Consider a spherical body of mass M and radius r, rolling down on an inclined plane without slipping. Let the angle of inclination of the plane is \(\theta\) with the horizontal (Seefig.)
It \(\omega\) be the angular velocity of the body, then the linear velocity of the body is

\(v=r\omega\)................1
The various forces acting on the body are:
(a) The weight (Mg) of the body in the vertically downward direction
(b) Normal reaction (R) of the surface of the plane on the body which acts vertically upward.
(c) The force of friction (F) which acts opposite to the direction of motion of the body. Resolve Mg into two components:
\(\rightarrow\) Mg cos 8 is the horizontal component, which is equal and opposite to the normal reaction
i.e, R=Mg cos \(\theta\) .............2
\(\rightarrow\) Mg sin 8 is the vertical component, which acts in the direction of the motion of the body
\(\therefore\) Net force acting in the direction of motion of the body = Mg sin\(\theta\) - F
Let a = acceleration of the body
Therefore, equation of motion of the body is given by,
Ma = Mg sin \(\theta\) - F ...........3
The external torque acting on the body is produced by the force of friction (F) and its lever arm is r (the radius of the spherical body) [The lines of action of Mg and R pass through the centre of mass of body and hence do not contribute anything to the torque about the centre of mass].
\(\therefore \tau=Fr\)..........4
If \(\alpha\) be the angular acceleration produced in the body whose moment of inertia is I, then
\(\tau=I\alpha\)............5
From eqns (iv) and (v), we get
\(Fr=I\alpha\ or\ F=I\frac{\alpha}{r}\)................6
Using eqn. (vi) in eqn. (iii), we get
\(\therefore Ma=Mgsin\theta-I\frac{\alpha}{r}\)
\(a=r\alpha\ or \alpha=\frac{a}{r}\)
\(\therefore Ma=Mgsin\theta-I\frac{a}{r^2}\Rightarrow a=gsin\theta-\frac{Ia}{Mr^2}\)
\(a(1+\frac{I}{Mr^2})=gsin\theta\)
\(a=\frac{gsin\theta}{1+(\frac{I}{Mr^2})}\)
\(\alpha=\frac{a}{r}=\frac{gsin\theta}{r[1+\frac{I}{Mr^2}]}\)
Substituting this value in eqn. (vi), we get
\(F=\frac{I}{r}.\frac{gsin\theta}{r[1+\frac{I}{Mr^2}]}=\frac{Igsin\theta}{r^2[1+\frac{I}{Mr^2}]}\)
This is the force of friction, required by the body to roll down an inclined plane without slipping.

Theorem of parallel axes: According to this A theorem, moment of inertia of a rigid body about any axis AB is equal to moment of inertia of the body about another axis KL passing through centre of mass Cof the body in a direction parallel to AB, plus the product of total mass M of the body and square of the perpendicular distance between the two parallel axes.
If h is perpendicular distance between the axes AB and KL, then
Suppose rigid body is made up of n particles m₁,m₂,...m_i,.....m_n at perpendicular distances r₁,r₂,...r_i,.....r_n respectively from the axis KL passing through centre of mass C of the body.
If r _i is the perpendicular distance of the particle of mass m_i from KL then
\({ I }_{ KL }=\sum _{ i }^{ }{ { m }_{ i }{ r }_{ i }^{ 2 } } \)                   ...(i)
The perpendicular distance of i^th particle from the axis
AB = (r_i+n)
or  I_AB = \(\sum _{ i }^{ }{ { m }_{ i } } ({ r }_{ 1 }+h)^{ 2 }\)
=\(\sum _{ i }^{ }{ { m }_{ i } } ({ r }_{ i }^{ 2 }+h^{ 2 }+2{ r }_{ i }h)\)
=   \(\sum _{ i }^{ }{ { m }_{ i }{ r }_{ i }^{ 2 } } +\sum _{ i }^{ }{ { m }_{ i }{ h }^{ 2 } } +2h\sum _{ i }^{ }{ { m }_{ i }{ r }_{ i } } \)    ...(ii)
As the body is balanced about the centre of mass, the algebraic sum of the moments of the weights of all particles about an axis passing through C must be zero.
\(\sum _{ i }^{ }{ ({ m }_{ i }g) } { r }_{ i }=0\quad or\quad 8\sum _{ i }^{ }{ { m }_{ i }{ r }_{ i } } \)
or     \(\sum _{ i }^{ }{ { m }_{ i }{ r }_{ i } } =0\)  ...(iii)
From equation (ii), we have
I_AB = \({ I }_{ AB }=\sum _{ i }^{ }{ { m }_{ i }{ r }_{ i }^{ 2 } } +\left( \sum _{ }^{ }{ { m }_{ i } } \right) { h }^{ 2 }+0\)
or  I_AB=I_KL + Mh²
where  I_KL =\(\sum _{ i }^{ }{ { m }_{ i }{ r }_{ i }^{ 2 } } \)  and M =\(\sum { { m }_{ i } } \)

(a) Using law of conservation of energy

\(\frac { 1 }{ 2 } mv^{ 2 }+\frac { 1 }{ 2 } I{ \omega }^{ 2 }\)=mgh
or  \(\frac { 1 }{ 2 } mv^{ 2 }+\frac { 1 }{ 2 } \left( \frac { 2 }{ 5 } m{ R }^{ 2 } \right) \frac { { v }^{ 2 } }{ { R }^{ 2 } } =mgh\)
or \(\frac { 7 }{ 10 } v^{ 2 }=gh\) or v =\(\sqrt { \frac { 10gh }{ 7 } } \)
Since h is same for both the inclined planes therefore v is the same
(b) 

\(l=\frac { 1 }{ 2 } \left( \frac { g\quad sin\quad \theta }{ 1+\frac { { K }^{ 2 } }{ { R }^{ 2 } } } \right) { t }^{ 2 }=\frac { g\quad sin\theta }{ 2\left( 1+\frac { 2 }{ 5 } \right) } { t }^{ 2 }=\frac { 5gsin\theta }{ 14 } { t }^{ 2 }\)
or \(t=\sqrt { \frac { 14l }{ 5g\quad sin\quad \theta } } \)
Now, sin \(\theta\)=\(\frac { h }{ l } \) or l =\(\frac { h }{ sin\theta } \)
∴ \(t=\frac { 1 }{ sin\quad \theta } \sqrt { \frac { 14h }{ 5g } } \)
Lesser the value of,\(\theta\) more will be t.
(c) Clearly, the solid sphere will take longer to roll down the plane with smaller inclination.

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.

We know that angular momentum \(\vec { l } \) of a particle having position vector \(\vec { r } \) and momentum \(\vec { p } \) is given by
\(\vec { l } \) = \(\vec { r } \)\(\times\)\(\vec { p } \)
But, \(\vec { r } \)= [x\(\vec { i} \)+ y\(\vec { j} \)+ z\(\vec { k} \)] where x,y,z are the components of
\(\vec { r } \)and\(\vec { p } \) = [p_x\(\vec { i} \)+ p_y\(\vec { j} \)+ p_z\(\vec { k} \)]
∴ \(\vec { l } \) = \(\vec { r } \)\(\times\)\(\vec { p } \) [x\(\vec { i} \)+ y\(\vec { j} \)+ z\(\vec { k} \)]\(\times\)[p_x\(\vec { i} \)+ p_y\(\vec { j} \)+ p_z\(\vec { k} \)]
or (lx\(\vec { i} \)+ ly\(\vec { j} \)+ lz\(\vec { k} \)) = \(\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ x & y & z \\ { p }_{ x } & { p }_{ y } & { p }_{ z } \end{matrix} \right| \)
 =(yp_z - zp_y)\(\vec { i} \) +(zp_x - xp_z)\(\vec { j} \)+(xp_y - yp_x)\(\vec { k} \)
From this relation, we conclude that 
lx= yp_z - zp_y, l_y= zp_x - xp_z and l_z = xp_y - yp_x
If the given particle moves only in the x - y plane, then z = 0 and p_z = 0 and hence,
\(\vec { l } \)=(xp_y-yp_x)\(\hat { k } \) ,which is only the .z-comonent of \(\vec { l } \)
It means that for a particle moving only in the x - y plane, the angular momentum has only the z-component

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.

\(v=\sqrt { \frac { 4 }{ 3 } gl\sin { \theta } } \).

KE is not conserved in process.
K_final > K_initial
Muscular work done by the man in folding his arms is converted into KE.

In this question we are given a solid uniform cone of radius R and height h, we need to find the center of mass from the vertex. We need to use the below diagram to understand the problem.
Let us consider the radius of the base of the cone to be R and height is h.
We will consider a small circular cross section of radius ‘z’ and thickness ‘dr’ from a distance ‘r’ from the vertex of the solid cone.
Consider two triangles ABC and AOD. These triangles are similar triangles, so we can write,
AB/AO=BC/OD
This can be expressed as,
r/h=z/R…… (1)
The volume of the small volume element considered is, dV=πz²dr,
which can be written in terms of r from equation (1). So we get,
\(d V=\frac{\pi R^2 r^2}{h^2} d r\) ...(2)
Center of mass for continuous mass distribution is given by the formula,
\(C . M=\frac{1}{M} \int r \rho \frac{\pi R^2 r^2}{h^2} d r \)
\( \Longrightarrow C . M=\frac{\pi R^2 \rho}{M h^2} \int r^3 d r\)
The limits of integration are from 0 to h, so
\(C . M=\frac{\pi R^2 \rho}{M h^2} \int_0^h r^3 d r\)
On integrating and applying the limits we get,
\(C . M=\frac{\pi R^2 \rho h^2}{4 M}\)
Substituting M as \(M=\rho V=\frac{\rho}{3 \pi R^2 h}\), we can write
\(\mathrm{C} \cdot \mathrm{M}=\frac{3 h}{4}\)
At height \(\frac { h }{ 4 } \) from the point, O. O is at the centre of the base of the cone

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