Let reactions on two knife edges be N₁ and N₂ respectively.
Then  N₁+ N₂ = W       ...(i)
and from principle of moments, taking moments at point A, we have
N₂ \(\times\)d = W\(\times\)x    ..(ii)
Equation (ii) leads
N₂ =\(\frac { Wx }{ d } \)
and substituting this value in (i), we get N₁ = W -N₂ = W-\(\frac { Wx }{ d } \) = W\(\left( 1-\frac { x }{ d } \right) \)

We know ፒ = Iα or \(\alpha =\frac { \tau }{ I } \)
 ∴ Angular acceleration produced in the cylinder is \(\alpha _{ c }=\frac { \tau }{ I_{ c } } \)
Similarly, acceleration produced in the sphere is
\(\alpha _{ S }=\frac { \tau }{ I_{ S } } \)
 ∴ \(\frac { \alpha _{ C } }{ \alpha _{ S } } =\frac { I_{ S } }{ I_{ C } } \)
Now \(I_{ S }=\frac { 2 }{ 3 } { MR }^{ 2 }\quad and\quad { I }_{ C }=\frac { 1 }{ 2 } { MR }^{ 2 }\)
 ∴ \(\frac { { \alpha }_{ C } }{ { \alpha }_{ S } } =\frac { 4 }{ 3 } \) or \({ \alpha }_{ C }=\frac { 4 }{ 3 } { \alpha }_{ S }\)
or \({ \alpha }_{ C }>{ \alpha }_{ S }\)
Thus cylinder will acquire greater speed than that of the sphere

We know that angular acceleration is defined as 
∴ \(\alpha =\frac { d\omega }{ d\theta } .\frac { d\theta }{ dt } =\omega \frac { d\omega }{ d\theta } \)                \(\left[ \because \omega =\frac { d\theta }{ dt } \right] \)
or α d \(\theta\) = ωdω
On integrating, we have
\(\int _{ 0 }^{ \theta }{ \alpha d\theta } =\int _{ { \omega }_{ 0 } }^{ \omega }{ \omega d\omega } \)
∴ \(\left[ \alpha \theta \right] _{ 0 }^{ \theta }=\left[ \frac { { \omega }^{ 2 } }{ 2 } \right] _{ { \omega }_{ 0 } }^{ \omega }\)
or ω(\(\theta\)-0) = \(\frac { { \omega }^{ 2 } }{ 2 } -\frac { { \omega }_{ 0 }^{ 2 } }{ 2 } \)
⇒ω²-ω₀²  = 2α\(\theta\) ,which is the requisite relation

Linear acceleration of a body rolling down an inclined plane
a =\(\frac { g\quad sin\theta }{ 1+\frac { 1 }{ M{ r }^{ 2 } } } \)
Linear acceleartion of ring =\(\frac { g\quad sin\theta }{ 1+\frac { M{ r }^{ 2 } }{ M{ r }^{ 2 } } } =\frac { g sin\theta }{ 2 } \)          [∵ I= Mr²]
Linear acceleration of disc =\(\frac { g\quad sin\theta }{ 1+\frac { M{ r }^{ 2 } }{ 2M{ r }^{ 2 } } } \) =\(\frac { g sin \theta }{ 1.5 } \)          [∵ I=\(\frac { 1 }{ 2 } \)Mr²]
Linear acceleration of sphere =\(\frac { g\quad sin\theta }{ 1+\frac { 2 }{ 5 } \frac { M{ r }^{ 2 } }{ M{ r }^{ 2 } } } \) =\(\frac { g sin\theta }{ 1.5 } \)     [∵I=\(\frac { 1 }{ 2 } \)Mr²]
Therefore, the sphere reaches first and ring the last.

Fraction of translational kinetic energy = \(\frac { \frac { 1 }{ 2 } { mv }^{ 2 } }{ \frac { 1 }{ 2 } { mv }^{ 2 }+\frac { 1 }{ 2 } I{ \omega }^{ 2 } } \)
\(=\frac { 1 }{ 1+\frac { { k }^{ 2 } }{ { r }^{ 2 } } } =\frac { { r }^{ 2 } }{ ({ k }^{ 2 }+r^{ 2 }) } \)

Let 0 be the centre of the disc and O' that of the hole.
To find the centre of mass, we use the fact that a body balances at this point, i.e the algebraic sum of the moments of the weights about the centre of gravity is zero. The weight W₁ of the disc acts at point O. The hole can be regarded as a negative weight W₂ acting at O'
If X is the distance of the centre of gravity of the combination from point 0, then
X = \(\frac { { W }_{ 1 }\times O+(-{ W }_{ 2 })\times 3 }{ { W }_{ 1 }+(-{ W }_{ 2 }) } \)
Also W₁= ρ\(\pi \)\(\times\)(6)² =36 ρ\(\pi \)
W₂=ρ\(\pi \)\(\times\)(1)² =ρ​​​​\(\pi \)
where ρ is the mass per unit area of the disc.
Substituting the values of W₁ and W₂, we get
\(X=\frac { -\rho \pi \times 3 }{ 36\rho \pi -\rho \pi } m=\frac { -3 }{ 35 } m\)
The negative sign indicates that the centre of gravity is to the left of the point O.

Kinetic energy of rotation =  \(\frac { 1 }{ 2 } I{ \omega }^{ 2 }=\frac { 1 }{ 2 } (I\omega )=\frac { 1 }{ 2 } L\omega \)
L= Iω is constant if moment of inertia (I) of the system changes. It means as I changes, then ω also changes to keep Iω= constant. Hence K.E. of rotation also changes with the change in I. In other words, rotation K.E. is not conserved.

(a) Moment of inertia of sphere about any diameter = MR²
Applying theorem of parallel axes
Moment of inertia of sphere about a tangent to the sphere =\(\frac { 2 }{ 5 } \)MR²+  M(R)² =\(\frac { 7 }{ 5 } \)MR²
(b) We are given, moment of inertia of the disc about any of its diameters =\(\frac { 1 }{ 4 } \)MR²
(i) Using theorem of perpendicular axes, moment of inertia of the disc about an axis passing through its centre and normal to the disc =2\(\times\)\(\frac { 1 }{ 4 } \)MR² +\(\frac { 1 }{ 2 } \)MR²
(ii) Using theorem axes, moment of inertia of the disc passing through a point on its edge and normal to the dies = \(\frac { 1 }{ 2 } \) MR² + MR² =\(\frac { 3 }{ 2 } \)MR²

Let the acceleration of the centre of mass of disc be a, then , Ma = F - f 
The angular acceleration of the disc is a =a/ R (if there is no sliding).
Then, \(\left( \frac { 1 }{ 2 } M{ R }^{ 2 } \right) \alpha \) = Rf \(\Rightarrow \) Ma =2f
Thus, f = F /3.Since, there is no sliding.
\(\Rightarrow \) f <\(\mu \) F\(\le \) 3\(\mu \) Mg

Angular displacement is
\(\Delta \theta =-2rad-(+6rad)=-8rad\)

Angular displacement is
\(\Delta \theta ={ \theta }_{ f }-{ \theta }_{ i }=-8rad-(-4rad)=-4rad\)

Angular displacement is
\(\Delta \theta ={ \theta }_{ f }-{ \theta }_{ i }=6-(-2)=8rad\)

(a) Given, mass of the cylinder m =10 kg
Radius, r = 15 cm = 0.15 m
Inclination of plane, \(\theta ={ 30 }^{ \circ }\) 
Coefficient of static friction, \({ \mu }_{ s }=0.25\).
Force of friction acting on the cylinder on the inclined plane,
\(F=\frac { 1 }{ 3 } mg\sin { \theta } =\frac { 1 }{ 3 } \times 10\times 9.8\times \sin { { 30 }^{ \circ } } \)
\(=\frac { 1 }{ 3 } \times 10\times 9.8\times \frac { 1 }{ 2 }\)
\( =16.3\quad N\)
(b) Force of friction acts perpendicular to the direction of displacement.
\(\therefore \)  Work done aganist friction during rolling
 \(W=Fscos{ 90 }^{ \circ }=0\).
(c) For rolling without slipping,
\(\mu =\frac { 1 }{ 3 } \tan { \theta } \)  
or     \(\tan { \theta } =3\mu =3\times 0.25=0.75\) 
\(=tan{ 36 }^{ \circ }{ 54 }^{ ' }\) 
or      \(\theta ={ 36 }^{ \circ }{ 54 }^{ ' }={ 37 }^{ \circ }\).

Given, mass of bullet (m) = 10 gm = 0.01 kg
Speed of bullet (v) = 500 m/s
Width of the door (l) = 1.0 m
Mass of the door (M) = 12kg
As bullet gets embedded exactly at the centre of the door, therefore its distance from the hinged end of the door,
r = \(\frac { l }{ 2 } =\frac { 1 }{ 2 } m\) 
Angular momentum transferred by the bullet to the door,
 \(L=mv\times r=0.01\times 500\times \frac { 1 }{ 2 } =2.5Js\) 
Moment of inertia of the door about the vertical axis at one of its end,   \(I=\frac { { Ml }^{ 2 } }{ 3 } =\frac { 12\times { \left( 1 \right) }^{ 2 } }{ 3 } =4kg-m^{ 2 }\) 
But angular momentum, \(L=I\omega \)  
 \(2.5=4\times \omega \) 
\(\omega =\frac { 2.5 }{ 4 } =0.625rad/s\).

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
\(=I\omega =0.625\times 100\)
\(\\ =62.5\ kg-{ m }^{ 2 }/s\)