Solution:

(i) Moment of inertia is inertial equivalent in rotational motion. Moment of inertia of a rigid body is defined as the sum of the products of the constituent masses and the squares of the perpendicular distance from the axis of rotation. If `m_1 m_2, ..... m_n`, are the masses at perpendicular distances `r_l> r_2, ..r_n` then, moment of inertia

`I = m_1r_(1)^(2) + m_2r_(2)^(2) + .... + m_n r_(n)^(2)`

`sum_(i=1)^(n) m_ir_(i)^(2)`

It is measured in `kgm^2` and has the
dimensions of `ML^2`

`"Parallel Axis theorem"`
The moment of inertia about an axis passing parallel to the axisthrough the centre of mass of a rigid body is the sum of the moment of inertia `(I_(cm)` of the body about the axis through centre of mass and the product of its mass M and square of the separation (a^2) between the parallel axis.

`i.e I = I_(cm)+Ma^2`

`"Perpendicular Axis theorem"`
If `I_x` and `I_y` are the moment of inertia about the `x` and `y-axis` of any rigid
mass, then the moment of inertia I.
about `z-axis` is the sum of `I_x` and `I_r`

(ii) Moment of in tertia of a uniform circular disc about its diameter : Using the theorem of perpendicular axes, we get `I_d + I_d =` Moment of inertia of the disc about perpendicular axis yoy'

i.e , `2I_d = 1/2 MR^2`

or `I_d = 1/4 MR^2`

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Solution:

We know `L = r x p r x mu`
Since `u = rw,`
we have `L = r x nromega) = .r2omega = Iomega`

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Solution:

Rigid body is one in which the separation between any two constituent masses remains constant. It can execute translatory and rotational motion. Equilibrium identifies the stability of any body. For equilibrium,

(i) `sumf = 0 (i.e.,)` the total unbalanced force should be zero.
(ii) `sumr = 0` (i.e.,) the net torque acting on the body should be zero.

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Solution:

Moment of inertia of a body about a given axis is defined as the sum of the products of masses of all the particles of body and squares of their respective perpendicular distances from axis of rotation. Moment of inertia depends on
(i) Distribution of Mass
(ii) Orientation of axis of rotation.
Diver does so, so that moment of inertia I of
his body decreases. As angular momentum
(I m) remains constant, therefore, angular
velocity w of his body increases.

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Solution:

We know `vect = vecr xxvecF`
If `vecr = xhati + yhatj +zhatk` and `vecF = F_xhati + F_Zhatk` we have

`t = t_xhati + t_y hatj +t_zhatk= | ((hati,hatj, hatk), (x, y , z) (F_x,F_y F_z))|`

`= hati (yF_z_zF_y) -hatj(x(F_z-zF_x)+hatk(xF_y_yF_x)`

Comparing the coefficients of i,] and k, we have

`t_2 = xF_y-yF_x`

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Solution:

Torque `vect = vecr xxvecF = vecr xxvec(dp)/(dt) (vecr xxvecp) = vec(dl)/(dt)`

If no external torque acts on a body, , `vect= 0`

then `vec(dL)/(dt) = 0` or `vecL` is conserved.

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Solution:

When the planet moves along the line joining the sun and the planet it sweeps some area given by by `A = 1/2r^2theta` where `theta` is the angular displacement.

`:. (dA)/(dt)=1/2r^2(dtheta)/(dt) = 1/2 r^2omega`

`(dA)/(dt) = 1/2m mr^2omega= L/(2m)`
Since no torque acts, angular momentum L

is a constant so `(dA)/(dt)` is a constant i.e the line joining the sun and the planet sweeps equal intervals of time

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Solution:

The moment of inertia about an axis parallel to the axis through the centre of mass is the sum of the moment of inertia about the axis through centre of mass and the product of the mass M and the square of the separation between the axes, i.e.,
`I= MI_(cm)+Ma^2`

Moment of inertia is the product of mass and the square of the separation from the axis. The moment of inertia of the system of n masses shown in the a figure about the parallel is

`I= m_1 (x_1+a)^2 + m_2 (a-x_2)^2 +m_3 (x_3-a)^2`

`+ .........`for n masses.

In genereal, `I = sum(i-1)^(n) m (x_x_i+a_i)^2`


if all the n masses are considered equal.

`I= sum(i-1)^(n) mx_(i)^(2) + sum(i-1)^(n) ma^(2)+2asum(i-1)^(n) mx_i`

`I= I_(cm)+Ma^2 :. summx_i = 0` if the centre of mass is considered as origain

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Solution:

Torque is the moment of force. It is written as the cross-product of the position vector of the point of application of the force and the force, i.e., `vect = vecr xxvecF`

the angle between `vecr` and `vecF` sin `theta,` where `theta` is It is measured in Nm and has dimensions torque is the product of the perpendicular distance between the axis of rotation and point of application of force in line with the force. It is the cause of rotational motion. To open the door, handle is to be rotated. So we need a torque. With less force one can exert more torque if the handle is at the edge.

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Solution:

According to perpendicular axis theorem, the sum of the moment of inertia about `x` and `y` axes is equal to the moment of inertia about z-axis. The mass m has co-ordinates `(x, y).`

The moment of inertia about x-axis,

`I_x = my^2`
about y - axis, `I_y = mx^2`
`I_x+I_y=m(x^2+y^2)`
`= M(sqrt(X^2+Y^2)^2`

`I_x+L_y= M`(r distance from z-axis))`^2`

`I_x+i_y+i_2`

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Solution:

Angular momentum `L = I omega`
Rotational `K.E., E_k = Iomega^2`


`E_k = 1/2 ((Iomega)^2)/(I) = (L^2)/(2I)`

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Solution:

Zero. Since r = r.LF and r.L = 0 for all points on the axis.

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Solution:

No external force acts. The centre of mass will follow its original path with every particle scattered.

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Solution:

Torque rotating a particle in xy plane is

`t = xE_y - yF_x`..(I)

`P_x = mv_x` and `P_y = mv_y` are `x, y` component
of linear momentum of body.

According to Newton's 2nd law of motion.

`F_x= (dP_x)/(dt) = (d)/(dt)(mvb_x) = (mdv_x)/(dt)`

`F_y = (dP_Y)/(dt) = (d)/(dt) (mv_y) = (mdvy)/(dt)`

Substituting in equation (i), we get

`t = xm(dv_y)/(dt) - ym (dv_x)/(dt)`

`t = m[x(dv_y)/(dt) - y (dv_x)/(dt)]`

`t = m (d)/(dt) (xv_y- yv_x)`

`t = (d)/(dt) (xmv_y - ymmu_x)`

`t = (d)/(dt) (xmv_y- ymu_x)`

`t = (d)/(dt) (xP_y- yP_x)`

Put `(xP_y - yP_x) = L`

`t = vec(dL)/(dt)`

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Solution:

Moment of inertia is the analogue of mass in rotational motion.
Let the body consists of particles of masses
`m_1 m_2, m_3` at perpendicular distances `r_1`
`r_2, r_3` respectively from axis of rotation.
If `v1_ v_2, v_3` . are the respective linear velocities
of particles, then

`u_1 = r_1 omega, v_2 = r_2omega, v_3 = r_2omega`

K.E of mass `m_1` is `1/2 m_1v_(1)^(2) = 1/2 m_1 (r_1omega)^2`

Similarly K.E of other particles of the body are

`1/2 m_2r_(2)^(2)omega^2 , 1/2 m_3r_(3)^(2) omega^2`

`= 1/2 |sum_(i=1)^(i=m) m_i r_(i)^(2) |omega^2`

put `summ_ir_(i)^(2) = I`

K.E of rotation `= 1/2 Iomega^2`

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Solution:

(i) To prove `L= Iomega`
We know,

`L= r xx p = r xx mv`

(ii) To prove `t= Ialpha`


We know `t = r F = rma`
`= mr^2alpha= Ialpha`

`vect = vecrxxvecF`

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Solution:

Consider a ring of mass M and radius r. Divide the ring into large number of small segments, each of length dl. For every segment dl, the moment of inertia is, m'r2 where m' is its mass.

`m' = (M)/(2pir) dl`

Moments of inertia `I ' = (M)/(2pir) dl`
Net moment of inertia for the ring about the perpendicular axis


`I = int I'dl = int_(0)^(2pir) (M)/(2pir) M/(2pir)r^2dl`
`I = int I'dl = int_(0)^(2pir) dl = (Mr)/(2pi) |l|_(0)^(2pi) = Mr^2`

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