If the orbits of the Moon and Earth lie on the same plane, during full Moon of every month, we can observe lunar eclipse. If this is so during new Moon we can observe solar eclipse. But Moon's orbit is tilted 5^o with respect to Earth's orbit. Due to this 5^o tilt, only during certain periods of the year, the Sun, Earth and Moon align in straight line leading to either lunar eclipse or solar eclipse depending on the alignment.

The weight of an object is defined as the downward force whose magnitude W is equal to the upward force that must be applied to the object to hold it at rest or at constant velocity relative to the Earth.

(i) Implies that the satellite is bound to the Earth and it cannot escape from the Earth
(ii) As h approaches ∝ the total energy tends to zero. Its physical meaning is that the satellite is completely free from the influence of Earth's gravity and is not bound to Earth at large distances.

The gravitational potential at a distance r due to a mass is defined as the amount of work required to bring unit mass from infinity to the distance r and it is denoted as V(r).
\(\mathrm{V}(\mathrm{r})=-\frac{\mathrm{Gm}}{r}\)

No, potential energy is not the property of a single object. But, it is the gravitational potential energy of the system consisting of two masses m₁ and m₂ separated by a distance r, is the gravitational potential energy difference of the system when the masses are separated by an infinite distance and by distance r.
\(\mathrm{U}(\mathrm{r}) =\mathrm{U}(\mathrm{r})-\mathrm{U}(\alpha)  \)
\(\therefore \mathrm{U}(\alpha) =0\)

Yes, the angular momentum of a planet is conserved.
During the orbitary motion of the planets around the Sun, the line of action of gravitational force passes through the axis, the external torque is zero. Hence the angular momentum is conserved.
Torque,  \( { \tau } =\frac { dL }{ dt }\)
If torque is zero then angular momentum (L) is constant i.e., it is conserved.

Newton's law of gravitation states that a particle of mass M₁ attracts any other particle of mass M₂ in the universe with an attractive force. The strength of this force of attraction was found to be directly proportional to the product of their masses and is inversely proportional to the square of the distance between them.
\(\vec{F}=-\frac{G M_{1} M_{2}}{r^{2}} \hat{r}\)

Translational velocity (V_TRANS) or velocity of center of mass, V_CM = 5 m s^-1
The radius is, R = 1.5 m and the angular velocity is, \(\omega\) = 3 rad s^-1
Rotational velocity, V_ROT = R\(\omega\)
V_ROT = 1.5 x 3
V_ROT = 4.5 ms^-1
As v_CM > R\(\omega\) (or) V_TRANS> R\(\omega\), It is not in pure rolling but sliding.

The concepts to form the integrand to find the moment of inertia could be borrowed from the earlier derivation. Now, the origin is fixed to the left end of the rod and the limits are to be taken from 0 to l.

\(I=\frac{M}{l}\int ^{1}_{0}{x^{2}dx}=\frac{M}{l}[\frac{x^{3}}{3}]^{1}_{0}=\frac{M}{l}[\frac{l^{3}}{3}]\)
\(I=\frac{1}{3}Ml^{2}\)

Law of conservation of angular momentum states that, when no external torque acts on the body, the net angular momentum of a rotating body remains constant.

Radius of gyration of an object is the perpendicular distance from the axis of rotation to an equivalent point mass, which would have the same mass as well as the same moment of inertia of the object.

The center of gravity of a body is the point at which the entire weight of the body acts irrespective of the position and orientation of the body.

A pair of forces which are equal in magnitude but opposite in direction and separated by a perpendicular distance so that their lines of action do not coincide that causes a turning effect is called a couple.

S.No	Gravitational potential	Gravitational potential energy
1.	It depends only on the source mass.	It depends upon two masses and the distance between them.
2.	It is a scalar	It is a scalar
3.	It's unit is J Kg-1.	Its unit is Joule.

The gravitational potential energy U(r) of a system of two masses m₁ and m₂ separated by a distance r as the amount of work done to bring the mass m₂ from infinity to a distance r assuming m₁ to be fixed in its position and is written as
\(\mathrm{U}(\mathrm{r})=-\frac{G m_{1} m_{2}}{r}\)

The gravitational field at a point which is at a distance r from m, is defined as the gravitational force experienced by unit mass placed at that point.
It's unit is N/kg or ms^-2.

Here \(\omega_1=40\)rpm, \(I_2=\frac{2}{5}I_1\)
By the principle of conservation of angular momentum,
\(I_1 \omega_1=I_2 \omega_2 \)or \(I_1 \times 40 =\frac{2}{5}I_1\omega_2 or \omega_2=100 rpm.\)
(ii) Initial kinetic energy of rotation
\(\frac{2}{5}I_12\omega^{2}_{1}=\frac{2}{5}I_1(40)^{2}=800 I_1\)
New kinetic energy of rotation
\(\frac{2}{5}I_2 \omega^{2}_{2}=\frac{1}{2}\times \frac{2}{3} I_1 (100)^{2}=2000 I_1\)
\(\frac{New K.E.}{Initial K.E.}=\frac{2000 I_1}{800 I_1}=2.5\)
Thus the child's new kinetic energy of rotation is 2.5 times its initial kinetic energy of rotation. This increase in kinetic energy is due to the internal energy of the child which he uses in folding his hands back from the out stretched position. 

2: 1

(x, y) = (0.54 m, 0.36 m)

It plays the same role in rotatory motion as the mass does in translatory motion.

Here M = 3 kg, R = 40 cm = 0.40 m, F = 30 N
Torque, \(\tau \)= F \(\times\) R = 30 \(\times\) 0.40 = 12 Nm
M.l. of the hollow cylinder about its own axis,
I = MR² = 3 \(\times\) (0.40)² = 0.48 kgm²
Angular acceleration, \(\alpha =\frac { \tau }{ I } \)= \(\frac {12}{0.48}\)= 25 rad s^-2
Linear acceleration, \(\alpha\) = R\(\alpha\) = 0.40\(\times\)25= 10 ms^-2

As shown in figure, the axis of rotation passes through m₁. The distances of mi, mi and m3 from the axis of rotation are \(O,{a\over 2}\) and \({a\over 2}\) respectively.
\(\therefore\) M.I of the system about the altitude through m₁ is 
\(I=m_1{r}_{1}^{2}+m_2{r}_{2}^{2}+m_3{r}_{3}^{2}\)
\(=m_1(0)^2+m_2\left( {a\over 2} \right)^2+m_2\left( {a\over 2} \right)^2\)
or \(I={a^2 \over 4}(m_2+m_1)\)

At minimum distance r₁ = 7 \(\times\) 10¹⁰m, velocity V₁= 6 \(\times\) 104 ms^-1
At maximum distance r₂ = 1.4 \(\times\) 10¹²m ,velocity V₂= ?
By conservation of angular momentum l₁ω₁ = I₂ω₂
or mr₁²ｘ\(\frac { v_{ 1 } }{ r_{ 1 } } \) = mr₂² ｘ\(\frac { v_{ 2 } }{ r_{ 2 } } \)
or v₁r₁ = v₂r₂
or v2 = \(\frac { v_{ 1 }r_{ 1 } }{ r_{ 2 } } \) = \(\frac { 6\times 10^{ -4 }\times 7\times { 10 }^{ 10 } }{ 1.4\times { 10 }^{ 12 } } \)= 3000 ms^-1

If fingers of right hand are kept along the position vector with palm facing the direction of the force and when the fingers are curled the thumb points to the direction of the torque.

(i) \({ X }_{ CM }=\frac { { m }_{ i }{ x }_{ i }+{ m }_{ 2 }{ x }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } \)
(ii) \({ X }_{ CM }=\frac { { m }_{ 2 }{ x }_{ 2 } }{ { m }_{ 1 }+{ m }_{ 2 } } \)
(iii) \({ m }_{ 1 }{ x }_{ 1 }={ m }_{ 2 }{ x }_{ 2 }\)

Given: diameter = 0.5m, Mass = 16 kg
Formula: Moment of inertia of solid disc
I = \(\frac { 1 }{ 2 } \)MR² = \(\frac { 1 }{ 2 } \times 16\times \left[ \frac { { 0.5 } }{ 2 } \right] ^{ 2 }\)= \(\frac { 1 }{ 2 } \)
Angular velocity \(\omega \) = \(\frac { 120\times 2\pi }{ 60 } \) = \(4\pi \)
And change in angular momentum \(\Delta L=I\omega -0=I\omega \)
Angular impulse \(\tau \times t=\Delta L\Rightarrow \tau =\frac { \Delta L }{ t } =\frac { 1 }{ 8 } \times \frac { 1 }{ 2 } \times 4\pi =\frac { \pi }{ 4 } Nm\quad \)

It is the combination of pure translational and pure rotational motion.
1. The rolling motion is the most commonly observed motion in daily life. The motion of wheel is an example of rolling motion.
2. Round objects like ring, disc, sphere etc. are most suitable for rolling. Let us study the rolling of a disc on a horizontal surface. Consider a point P on the edge of the disc.
3. While rolling, the point undergoes translational motion along with its center of mass and rotational motion with respect to its center of mass.

1. Greater the mass concentrated away from the axis, greater the moment of inertia.
2. Moment of inertia of a body about an axis of rotation resists a change in it. In rotational motion, moment of inertia increases with increase in torque to change it's rotation.

Sum of the clockwise moments is equal to sum of the anticlockwise moments when a body is in rotational equilibrium or algebraic sum of moments at any point is zero.

The linear acceleration along the inclined plane can be computed by
\(a=\frac{g sin\theta}{1+\frac{K^{2}}{R^{2}}}\)
For a thin uniform circular ring, axis passing through its center is I = MR².
\(\therefore K^{2}=R^{2} \Rightarrow \frac{K^{2}}{R^{2}}=1\)
And the angle of inclination, \(\theta=45 ^{0}\)
\(\Rightarrow (sin 45^{0}=\frac{1}{\sqrt{2}})\)
Hence, \(a=\frac{g+\frac{1}{\sqrt{2}}}{1+1}\)
\(a=\frac{g}{2\sqrt{2}}\)ms^-2.