Is it a perfect blend of Mechanics and Calculus?

I have already asked you all in the question to try this before attempting this question.

So here I am not writing the equations which can be found by using FBD's and energy conservation. But I am uploading FBD's which are quite helpful.

Let $\theta$ be the angle with the vertical at which the ball is slipping and rotating together. But the CoM of the sphere still performs circular motion about the edge. Let the tangential velocity and the acceleration of the CoM at that particular angle $\theta$ be $v$ and $a$ respectively. Let $\omega$ be the angular velocity of the CoM of the ball about the edge.

So I have 2 equations-

$mgcos\theta -N=m\frac { { v }^{ 2 } }{ r }$ .............(1)

and $\\ mgsin\theta -\mu N=ma$

From eq(1) $mgcos\theta -m\frac { { v }^{ 2 } }{ r }= N$

or and $\\ mgsin\theta -\mu (mgcos\theta -m\frac { { v }^{ 2 } }{ r })=ma$

So $\\ mgsin\theta -\mu (mgcos\theta -m\frac { { v }^{ 2 } }{ r })=m\frac{dv}{dt}$

or $\\ mgsin\theta -\mu (mgcos\theta -m\frac { { v }^{ 2 } }{ r })=m\frac{dv}{d \theta} \omega$

But $\omega=\frac { v }{ r }$

NOTE--Here v is the tangential velocity of the CoM at angle $\theta$ and $\omega$ is the angular velocity of the CoM of the ball about the edge and the ball is doing circular motion about the edge. So we can use this equation $v=r \omega$

So $mgsin\theta -\mu (mgcos\theta -m\frac { { v }^{ 2 } }{ r } )=m\frac { vdv }{ rd\theta }$

Let ${ v }^{ 2 }=x$ So $2vdv=dx$

So $gsin\theta -\mu (gcos\theta -\frac { x }{ r } )=\frac { dx }{ 2rd\theta }$

$\frac { dx }{ 2rd\theta } =gsin\theta -\mu gcos\theta +\mu \frac { x }{ r }$

$\frac { dx }{ d\theta } -2\mu x+2rg(\mu cos\theta -sin\theta )=0$

Here $IF=e^{\int-d\theta}=e^{-\theta}$

$x{ e }^{ -\theta }=\int { { e }^{ -\theta } } \left\{ 2rg(sin\theta -\mu cos\theta ) \right\} d\theta$

$x{ e }^{ -\theta }=2rg\int { { e }^{ -\theta } } \left\{ (sin\theta -\mu cos\theta ) \right\} d\theta$

$x{ e }^{ -\theta }=2rg\int { { e }^{ -\theta } } \left\{ (sin\theta -\mu cos\theta ) \right\} d\theta$

$x{ e }^{ -\theta }=2rg\int { { e }^{ -\theta } } sin\theta \quad d\theta -\mu \int { { e }^{ -\theta }cos\theta } d \theta$

$x{ e }^{ -\theta }=2rg\left\{ -\frac { { e }^{ -\theta }\left( sin\theta +cos\theta \right) }{ 2 } -\frac { 1 }{ 2 } \frac { { e }^{ -\theta }\left( sin\theta -cos\theta \right) }{ 2 } \right\}$

$x{ e }^{ -\theta }=-rg{ e }^{ -\theta }\left\{ \frac { \left( sin\theta +cos\theta \right) }{ 1 } +\frac { \left( sin\theta -cos\theta \right) }{ 2 } \right\}$

$x{ e }^{ -\theta }=-rg{ e }^{ -\theta }\left\{ \frac { 3sin\theta +cos\theta }{ 2 } \right\}+C$

$x=-rg\left\{ \frac { 3sin\theta +cos\theta }{ 2 } \right\} +{ Ce }^{ \theta }$

We can find the value of C by finding x at angle 43.82 deg.

So

$\frac { 7 }{ 10r } { v }^{ 2 }=g(1-cos\theta )$ I got this from equation energy conservation-

Putting g=9.8 $\theta=41.82$ I got $v^{2}=3.56r$

Now the the solution differential equation was

$x=-rg\left\{ \frac { 3sin\theta +cos\theta }{ 2 } \right\} +{ Ce }^{ \theta }.$

where $x=v^{2}$

So at $\theta=41.82$ $v^{2}$ is $3.56r$

So putting values I got

$3.56r=-rg\left\{ \frac { 3sin(41.82)+cos(41.82) }{ 2 } \right\} +{ 2.72 }^{ 0.73 }C$

So the value of $C=8.20r$

So ${ v }^{ 2 }=-rg\left( \frac { (3sin\theta +cos\theta ) }{ 2 } \right) +8.20r{ e }^{ \theta }$

or $\frac { { v }^{ 2 } }{ r } =-g\left( \frac { (3sin\theta +cos\theta ) }{ 2 } \right) +8.20{ e }^{ \theta }$

but $\frac { { v }^{ 2 } }{ r } =gcos\theta$ (got this relation from the equation for the moment at which ball looses contact with the cliff)

or $gcos\theta =-g\left( \frac { (3sin\theta +cos\theta ) }{ 2 } \right) +8.20{ e }^{ \theta }$

Now according to WolframAlpha the answer is $52.71$ .

Extremely poor solution Cos (theta)=10/17