Angular momentum of a body is measured with respect to certain origin.

So, a body in translatory motion can have angular momentum.
It will be zero, if origin lies on the line of motion of particle.

If geometrical centre of a body lies outside it, then centre of mass of body lies outside the body. 
Centre of mass of a L-shaped lamina lies outside it on the line of symmetry.

\(I=3.2\times { 10 }^{ -8 }kg-{ m }^{ 2 }\).

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

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.

The torque applied on the nut by the wrench is equal to the force multiplied by the perpendicular distance from the axis of rotation. Hence, to increase torque a wrench of longer arm is preferred.

Moment of force = force x the perpendicular distance of the line of action of force from the axis of rotation. Hence, the moment of force about a point decreases if the line of action of the force moves towards that point.

For any planar solid, centre of mass always lies at its geometrical centre.
Geometrical centre of a triangle is intersection point of its media.
So, for any given triangular lamina
Centre of mass is at its centroid, point of intersection is media.

If geometrical centre of a body lies outside it, then centre of mass of body lies outside the body. Centre of mass of a L-shaped lamina lies outside it on the line of symmetry.

When a cat falls to ground from a height, it stretches its body alongwith the tail so that its moment of inertia become high. Since,\(I\omega \) is to remain constant, the value of angular speed \(\omega \) decreases and therefore the cat is able to land on the ground gently.

M = 20 Kg, \(\omega =100rad/s,R=0.25m\).
Moment of inertia of cylinder about its own axis
\(=\frac { 1 }{ 2 } MR^{ 2 }=\frac { 1 }{ 2 } \times 20\times { \left( 0.25 \right) }^{ 2 }\)    
Rotational KE \(=\frac { 1 }{ 2 } { I\omega }^{ 2 }\) 
 \(=\frac { 1 }{ 2 } { I\omega }^{ 2 }=\frac { 1 }{ 2 } \times 0.625\times { \left( 100 \right) }^{ 2 }=3125J\).

Centre of mass and centre of gravity are two different concepts. But if g goes not vary from one part of body to other than CG and CM coincides.
So, when vertical height of the object is very small compared to a radius of earth, we call object small, otherwise, we call it extended. In above context, building and pond are small objects and a deep lake and a mountain are large extend objects.

All mass of a hollow cylinder at a distance R from axis of rotation. Whereas in case of a sphere, most of mass lies at a distance less than R from axis of rotation. As moment of inertia is \(\sum M_{ i }{ R }_{ i }^{ 2 }\), so sphere as a lower value of moment of inertia.

Moment of inertia of a sphere is less than that of a cube of same mass.

Concentration of mass at the rim increases the moment of inertia and thereby brings uniform motion.

Zero

As torque = r \(\times \) F
That means torque produced by force is in a plane perpendicular plane containing rand F.
So, if door is in xy-plane, torque produced by weight is in ± z-direction,

It is never about an axis passing through y-direction.

(a) Moment of inertia of man and platform system 
I_i = 7.6 kg-m²
Change in moment of inertia of man and platform system when he stretches his hands to a distance of 90 cm = \(2\times mr^{ 2 }=2\times 5\times { \left( 0.9 \right) }^{ 2 }\) 
= 8.1 kg-m²
I_i = I + 8.1 = 7.6 + 8.1 = 15.7 kg-m²
Initial angular velocity,\({ \omega }_{ i }=30rpm\) 
Initial angular momentum of system,
\({ L }_{ i }=I_{ i }{ \omega }_{ i }=15.7kg-{ m }^{ 2 }\times 30rpm\) 
When man folds his hands to a distance of 20 cm,
Moment of inertia of man =\(2\times { mr }^{ 2 }=2\times 5\times { \left( 0.2 \right) }^{ 2 }\) 
 \(=0.4kg-{ m }^{ 2 }\)  
So, final moment of inertia of man and platform system
= 7.6 + 0.4 = 8 kg-m²
Final angular momentum of system
 \({ L }_{ f }=I_{ f }{ \omega }_{ f }=8\times { \omega }_{ f }\)   
Equating initial and final values
\({ L }_{ i }={ L }_{ f }\) 
\(\Rightarrow { \omega }_{ f }=\frac { 15.7\times 30 }{ 8 } \) 
= 58.88 rpm.
(b) KE is not conserved in process.
K_final > K_initial
Muscular work done by the man in folding his arms is converted into KE.

Thus, the force of friction \({ \mu }_{ k }\)mg produces an acceleration a in the centre of mass. So, the equation of motion for centre of mass is
\({ \mu }_{ k }\) mg = ma
The torque due to force of friction is \({ \mu }_{ k }mg\times R\). It produces angular retardation given by
\({ \mu }_{ k }mgR=-I\alpha \)   
Rolling begins when
\(v=R\omega \)     
But       v = 0 + at = \({ \mu }_{ k }gt\) 
and \(\omega ={ \omega }_{ 0 }+\alpha t={ \omega }_{ 0 }-\frac { { \mu }_{ k }mgR }{ I } t\)    [using Eq. (ii)]
or   \(\frac { v }{ R } ={ \omega }_{ 0 }-\frac { { \mu }_{ k }mgR }{ I } t\Rightarrow \frac { { \mu }_{ k }gt }{ R } ={ \omega }_{ 0 }-\frac { { \mu }_{ k }mgRt }{ I } x = {-b \pm \sqrt{b^2-4ac} \over 2a}\) 
or  \(\frac { { \mu }_{ k }gt }{ R } \left[ 1+\frac { { mR }^{ 2 } }{ I } \right] ={ \omega }_{ 0 }\quad or\quad t=\frac { R{ \omega }_{ 0 } }{ { \mu }_{ k }g\left[ 1+\frac { { mR }^{ 2 } }{ I } \right] } \) 
For a disc, \(I={ mR }^{ 2 }/2\quad \) 
\(\therefore t=\frac { R{ \omega }_{ 0 } }{ { 3\mu }_{ k }g } =\frac { 0.01\times 10\pi }{ 3\times 0.2\times 9.8 } =0.53s\) 
For a ring, \(I={ mR }^{ 2 }\) 
\(\therefore \ t=\frac { R{ \omega }_{ 0 } }{ { 2\mu }_{ k }g } =\frac { 0.01\times 10\pi }{ 3\times 0.2\times 9.8 } =0.80s\).

(i) a = 6 x 10^-2 m/s²
(ii) \(\alpha \) = 1.20 rad/s²

To roll a disc, one require a linear velocity which can be provided only by a tangential force. As frictional force is the only tangential force in this case, so it is necessary for the rolling of the disc. Initially, friction will be kinetic. 
(i) As frictional force at B opposes the angular velocity of B.So, frictional force is in the forward direction, the sense of frictional torque is such as to oppose the angular motion and produce some linear motion, so that the condition of pure rolling (v_CM = R_w) should be fulfilled. 
(ii) After pure rolling starts there will be no need of friction, so friction force will become zero. 

Total mass of the car = 1800kg
Let m and (900-m)kg be the masses of each front wheel and each back wheel, respectively.
Distance of centre of gravity from the back axle = 1.80-1.05 = 0.75m
Taking torque about centre of gravity,
\(m\times 1.05=(900-m)\times 0.75\)
\(\\ 1.05m+0.75m=900\times 0.75\)
\(1.80m=900\times 0.75\)
\(\\ m=\frac { 900\times 0.75 }{ 1.80 } =375kg\)
\( \therefore \ (900-m)=900-375=525kg\)
\(\\ \therefore \ Weight\ of\ each\ front\ wheel(w_{ 1 })={ m }_{ 1 }g\)
\({ (w }_{ 1 })=375\times 9.8=3675N\)
Forced exerted by the level ground on each front wheel.
= force exerted by each front wheel on the level ground \({ (w }_{ 1 })=3675N\)
Weight of each back wheel (\({ w }_{ 2 }\))=525 x 9.8=  5145N
Forced exerted by the level ground on each back wheel = 5145N.

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
\(\alpha =\frac { F }{ m } =\frac { 30 }{ 3 } =10m/{ s }^{ 2 }\).