Ball, Don't Fall!

The following diagram show's all the forces being applied on the sphere when it is making an angle $\theta$ with the horizontal.

Slipping

I will be taking direction along the normal as x-axis and direction anti parellel to friction as y-axis.

Till the moment the ball starts slipping we can assume the whole sphere to be rotating about the tip.

First thing we would be calculating is the angular velocity of the ball, it can be easy calculated by using work energy theorom.

Work done by gravity = Change in Kinetic energy.

$mgR(1-cos(\theta))=\frac{1}{2}{I}_{tip}{\omega}^{2}$

${I}_{tip}={I}_{com}+m{R}^{2}=\frac{7}{5}m{R}^{2}$

We get the expression of $\omega$ as :

$\omega = \sqrt{\frac{10g(1-cos(\theta))}{7R}}$

Now we will find angular acceleration of the ball :

$mgsin(\theta)R=\frac{7}{5}m{R}^{2}\alpha$

$\Rightarrow \alpha = \frac{5gsin(\theta)}{7R}$

$\Rightarrow {a}_{y}=\alpha R=\frac{5gsin(\theta)}{7}$

There is no acceleration of ball in x-axis. Hence writing equation of equilibrium in x-direction we get :

$mgcos(\theta)=N+m{\omega}^{2}R$

Putting the expression of $\omega$ we get :

$N=\frac{mg(17cos(\theta)-10)}{7}$

Also using force equation in y direction we get :

$mgsin(\theta)-f=m{a}_{y}$

Putting the expression of ${a}_{y}$ we get :

$f=\frac{2mgsin(\theta)}{7}$

At the time of slipping we have :

$\eta N = f$

Where $\eta$ is the co-efficient of friciton :

Putting the expression of $N,f$ and $\eta=0.5$ we get :

$\large \frac{mg(17cos(\theta)-10)}{14}=\frac{2mgsin(\theta)}{7}$

Solving $\theta$ comes out to be :

$\theta = { tan }^{ -1 }(\frac { \sqrt { 205 } }{ 10 } )-{ tan }^{ -1 }(\frac { 4 }{ 17 } )$

The situation

figure is in rotating frame

Slipping begins when the acceleration of the point of contact of the ball with the edge is no longer at rest,,

That is the superposition of rotational and translational accelerations about the com for the contact point is no longer 0

or mathematically $a\ge \alpha R$

now in the limiting case,, friction is just able to support pure rolling about edge or it is maximum,

Also we know that only friction provides the torque about com,,

and that balls centre of mass only moves perpendicular to the direction of normal reaction,, and the only forces along it are the component of gravity along it (sin component) and friction acting opposite to it giving the ball acceleration along it.

thus we have

$gsin\theta \quad -\quad \frac { f }{ m } =\quad a\\ \\ fR=I\alpha \\ \\ f=\mu N=\mu (mgcos\theta -m{ w }^{ 2 }r)$

and finally through energy conservation we have

$2mgR(1-cos\theta )={ I }_{ about\quad edge }{ w }^{ 2 }$

Solving all these we get two values for the angle as

$\theta =41.83(approx)\quad and\quad \theta =68.344$

the latter is rejected obviously ( can be shown from the fact that had the ball not slipped at 41.83 degrees it would have lost contact at somewhere near 53 degrees, and so theres no way it can slip at an angle greater than that)

thus we get answer

Mine looks cumbersome.