L is conserved. If the earth shrinks, duration of the day decreases.

To develop torque with less force being applied.

Muscular energy to mechanical energy.

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.

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 \(=\frac { 1 }{ 2 } { I\omega }^{ 2 }\) 
 \(=\frac { 1 }{ 2 } { I\omega }^{ 2 }=\frac { 1 }{ 2 } \times 0.625\times { \left( 100 \right) }^{ 2 }=3125J\).

Molten ice from poles into ocean and so mass is going away from axis of rotation. So, moment of inertia of earth increases and to conserve angular momentum, angular velocity (\(\omega \)) decreases. So, time period of rotation increases (\(T=2\pi /\omega \) ). So, net effect of global warming is increasing in the duration of day.

Let a parallelopiped be formed on the three vectors
 \(\vec { OA } \)= \((\vec { a } ),\) \(\vec { OB} \) = \(\vec { b } \) and  \(\vec { OC } \)=\(\vec { c } \)
Now, \(\vec { b } \) \(\times\)\(\vec { c } \)= bc sin 90^o \(\hat { n } \)=bc \(\hat { n } \)
where \(\hat { n } \) is unit vector along \(\vec { OA } \) perpendicular to the plane containing \(\vec { b } \) and \(\vec { c } \)
Now \(\vec { a } \)(\(\vec { b } \) \(\times\)\(\vec { c } \)) = (a) (bc) cos 0^o
= a b c
which is equal in magnitude to the volume of the parallelopiped.

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) We shall use ω = ω₀ + αt
ω₀ = initial angular speed in rad/s
= 2π × angular speed in rev/s
\(=\frac{2 \pi \times \text { angular speed in rev } / \mathrm{min}}{60 \mathrm{~s} / \mathrm{min}} \)
\(=\frac{2 \pi \times 1200}{60} \mathrm{rad} / \mathrm{s} \)
\(=40 \pi \mathrm{rad} / \mathrm{s}\)
Similarly \(\omega=\) final angular speed in rad / s
\( =\frac{2 \pi \times 3120}{60} \mathrm{rad} / \mathrm{s} \)
\(=2 \pi \times 52 \mathrm{rad} / \mathrm{s} \)
\(=104 \pi \mathrm{rad} / \mathrm{s}\)
Therefore Angular acceleration
\(\alpha=\frac{\omega-\omega_0}{t} =4 \pi \mathrm{rad} / \mathrm{s}^2\)
The angular acceleration of the engine \(=4 \pi \mathrm{rad} / \mathrm{s}^2\)
(ii) The angular displacement in time t is given by
\( \theta=\omega_0 t+\frac{1}{2} \alpha t^2\)
\(=  \left(40 \pi \times 16+\frac{1}{2} \times 4 \pi \times 16^2\right) \mathrm{rad}\)
\(=  (640 \pi+512 \pi) \mathrm{rad} \)
\(=  1152 \pi \mathrm{rad}\)
Number of revolutions \(=\frac{1152 \pi}{2 \pi}=576\)

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

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

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, \({ \mu }_{ s }=0.25\).
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    \(\theta ={ 36 }^{ \circ }{ 54 }^{ ' }={ 37 }^{ \circ }\).

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

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