Hemispherical bowl For "Thirsty Crows"

The interesting thing about this problem is that we dont have to locate the distance of contact points from the centre of mass OR consider the torques of the normal reactions simply because this is a hemisphere, we will soon see why so.

Let us consider the equllibrium of the hemisphere,

we have the following forces

1) Normal reaction of floor (lets call it N1)

2) Normal reaction of wall (N2)

3) Friction at ground (f1)

4) friction at wall (f2) (let us assume it is upward)

5) Gravity through COM located at R/2

Now we have using translation equllibirum condition

$mg={ N }_{ 1 }+\mu { N }_{ 2 }\\ \mu { N }_{ 1 }={ N }_{ 2 }\\$

solving them we get

${ N }_{ 2 }=\frac { mg\mu }{ 1+{ \mu }^{ 2 } } \\$

now the interesting part occurs,, let us consider the rotational equllibrium about the centre of hemisphere (not COM) (the centre of base instead)

and why ? because the torques due to normal reactions about that point is 0 because all lines normal to the surface of a sphere pass through the centre, and so we dont have to care about the points of contact and their location, also the frictional forces are perpendicular to the radii so it is easy to evaluate their torques

Now, about the COM, we have to consider torques due to gravity and friction

so we have

$mg\frac { R }{ 2 } sin\theta =\left( { f }_{ 1 }+{ f }_{ 2 } \right) R=(\mu )({ N }_{ 1 }+{ N }_{ 2 })R=\mu (1+\mu )({ N }_{ 2 })R=\frac { mg\mu (1+\mu ) }{ 1+{ \mu }^{ 2 } } R\\$

or $sin\theta =2\frac { \mu (1+\mu ) }{ 1+{ \mu }^{ 2 } } \\$

thus we get the answer

don't take torques about c.o.m rather centre of the sphere should be taken as the reference point since normal reactions will pass through that and will make it easy to do so ! $$ and as usual the torques and normal forces balance for transitional and rotational equilibrium ! you get a=2, b=1 , y=2 , d=1 ! nice problem !