Ball, Don't Fall! - Part 2

Using energy conservation between $\theta=0$ and $\theta=30$ In ICR frame

$mgr(1-\cos { \theta } )\quad =\quad \cfrac { 1 }{ 2 } { I }_{ bottom }{ \omega }^{ 2 }\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \cfrac { 7 }{ 10 } m{ r }^{ 2 }{ \omega }^{ 2 }\\ \\ r{ \omega }^{ 2 }\quad =\quad \cfrac { 10g(1-\cos { \theta } ) }{ 7 } \quad \quad .\quad .\quad .\quad .\quad .\quad .\quad .\quad \quad (1)$ .

Now writing equation in Common normal direction of sphere

$mg\cos { \theta } \quad -\quad N\quad =\quad m{ r }{ \omega }^{ 2 }\\ \\ f\quad =\quad \mu N\quad =\quad \mu g(mg\cos { \theta } \quad -\quad m{ r }{ \omega }^{ 2 })\quad \quad \quad .....\quad (2)$ .

Now using torque equation about COM .

$\tau \quad =\quad fr\quad =\cfrac { 2 }{ 5 } m{ r }^{ 2 }\alpha \quad \quad \quad .......\quad (3)\\ \\ r\alpha \quad =\quad \cfrac { 5\mu }{ 2 } (g\cos { \theta } \quad -\quad { r }{ \omega }^{ 2 })\quad ......(4)\\$ .

Now writing equation along common tangent :

$mg\sin { \theta } \quad -\quad f\quad =\quad m{ a }_{ com }\quad \quad ......\quad (5)\\ \\ \because \quad { a }_{ com }\quad =\quad r\alpha \quad \quad \quad \quad ........(6)\\$ .

Now By putting $\theta=30$ . and using all equations we get

$\boxed { \mu \quad =\quad \frac { 2 }{ 17\sqrt { 3 } \quad -\quad 20 } }$ .

during motion there are two components of acceleration,tangential and radial

mg $\sin \theta$ - f = ma ...(1)

mg $\cos \theta$ - N = $m \cdot \frac {v^2}{(R+r)}$ ...(2)

also,by applying torque equation about center of ball

fr= $I \cdot \frac {a}{(r)}$ ...(3)

Also,

a= $v \cdot \frac {dv}{ds}$

where ds= $(R+r) \cdot d \theta$

Therefore,

$v \cdot dv$ = $a(R+r) \cdot d \theta$ ....(4)

from (3) and (1),

f= $\frac {mg \sin \theta}{(1+ \frac {mr^2}{I})}$ ....(5)

Now,from (3),(4) and (5),

$v \cdot dv$ = $\frac {mg \sin \theta}{(I+mr^2)} \cdot r^2(R+r) \cdot d \theta$

we then integrate this and then substitute $v^2$ in (2),from this we get N

Then for limiting condition, we use,

$f= \mu N$ ....(6)

we substitute the value of N and f in (6) to get coefficient of friction

we then substitute all the values to get the answer.