In all the four cases, as the mass density is uniform, centre of mass is located at their respective geometrical centres.
No, it is not necessary that the centre of mass of a body should lie on the body. For example, in case of a circular ring, centre of mass is at the centre of the ring, where there is no mass.

1.9 km-m/s

\(2{ MR }^{ 2 }\).

\(\frac { 5 }{ 4 } { MR }^{2}\)

The bottom of a ship is made heavy so that its centre of gravity remains low. This ensures the stability of its equilibrium.

For a given force, torque can be increased if the perpendicular distance of the point of application of the force from the axis of rotation is increased.
Hence, the handle put near the circumference produces maximum torque.

Centre of mass of a body lies towards region of heavier mass. So, if bat is cut through its centre of mass, both parts are not equal masses.

The moment of inertia \(\left( I=\frac { 1 }{ 2 } MR^{ 2 } \right) \) of the earth about its own axis will become one-fourth and so its angular velocity will become four times\(\left( L=I\omega =constant \right) \). Hence, the time period will reduce to one-fourth \(\left( T=2\pi /\omega \right) \), i.e. 6 hours.

No, the moment of inertia greatly depends on the distribution of mass about the axis of rotation.

Here\(\vec { r_{ i } } =\vec { r'_{ i } } +\vec { R } +R\) and \(\vec { V_{ i } } =\vec { V'_{ i } } +\vec { V } \)
where\(\vec { r'_{ i } } \) and \(\vec { \upsilon '_{ i } } \) denote the radius vector and velocity of the i^th, particle referred to centre of mass O' as the new origin and\(\vec { V } \) is the velocity of centre of mass relative to O.
Kinetic energy of the system of particles
K = \(\frac { 1 }{ 2 } \sum { { m }_{ i } } { v }_{ i }^{ 2 }\)
= \(\frac { 1 }{ 2 } \sum { { m }_{ i } } \vec { { v }_{ i } } .\vec { { v }_{ i } } \)
= \(\frac { 1 }{ 2 } \sum { { m }_{ i } } \left( { V }_{ i }^{ '2 }+{ V }^{ 2 }+2\vec { { V }_{ i }' } .\vec { V } \right) \)
= \(\frac { 1 }{ 2 } \sum { { m }_{ i }{ V }_{ i }^{ '2 } } +\frac { 1 }{ 2 } \sum { { m }_{ i }{ V }_{ i }^{ '2 } } +\sum { { m }_{ i }\vec { { V }_{ i } } } .\vec { V } \)
= \(\frac { 1 }{ 2 } { MV }^{ 2 }+K'\)
Where M=\(\sum { { m }_{ i } } \)
=total mass of the system
K' = \(\frac { 1 }{ 2 } \sum _{ i }^{ }{ { m }_{ i } } { v }_{ i }^{ '2 }\)
= kinetic energy of motion about the centre of mass
or \(\frac { 1 }{ 2 } \)Mv² kinetic energy of motion of centre of mass
since \(\sum _{ i }^{ }{ { m }_{ i } } \vec { { V }'_{ i } } .\vec { V } =\sum { { m }_{ i } } \frac { d\vec { { r }_{ i } } }{ dt } .\vec { V } \)
=\(\frac { d }{ dt } \left( \sum { { m }_{ i } } \vec { r'_{ i } } \right) .\vec { V } =\frac { d }{ dt } \left( M\vec { R } .\vec { V } \right) \)
= 0

Suppose the equilateral triangle lies in the xy-plane with mass 3 kg at the origin.
Let (x, y) be. the coordinates of centre of mass.

Clearly, AB=\(\sqrt { { (OB) }^{ 2 }-{ (OA) }^{ 2 } } =\sqrt { { (1) }^{ 2 }-{ \left( \frac { 1 }{ 2 } \right) }^{ 2 } } =\frac { \sqrt { 3 } }{ 2 } \)m
Now,  x₁= 0, x₂ = 1m, x₃ = OA = 0.5 m
m₁ = 3kg, m₂ = 4 kg, m₃ = 5 kg
\(\therefore x=\frac { { m }_{ 1 }{ x }_{ 1 }+{ m }_{ 2 }{ x }_{ 2 }+m_{ 3 }{ x }_{ 3 } }{ { m }_{ 1 }+{ m }_{ 2 }+{ m }_{ 3 } }\)
\( \\ =\frac { 3\times 0+4\times 1+5\times 0.5 }{ 3+4+5 } \)
 = \(\frac { 6.5 }{ 12 } \) = 0.54m
Again,   y₁ = 0, y₂ = 0, y₃ = AB = \(\frac { \sqrt { 3 } }{ 2 } \),
\(\therefore y=\frac { { m }_{ 1 }{ y }_{ 1 }+{ m }_{ 2 }{ y }_{ 2 }+m_{ 3 }{ y }_{ 3 } }{ { m }_{ 1 }+{ m }_{ 2 }+{ m }_{ 3 } }\)
\( \\ =\frac { 3\times 0+4\times 0+5\times \left( \frac { \sqrt { 3 } }{ 2 } \right) }{ 3+4+5 } \)
=\(\frac { 5\times \sqrt { 3 } }{ 2\times 12 } \) = 0.36 m
Thus, the coordinate of centre of mass are (0.54 m, 0.36 m).

\(K=F/\sqrt { 3 } \).

Centre of mass of a sphere lies at its geometrical centre.

Let mass of cimet be M and its angular speed be \(\omega \)  when situated at a distance r from the sun, then its angular momentum
L = I \(\omega \)  = Mr² \(\omega \)
If v be the linear speed, then L = Mr² \(\omega \)  = Mrv 
In accordance with conservation law of angular momentum, we can write that
mr₁v₁ = mr₂v₂
\(\therefore \quad { v }_{ 2 }=\frac { { r }_{ 1 }{ v }_{ 1 } }{ { r }_{ 2 } } =\frac { 7\times{ 10 }^{ 10 }\times60 }{ 1.4\times{ 10 }^{ 13 } } =0.3\ Km/s\ or\ 300m/s\)

Moment of inertia of cylinder about its own axis = \(=\frac { 1 }{ 2 } MR^{ 2 }\)
\(=\frac { 1 }{ 2 } \times 20\times { \left( 0.25 \right) }^{ 2 }kg-m^{2}\)    
= 0.625 kg-m²
Kinetic energy of rotating cylinder
 \(=\frac { 1 }{ 2 } { I\omega }^{ 2 }=\frac { 1 }{ 2 } (0.625) { \left( 100 \right) }^{ 2 }J=3125J\).
Angular momentum of cylinder about its own axis
\(=I\omega =0.625\times 100\)
\(\\ =62.5\ kg-{ m }^{ 2 }/s\)

Torque on cylinder, \(\tau =force\times radius\) 
\(=30\times 40=12N-m\)  
Moment of inertia of hollow cylinder about its axis
\(I={ MR }^{ 2 }=3\times { \left( 0.4 \right) }^{ 2 }=0.48kg-{ m }^{ 2 }\)     
Also,    \(\tau =I\alpha \Rightarrow \alpha \frac { \tau }{ I } \) 
\(\therefore \alpha =\frac { 12 }{ 0.48 } =25{ s }^{ -2 }\)    
Linear acceleration of rope
\(\alpha =\frac { F }{ m } =\frac { 30 }{ 3 } =10m/{ s }^{ 2 }\).              

\(Given,\ { m }_{ 1 }=1kg,{ m }_{ 2 }=2kg\)
\( { x }_{ 1 }=1m,{ x }_{ 2 }=-1m\)
\( { y }_{ 1 }=2m,{ y }_{ 2 }=3m\)
\(\\ \therefore \quad { x }_{ CM }=\frac { { m }_{ 1 }{ x }_{ 1 }+{ m }_{ 2 }{ x }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } =\frac { 1\times 1+2\times -1 }{ 1+2 } \)
\(=\frac { 1-2 }{ 3 } =\frac { -1 }{ 3 } =-0.33\)
\(\\ and\ { y }_{ CM }=\frac { { m }_{ 1 }{ y }_{ 1 }+{ m }_{ 2 }{ y }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } =\frac { 1\times 2+2\times 3 }{ 1+2 } \)
\(\\=\frac { 2+6 }{ 3 } =\frac { 8 }{ 3 } =2.66\)
Thus, the coordinates of the centre of mass are (-0.33, 2.66).

Let from a bigger uniform disc of radius R with centre O a smaller circular hole of radius\(\frac { R }{ 2 } \) with its centre at O₁ (where OO₁= \(\frac { R }{ 2 } \)) is cut out. Let centre of gravity or the centre of mass of remaining flat body be at O₂ where OO₂ = x. If σ be mass per unit area, then mass of whole disc M₁ = \(\pi\)R²σ and mass of cut out part
M₂ = \(\pi\)\(\left( \frac { R }{ 2 } \right) ^{ 2 }\) σ =\(\frac { 1 }{ 4 } \) \(\pi\)R²σ = \(\frac { { M }_{ 1 } }{ 4 } \)
∴ x = \(\frac { { M }_{ 1 }\times (0)-{ M }_{ 2 }({ OO }_{ 1 }) }{ { M }_{ 1 }-{ M }_{ 2 } } =\frac { 0-\frac { { M }_{ 1 } }{ 4 } \times \frac { R }{ 2 } }{ { M }_{ 1 }-\frac { { M }_{ 1 } }{ 4 } } =-\frac { R }{ 6 } \)
i.e O₂ is at distance \(\frac { R }{ 6 } \)from centre of disc on diametrically opposite side to centre of hole.

\( f \times R=\left(\frac{M R^2}{2}\right) \alpha \)
\( f=\frac{M R \alpha}{2} \)
\( F-f=M a_{c m} \)
\( a_{c m}=\frac{F-f}{M}\)
For pure rolling \(\mathrm{a}_{\mathrm{cm}}=\mathrm{R} \alpha\)
\(f=\frac{M}{2}\left(a_{c m}\right)=\frac{M}{2}\left(\frac{F-f}{M}\right)\)
so, on rearranging above equation, we get
F = 3f
\({ F }_{ max }=\mu mg ={ F }_{ max }=3\mu mg\).