad \longrightarrow \quad (2)\\ \\ \\ { V }_{ 1 }^{ 2 }\quad =\quad \cfrac { 8gR }{ 3 } (1-\cos { \theta } )\cos ^{ 2 }{ \theta } \quad \longrightarrow \quad (3)\\ \\ \\ { V }_{ 2 }^{ 2 }\quad =\quad \cfrac { 8gR }{ 3 } (1-\cos { \theta } )\sin ^{ 2 }{ \theta } \quad \longrightarrow \quad (4)$ .

Now using maxima concept on velocity of cylinder - 1 ( Lower Cylinder )

$\cfrac { d({ V }_{ 1 }^{ 2 }) }{ d\theta } \quad =\quad 0\\ \\ \boxed { \theta \quad =\quad \cos ^{ -1 }{ \cfrac { 2 }{ 3 } } } \\ \\ Or\\ \quad \quad \\ \quad \quad \quad \quad \theta \quad =\quad 0\quad \quad \quad \times \quad (rejeccted\quad )\\ \quad \quad \quad \quad \quad \\ Or\\ \quad \quad \quad \quad \theta \quad =\quad \cfrac { \pi }{ 2 } \quad \quad \times \quad (\quad rejected\quad )$ .

So again cylinder will loose contact at the same hight as that of in 1st Part of this question ( when no friction ) I rejected other values of angle because velocity of lower cylinder never be zero. And Now Further Motion is independent

So we get maximum velocity of lower block

$\boxed { { V }_{ 1,max }\quad =\quad \sqrt { \cfrac { 32gR }{ 81 } } }$ .

Now for Calculating Motional emf See similiar type This Question Solution Mechatism 2