Angular Momentum and Angular Impulse Questions in English

(C) Angular momentum is defined as the cross product of position vector $\vec{r}$ and linear momentum $\vec{p}$,given by $\vec{L} = \vec{r} \times \vec{p}$.
Axial vectors are vectors that represent rotational effects and their direction is determined by the right-hand rule,which is perpendicular to the plane of rotation.
Since angular momentum describes rotational motion,it is classified as an axial vector (or pseudovector).
Therefore,the correct option is $C$.

(B) We know that the angular momentum is given by $\vec{L} = \vec{r} \times \vec{p}$.
In terms of components,this is:
$\vec{L} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ x & y & z \\ p_x & p_y & p_z \end{vmatrix}$
Since the motion is in the $x-y$ plane,$z = 0$ and $p_z = 0$. Thus,$\vec{L} = \hat{k}(x p_y - y p_x)$.
Given that the particle moves parallel to the $X$-axis at a distance $b$,its position is $(x, y) = (vt, b)$ and its momentum is $\vec{p} = (mv, 0)$.
Substituting these values:
$\vec{L} = \hat{k}(vt \cdot 0 - b \cdot mv) = -mvb\,\hat{k}$.

(A) The angular momentum $\vec L$ is given by the cross product of position vector $\vec r$ and momentum vector $\vec p$:
$\vec L = \vec r \times \vec p = \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & 2 & -1 \\ 3 & 4 & -2 \end{vmatrix}$
$= \hat i(2(-2) - (-1)(4)) - \hat j(1(-2) - (-1)(3)) + \hat k(1(4) - 2(3))$
$= \hat i(-4 + 4) - \hat j(-2 + 3) + \hat k(4 - 6)$
$= 0\hat i - 1\hat j - 2\hat k$
$= -\hat j - 2\hat k$
Since the angular momentum vector $\vec L = 0\hat i - 1\hat j - 2\hat k$ has no component along the $x$-axis (the coefficient of $\hat i$ is $0$),the vector $\vec L$ lies in the $yz$-plane.
Therefore,the angular momentum is perpendicular to the $x$-axis.

(B) The radius vector $\vec{r}$ is the displacement between the two points: $\vec{r} = \vec{r_2} - \vec{r_1} = (2\hat{i} - 3\hat{j} + \hat{k}) - (2\hat{i} + \hat{j} + \hat{k}) = -4\hat{j}$.
Linear momentum $\vec{p} = 2\hat{i} + 3\hat{j} - \hat{k}$.
Angular momentum $\vec{L} = \vec{r} \times \vec{p}$.
$\vec{L} = (-4\hat{j}) \times (2\hat{i} + 3\hat{j} - \hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & -4 & 0 \\ 2 & 3 & -1 \end{vmatrix}$.
Calculating the determinant: $\vec{L} = \hat{i}((-4)(-1) - (0)(3)) - \hat{j}((0)(-1) - (0)(2)) + \hat{k}((0)(3) - (-4)(2))$.
$\vec{L} = \hat{i}(4) - \hat{j}(0) + \hat{k}(8) = 4\hat{i} + 8\hat{k}$.
Wait,re-evaluating the cross product: $\vec{L} = (-4\hat{j}) \times (2\hat{i}) + (-4\hat{j}) \times (3\hat{j}) + (-4\hat{j}) \times (-\hat{k}) = -8(\hat{j} \times \hat{i}) - 12(\hat{j} \times \hat{j}) + 4(\hat{j} \times \hat{k}) = -8(-\hat{k}) - 0 + 4(\hat{i}) = 4\hat{i} + 8\hat{k}$.
Given the options,there appears to be a sign convention difference in the problem statement or options. Based on standard vector cross product rules,the result is $4\hat{i} + 8\hat{k}$. If we assume $\vec{r} = \vec{r_1} - \vec{r_2} = 4\hat{j}$,then $\vec{L} = (4\hat{j}) \times (2\hat{i} + 3\hat{j} - \hat{k}) = 8(\hat{j} \times \hat{i}) + 4(\hat{j} \times \hat{k}) = -8\hat{k} + 4\hat{i} = 4\hat{i} - 8\hat{k}$.

(B) The formula for angular momentum $L$ is given by $L = r \times p$,where $r$ is the position vector and $p$ is the linear momentum.
Dimensional analysis: $L = [L] \times [M L T^{-1}] = [M L^2 T^{-1}]$.
In $SI$ units,this corresponds to $kg \cdot m^2/s$.
Since $1 \text{ Joule} = 1 \text{ kg} \cdot m^2/s^2$,we can write $1 \text{ Joule} \cdot s = (1 \text{ kg} \cdot m^2/s^2) \times s = 1 \text{ kg} \cdot m^2/s$.
Therefore,the unit of angular momentum is $Joule \cdot s$.

(D) The angular momentum $\vec L$ is defined as $\vec L = \vec r \times \vec p$.
For a particle moving in a circular path,the position vector $\vec r$ and linear momentum $\vec p$ lie in the plane of the circle.
According to the right-hand rule,the direction of $\vec L$ is perpendicular to the plane of rotation.
Since the particle continues to move in the same circular path,the plane of rotation does not change.
Therefore,the direction of the angular momentum vector remains constant,even if its magnitude changes due to the decreasing speed of the particle.
Thus,option $(D)$ is correct.

(B) The angular momentum $L_{ang}$ of a particle about a point $O$ is defined as the cross product of its position vector $\vec{r}$ and its linear momentum $\vec{p} = m\vec{v}$.
Mathematically,$L_{ang} = |\vec{r} \times \vec{p}| = |\vec{r} \times m\vec{v}| = m v r \sin(\theta)$.
Here,$r \sin(\theta)$ represents the perpendicular distance from the point $O$ to the line of motion of the particle,which is given as $l$ in the figure.
Therefore,the angular momentum is $L_{ang} = m v l$.

(C) The rotational kinetic energy $E$ is given by the formula $E = \frac{1}{2} I \omega^2 = \frac{L^2}{2I}$,where $L = I \omega$ is the angular momentum.
From this,we can express the angular momentum as $L^2 = 2EI$,or $L = \sqrt{2EI}$.
For body $A$,$L_A = \sqrt{2 E_A I_A}$.
For body $B$,$L_B = \sqrt{2 E_B I_B}$.
The ratio of angular momenta is $\frac{L_A}{L_B} = \sqrt{\frac{2 E_A I_A}{2 E_B I_B}} = \sqrt{\frac{E_A}{E_B} \cdot \frac{I_A}{I_B}}$.
Given $I_A = I_B/4$,we have $\frac{I_A}{I_B} = \frac{1}{4}$.
Given $E_A = 100 E_B$,we have $\frac{E_A}{E_B} = 100$.
Substituting these values: $\frac{L_A}{L_B} = \sqrt{100 \cdot \frac{1}{4}} = \sqrt{25} = 5$.

(C) Initial angular momentum of the system about the pivot point is due to the bullet: $L_i = Mv(L/2)$.
After the collision,the bullet is embedded in the rod,so the system rotates with angular velocity $\omega$. The final moment of inertia of the system is $I_f = I_{\text{rod}} + I_{\text{bullet}} = \frac{ML^2}{12} + M(L/2)^2 = \frac{ML^2}{12} + \frac{ML^2}{4} = \frac{ML^2 + 3ML^2}{12} = \frac{4ML^2}{12} = \frac{ML^2}{3}$.
By the principle of conservation of angular momentum,$L_i = L_f$,so $Mv(L/2) = I_f \omega$.
Substituting the values: $\frac{MvL}{2} = \left(\frac{ML^2}{3}\right) \omega$.
Solving for $\omega$: $\omega = \frac{MvL}{2} \cdot \frac{3}{ML^2} = \frac{3v}{2L}$.

(A) The angular momentum $L$ of a rotating body is given by the formula $L = I\omega$,where $I$ is the moment of inertia and $\omega$ is the angular velocity.
For a solid cylinder rotating about its own axis,the moment of inertia is $I = \frac{1}{2}MR^2$.
Given: Mass $M = 2\ kg$,Radius $R = 0.2\ m$,Angular velocity $\omega = 3\ rad/s$.
Substituting the values:
$I = \frac{1}{2} \times 2\ kg \times (0.2\ m)^2 = 1 \times 0.04 = 0.04\ kg\cdot m^2$.
Now,calculate the angular momentum:
$L = I\omega = 0.04\ kg\cdot m^2 \times 3\ rad/s = 0.12\ J-s$.
Therefore,the angular momentum of the cylinder before collision is $0.12\ J-s$.

(B) The rotational kinetic energy $K$ of a body is related to its angular momentum $L$ and moment of inertia $I$ by the formula $K = \frac{L^2}{2I}$.
Given that the angular momentum $L$ is the same for both bodies,the kinetic energy is inversely proportional to the moment of inertia,i.e.,$K \propto \frac{1}{I}$.
Since $I_1 > I_2$,it follows that $K_1 < K_2$.
Therefore,the second body (with the smaller moment of inertia) will have a higher kinetic energy.

(A) The angular momentum $L$ of a rigid body rotating about a fixed axis is given by the relation $L = I\omega$,where $I$ is the moment of inertia of the body about the axis of rotation.
For a rigid body,the moment of inertia $I$ remains constant.
Therefore,$L \propto \omega$.
This linear relationship implies that the graph of $L$ versus $\omega$ is a straight line passing through the origin with a slope equal to the moment of inertia $I$.

(D) The angular momentum of a particle about an origin $O$ is given by $\vec{L} = \vec{r} \times \vec{p} = m(\vec{r} \times \vec{v})$. The magnitude is $L = mvr \sin \theta = mv d$,where $d$ is the perpendicular distance from the origin to the line of motion.
$1$. For motion along $OA$,the line of motion passes through the origin $O$,so $d = 0$ and $L = 0$. Statement $A$ is correct.
$2$. For the line $DE$,the perpendicular distance from $O$ to the line $DE$ is constant $(1 \ m)$. Thus,$L$ is constant at all points on $DE$. Statement $B$ is correct.
$3$. At $B$,the perpendicular distance is $1 \ m$ and the velocity is in the $+y$ direction. At $D$,the perpendicular distance is $1 \ m$ and the velocity is in the $+x$ direction. The angular momentum is a vector quantity. The direction of $\vec{L}$ at $B$ is $\vec{r} \times \vec{v} = (1\hat{i}) \times (v\hat{j}) = mv\hat{k}$. The direction of $\vec{L}$ at $D$ is $\vec{r} \times \vec{v} = (1\hat{j}) \times (v\hat{i}) = -mv\hat{k}$. Thus,they have the same magnitude but opposite directions. Statement $C$ is correct.
$4$. As the particle moves towards the line $BC$,the perpendicular distance $d$ from the origin $O$ to the line of motion remains constant. Therefore,the angular momentum does not change. Statement $D$ is incorrect.

(D) The angular momentum $\vec{L}$ of a particle about a point $O$ is given by the cross product of its position vector $\vec{r}$ and its linear momentum $\vec{p} = m\vec{v}$.
Mathematically,$\vec{L} = \vec{r} \times \vec{p} = \vec{r} \times (m\vec{v}) = m(\vec{r} \times \vec{v})$.
The magnitude of the angular momentum is given by $L = mrv \sin(\theta)$,where $\theta$ is the angle between the position vector $\vec{r}$ and the velocity vector $\vec{v}$.
From the geometry of the figure,the perpendicular distance from the origin $O$ to the line of motion $PC$ is $\ell$.
Therefore,the magnitude of the angular momentum about $O$ is $L = m v \ell$.

(C) The angular momentum $\vec{L}$ of a particle with respect to the origin is given by $\vec{L} = \vec{r} \times \vec{p}$,where $\vec{p} = M\vec{v}$.
Since the particle moves parallel to the $X$-axis with constant velocity $\vec{v} = v\hat{i}$,its position vector is $\vec{r} = x\hat{i} + y\hat{j}$.
Thus,$\vec{L} = (x\hat{i} + y\hat{j}) \times (Mv\hat{i}) = Mv(x(\hat{i} \times \hat{i}) + y(\hat{j} \times \hat{i}))$.
Since $\hat{i} \times \hat{i} = 0$ and $\hat{j} \times \hat{i} = -\hat{k}$,we get $\vec{L} = -Mvy\hat{k}$.
Here,$M$,$v$,and $y$ (the perpendicular distance from the $X$-axis) are all constant.
Therefore,the angular momentum $\vec{L}$ remains constant.

(D) The angular momentum $\vec{L}$ of a particle is defined by the cross product $\vec{L} = \vec{r} \times \vec{p} = \vec{r} \times (m\vec{v})$.
Since the position vector $\vec{r}$ lies in the plane of rotation and the linear velocity vector $\vec{v}$ is tangent to the circular path (also in the plane of rotation),the cross product $\vec{r} \times \vec{v}$ results in a vector perpendicular to both $\vec{r}$ and $\vec{v}$.
By the right-hand rule,this direction is perpendicular to the plane of rotation,which corresponds to the axis of rotation.

(C) The angular momentum $L$ of a particle about the origin is given by the formula $L = |\vec{r} \times \vec{p}| = r p \sin \theta$.
From the geometry of the motion,the perpendicular distance $d$ from the origin to the line of motion is $d = r \sin \theta$.
Therefore,the angular momentum can be expressed as $L = p \cdot d = (mv) \cdot d$.
Given: mass $m = 1 \ kg$,velocity $v = 2 \ ms^{-1}$,and perpendicular distance $d = 12 \ cm = 0.12 \ m$.
Substituting these values into the formula:
$L = (1 \ kg) \times (2 \ ms^{-1}) \times (0.12 \ m) = 0.24 \ kg \cdot m^2s^{-1}$ (or $Js$).
Thus,the angular momentum is $0.24 \ Js$.

(D) The linear momentum $\vec{p}$ is given by $\vec{p} = m\vec{v} = 0.01 \ kg \times 5\hat{i} \ m/s = 0.05\hat{i} \ kg \cdot m/s$.
The angular momentum $\vec{L}$ about the origin is given by the cross product $\vec{L} = \vec{r} \times \vec{p}$.
Substituting the values: $\vec{L} = (10\hat{i} + 6\hat{j}) \times (0.05\hat{i})$.
Using the distributive property of the cross product: $\vec{L} = (10\hat{i} \times 0.05\hat{i}) + (6\hat{j} \times 0.05\hat{i})$.
Since $\hat{i} \times \hat{i} = 0$ and $\hat{j} \times \hat{i} = -\hat{k}$,we get:
$\vec{L} = 0 + 6 \times 0.05 \times (-\hat{k}) = 0.3 \times (-\hat{k}) = -0.3\hat{k} \ J \cdot s$.

(D) The centripetal force $F$ is given by $F = \frac{mv^2}{r}$.
Since angular momentum $L = mvr$,we can write $v = \frac{L}{mr}$.
Substituting this into the force equation:
$F = \frac{m}{r} \left( \frac{L}{mr} \right)^2 = \frac{m}{r} \cdot \frac{L^2}{m^2 r^2} = \frac{L^2}{mr^3}$.
Rearranging for $L^2$:
$L^2 = F \cdot m \cdot r^3$.
Given $m = 2 \ kg$,$r = 2 \ m$,and $F = 100 \ N$:
$L^2 = 100 \times 2 \times (2)^3 = 100 \times 2 \times 8 = 1600$.
Taking the square root:
$L = \sqrt{1600} = 40 \ J \cdot s$.

(C) The angular momentum of a rolling body about a point on the surface is the sum of the angular momentum due to translation and the angular momentum due to rotation about its center of mass.
$L = L_{\text{translation}} + L_{\text{rotation}}$
$L = MvR + I_c\omega$
Since the disc is rolling without slipping,the linear velocity of the center of mass is $v = R\omega$.
The moment of inertia of a disc about its center is $I_c = \frac{1}{2}MR^2$.
Substituting these values:
$L = M(R\omega)R + (\frac{1}{2}MR^2)\omega$
$L = MR^2\omega + \frac{1}{2}MR^2\omega$
$L = \frac{3}{2}MR^2\omega$

(A) The moment of inertia of the cylinder about its axis is $I = \frac{1}{2}MR^2 = \frac{1}{2} \times 5 \times (0.3)^2 = 0.225 \ kg \ m^2$.
An angular impulse $J = 3 \ kg \ m^2 s^{-1}$ is applied at $t = 0, 4, 8, 12, 16, 20, 24, 28 \ s$.
Total number of impulses applied up to $30 \ s$ is $8$ (at $t = 0, 4, 8, 12, 16, 20, 24, 28$).
Total angular momentum $L = \sum J = 8 \times 3 = 24 \ kg \ m^2 s^{-1}$.
Since $L = I\omega$,the angular speed is $\omega = \frac{L}{I} = \frac{24}{0.225} = 106.67 \ rad/s \approx 106.7 \ rad/s$.

(C) The moment of inertia of a solid cylinder about its axis is given by:
$I = \frac{1}{2} M R^2$
Substituting the given values ($M = 20 \ kg$,$R = 0.25 \ m$):
$I = \frac{1}{2} \times 20 \times (0.25)^2 = 10 \times 0.0625 = 0.625 \ kg \cdot m^2$
The angular momentum $L$ is given by the product of the moment of inertia and angular velocity $(\omega)$:
$L = I \omega$
Given $\omega = 100 \ rad \cdot s^{-1}$:
$L = 0.625 \times 100 = 62.5 \ J \cdot s$

(A) The angular momentum $\vec{L}$ is given by the cross product of the position vector $\vec{r}$ and the momentum vector $\vec{p}$.
$\vec{L} = \vec{r} \times \vec{p} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -1 \\ 3 & 4 & -2 \end{vmatrix}$
$\vec{L} = \hat{i}(2(-2) - (-1)(4)) - \hat{j}(1(-2) - (-1)(3)) + \hat{k}(1(4) - 2(3))$
$\vec{L} = \hat{i}(-4 + 4) - \hat{j}(-2 + 3) + \hat{k}(4 - 6)$
$\vec{L} = 0\hat{i} - 1\hat{j} - 2\hat{k} = -\hat{j} - 2\hat{k}$.
For a vector to be perpendicular to an axis,their dot product must be zero.
The $X$-axis is represented by the unit vector $\hat{i} = (1\hat{i} + 0\hat{j} + 0\hat{k})$.
Dot product of $\vec{L}$ and $\hat{i}$ is: $(0)(1) + (-1)(0) + (-2)(0) = 0$.
Since the dot product is zero,the angular momentum is perpendicular to the $X$-axis.

(B) By the principle of conservation of angular momentum about the center of mass of the rod:
Initial angular momentum $L_i = (2m)(v)(a) + (m)(2v)(2a) = 2mav + 4mav = 6mav$.
Final moment of inertia $I_f = I_{rod} + I_{mass1} + I_{mass2} = \frac{(8m)(6a)^2}{12} + (2m)(a)^2 + (m)(2a)^2$.
$I_f = \frac{8m \cdot 36a^2}{12} + 2ma^2 + 4ma^2 = 24ma^2 + 2ma^2 + 4ma^2 = 30ma^2$.
Using $L_i = I_f \omega$:
$6mav = 30ma^2 \omega$.
$\omega = \frac{6mav}{30ma^2} = \frac{v}{5a}$.

(D) The angular momentum $L$ of a particle about a point $P$ is given by the cross product of the position vector $\vec{r}$ and the linear momentum $\vec{p} = m\vec{v}$.
Mathematically,$\vec{L} = \vec{r} \times \vec{p} = \vec{r} \times (m\vec{v})$.
The magnitude is given by $L = r p \sin \theta$,where $\theta$ is the angle between the position vector $\vec{r}$ and the velocity vector $\vec{v}$.
In this case,the particle is moving along the line $PC$,and point $P$ lies on this same line.
Therefore,the position vector of the particle with respect to point $P$ is collinear with the velocity vector $\vec{v}$.
This means the angle $\theta$ between the position vector and the velocity vector is $0^\circ$ (or $180^\circ$).
Since $\sin(0^\circ) = 0$,the angular momentum $L = mvr \sin(0^\circ) = 0$.

(B) The angular momentum $L$ of a particle about an origin $O$ is given by the cross product of its position vector $\vec{r}$ and its linear momentum $\vec{p}$,i.e.,$\vec{L} = \vec{r} \times \vec{p}$.
The magnitude of angular momentum is given by $L = p \cdot d$,where $p$ is the magnitude of linear momentum and $d$ is the perpendicular distance from the origin $O$ to the line of motion of the particle.
Since the particle is moving along a straight line $AB$,the perpendicular distance $d$ from the origin $O$ to the line $AB$ is the same for all points on the line.
Assuming the particle moves with a constant velocity $v$,its linear momentum $p = mv$ remains constant throughout the motion.
Therefore,at any point on the line $AB$,the angular momentum $L = mv \cdot d$ is constant.
Thus,$L_A = L_B$.

(B) The mass of the ring is $m = 10 \ kg$ and the radius is $r = \text{diameter}/2 = 0.4/2 = 0.2 \ m$.
The moment of inertia of a ring about its axis is $I = mr^2 = 10 \times (0.2)^2 = 10 \times 0.04 = 0.4 \ kg \cdot m^2$.
The angular frequency $\omega$ is given by $\omega = 2\pi n$,where $n$ is the frequency in revolutions per second.
Given $n = 2100 \ \text{rpm} = 2100/60 \ \text{rev/s} = 35 \ \text{rev/s}$.
Thus,$\omega = 2 \times \pi \times 35 = 70\pi \ \text{rad/s}$.
The angular momentum $L$ is given by $L = I\omega$.
$L = 0.4 \times 70\pi = 28\pi \ \text{kg} \cdot m^2/s$.
Using $\pi \approx 22/7$,$L = 28 \times (22/7) = 4 \times 22 = 88 \ \text{kg} \cdot m^2/s$.

(D) The force acting on the point mass due to the string is radial,so the angular momentum remains conserved even when the string is pulled.
Let the mass of the particle be $m$,the initial radius be $r_1 = 2 \ m$,the initial linear velocity be $v_1 = 4 \ m/s$,and the initial angular velocity be $\omega_1 = v_1 / r_1 = 4 / 2 = 2 \ rad/s$.
Let the final radius be $r_2 = 1 \ m$,the final linear velocity be $v_2$,and the final angular velocity be $\omega_2$.
By the principle of conservation of angular momentum:
$L_1 = L_2$
$m r_1 v_1 = m r_2 v_2$
$r_1 v_1 = r_2 v_2$
Substituting the values:
$2 \times 4 = 1 \times v_2$
$v_2 = 8 \ m/s$
The new angular velocity is:
$\omega_2 = v_2 / r_2 = 8 / 1 = 8 \ rad/s$
The ratio of the final kinetic energy $(K_2)$ to the initial kinetic energy $(K_1)$ is:
$K_2 / K_1 = (\frac{1}{2} m v_2^2) / (\frac{1}{2} m v_1^2)$
$K_2 / K_1 = v_2^2 / v_1^2$
$K_2 / K_1 = (8)^2 / (4)^2 = 64 / 16 = 4$
Thus,the ratio is $4:1$.

(B) The angular momentum $L$ of a particle about the origin is given by $L = mvd$,where $m$ is the mass,$v$ is the velocity,and $d$ is the perpendicular distance from the origin to the line of motion.
The equation of the line is $2x - y + 4 = 0$.
The perpendicular distance $d$ from the origin $(0, 0)$ to the line $Ax + By + C = 0$ is given by $d = \frac{|C|}{\sqrt{A^2 + B^2}}$.
Here,$A = 2$,$B = -1$,and $C = 4$.
$d = \frac{|4|}{\sqrt{2^2 + (-1)^2}} = \frac{4}{\sqrt{4 + 1}} = \frac{4}{\sqrt{5}} \ m$.
Given mass $m = 5 \ kg$ and velocity $v = 3\sqrt{5} \ m/s$.
$L = mvd = 5 \times (3\sqrt{5}) \times \left(\frac{4}{\sqrt{5}}\right)$.
$L = 5 \times 3 \times 4 = 60 \ kg \ m^2 \ s^{-1}$.

(C) The only force acting on the projectile is its weight $mg$ acting downwards.
Taking point $P$ as the origin,the torque $\tau$ due to gravity at any time $t$ is given by $\tau = \vec{r} \times \vec{F} = x(t) \cdot mg$,where $x(t) = (v_0 \cos 45^{\circ})t = \frac{v_0}{\sqrt{2}}t$.
Thus,$\tau = mg \cdot \frac{v_0}{\sqrt{2}}t$.
The change in angular momentum $\Delta L$ is equal to the angular impulse:
$\Delta L = \int_{0}^{t} \tau \, dt = \int_{0}^{t} \frac{mgv_0}{\sqrt{2}} t \, dt = \frac{mgv_0}{\sqrt{2}} \left[ \frac{t^2}{2} \right]_{0}^{t} = \frac{mgv_0 t^2}{2\sqrt{2}}$.
Since the initial angular momentum at $P$ is zero (the line of action of the initial velocity passes through $P$),the angular momentum at time $t = \frac{v_0}{g}$ is:
$L = \frac{mgv_0}{2\sqrt{2}} \left( \frac{v_0}{g} \right)^2 = \frac{mgv_0^3}{2\sqrt{2}g^2} = \frac{mv_0^3}{2\sqrt{2}g}$.

(A) According to the principle of conservation of angular momentum,the total angular momentum before the impact is equal to the total angular momentum after the impact.
Initial angular momentum of the cylinder = $I\omega$
Initial angular momentum of the particle about the axis of the cylinder = $mvR$
Total initial angular momentum $L_1 = I\omega + mvR$
After the particle sticks to the rim,the new moment of inertia of the system becomes $I' = I + mR^2$.
Let the new angular velocity be $\omega'$.
Total final angular momentum $L_2 = (I + mR^2)\omega'$
Equating $L_1 = L_2$:
$I\omega + mvR = (I + mR^2)\omega'$
$\omega' = \frac{I\omega + mvR}{I + mR^2}$

(A) The angular momentum $L$ of a particle with respect to an origin $O$ is given by the formula $L = r \times p = r \times (mv),$ where $r$ is the position vector of the particle and $p$ is its linear momentum.
This can also be expressed as $L = m v d,$ where $d$ is the perpendicular distance from the origin $O$ to the line of motion of the particle.
Since the particle is moving along a straight line $AB,$ the perpendicular distance $d$ from the origin $O$ to this line remains constant at all points on the line.
Therefore,the angular momentum $L$ is constant for all points on the line $AB.$
Thus,$L_A = L_B.$

(A) The angular momentum $\vec{L}$ of a particle is defined by the cross product of its position vector $\vec{r}$ and its linear momentum $\vec{p} = m\vec{v}$,given by $\vec{L} = \vec{r} \times \vec{p}$.
Since $\vec{r}$ and $\vec{v}$ both lie in the plane of rotation,their cross product $\vec{L}$ must be perpendicular to both $\vec{r}$ and $\vec{v}$ according to the right-hand rule.
Therefore,the angular momentum vector is directed along a line perpendicular to the plane of rotation.

(A) Angular momentum is defined by the cross product of position vector and linear momentum: $\overrightarrow{L} = \overrightarrow{r} \times \overrightarrow{p}$.
Since it is defined by a cross product of two vectors and its direction is determined by the right-hand rule along the axis of rotation,it is classified as an axial vector.

(B) According to the principle of angular momentum, the rate of change of angular momentum $(\vec{L})$ of a system is equal to the net external torque $(\vec{\tau}_{ext})$ acting on it.
Mathematically, $\vec{\tau}_{ext} = \frac{d\vec{L}}{dt}$.
If $\vec{\tau}_{ext} = 0$, then $\frac{d\vec{L}}{dt} = 0$, which implies $\vec{L}$ is constant.
Therefore, the angular momentum of a system changes if and only if a non-zero net external torque acts on the system.

(D) The moment of linear momentum $\vec{p}$ about a point is defined as the cross product of the position vector $\vec{r}$ and the linear momentum $\vec{p}$.
Mathematically,it is expressed as $\vec{L} = \vec{r} \times \vec{p}$.
This quantity is known as angular momentum.
Therefore,the correct option is $D$.

(D) The angular momentum $\vec{L}$ of a particle is defined by the cross product of its position vector $\vec{r}$ and its linear momentum $\vec{p}$,given by $\vec{L} = \vec{r} \times \vec{p}$.
According to the right-hand rule for the cross product,the vector $\vec{L}$ is perpendicular to both $\vec{r}$ and $\vec{p}$.
Since both $\vec{r}$ and $\vec{p}$ lie in the plane of rotation,their cross product $\vec{L}$ must be perpendicular to the plane of rotation.
The line perpendicular to the plane of rotation passing through the fixed point is the axis of rotation.
Therefore,the direction of the angular momentum is along the axis of rotation.

(D) The angular momentum $(L)$ of a rotating rigid body is defined as the product of its moment of inertia $(I)$ and its angular velocity $(\omega)$.
Mathematically,this is expressed as:
$L = I \omega$
Therefore,the correct option is $(D)$.

(B) The moment of inertia $I$ of a ring about its axis is given by $I = mR^2$.
Given mass $m = 10 \ kg$ and diameter $d = 0.4 \ m$,the radius $R = d/2 = 0.2 \ m$.
So,$I = 10 \times (0.2)^2 = 10 \times 0.04 = 0.4 \ kg \cdot m^2$.
The angular velocity $\omega$ in $rad/s$ is given by $\omega = 2\pi n$,where $n$ is the frequency in $rev/s$.
Given $n = 2100 \ rpm = 2100/60 \ rev/s = 35 \ rev/s$.
Thus,$\omega = 2 \times \pi \times 35 = 70\pi \ rad/s$.
The angular momentum $L$ is given by $L = I\omega$.
$L = 0.4 \times 70\pi = 28\pi \ kg \cdot m^2/s$.
Using $\pi \approx 22/7$,$L = 28 \times (22/7) = 4 \times 22 = 88 \ kg \cdot m^2/s$.

(D) The angular momentum $L$ is defined as the cross product of the position vector $r$ and the linear momentum vector $p$,given by $L = r \times p$.
By the definition of the cross product,the resulting vector $L$ is always perpendicular to the plane containing both vectors $r$ and $p$.
Since both the position vector $r$ and the linear momentum vector $p$ lie in the orbital plane,the angular momentum vector $L$ must be perpendicular to the orbital plane.

(A) The rotational kinetic energy $E$ is given by $E = \frac{L^2}{2I}$,where $L$ is the angular momentum and $I$ is the moment of inertia.
Rearranging for $L$,we get $L = \sqrt{2EI}$.
Given $E = 10 \ J$ and $I = 8 \times 10^{-7} \ kg \ m^2$.
Substituting the values: $L = \sqrt{2 \times 10 \times 8 \times 10^{-7}}$.
$L = \sqrt{160 \times 10^{-7}} = \sqrt{16 \times 10^{-6}}$.
$L = 4 \times 10^{-3} \ kg \ m^2/s$.

(C) The angular momentum $L$ of a rotating body is defined as the product of its moment of inertia $I$ and its angular velocity $\omega$.
Mathematically,this is expressed as:
$L = I \omega$
Therefore,the correct option is $C$.

(D) The angular momentum $L$ of a rotating body is given by $L = I \omega$.
For a solid sphere (Earth) rotating about its axis,the moment of inertia is $I = \frac{2}{5} M R^2$.
The angular velocity $\omega$ in terms of the time period $T$ is $\omega = \frac{2 \pi}{T}$.
Substituting these values into the formula for angular momentum:
$L = \left( \frac{2}{5} M R^2 \right) \times \left( \frac{2 \pi}{T} \right) = \frac{4 \pi M R^2}{5 T}$.

(A) The angular momentum $\vec{L}$ of a particle is defined as the cross product of its position vector $\vec{r}$ and its linear momentum $\vec{p}$,given by $\vec{L} = \vec{r} \times \vec{p}$.
By the definition of the cross product,the resulting vector $\vec{L}$ is always perpendicular to the plane containing the vectors $\vec{r}$ and $\vec{p}$.
Since the particle moves in a plane of rotation,both $\vec{r}$ and $\vec{p}$ lie within this plane.
Therefore,the angular momentum vector $\vec{L}$ is always perpendicular to the plane of rotation.

(A) The impulse $J = Mv$ is applied at one end of the rod. This impulse provides an angular momentum about the center of mass $(CM)$ of the system.
The center of mass of the two-ball system is at the midpoint of the rod.
The angular momentum $L_{CM}$ imparted by the impulse $J$ about the center of mass is given by $L_{CM} = J \times r$,where $r = L/2$ is the distance from the center of mass to the point of application of the impulse.
$L_{CM} = (Mv) \times (L/2) = \frac{MvL}{2}$.
The moment of inertia $I$ of the system about an axis passing through the center of mass and perpendicular to the rod is $I = M(L/2)^2 + M(L/2)^2 = 2M(L^2/4) = \frac{ML^2}{2}$.
Using the relation $L_{CM} = I\omega$,we have:
$\frac{MvL}{2} = \left( \frac{ML^2}{2} \right) \omega$.
Solving for $\omega$:
$\omega = \frac{MvL/2}{ML^2/2} = \frac{v}{L}$.

(B) The angular momentum $L$ of a particle about the origin is given by $L = r \times p$,where $r$ is the position vector and $p$ is the linear momentum.
Alternatively,the magnitude of angular momentum is $L = p \times d$,where $d$ is the perpendicular distance from the origin to the line of motion.
Here,the mass $m$ moves with constant velocity $v$,so the linear momentum $p = mv$ is constant.
The perpendicular distance $d$ from the origin to the line of motion is constant and equal to $a$.
Therefore,the angular momentum $L = mv \times a = mva$,which is a constant value.

(B) The angular momentum $L$ of a particle about the origin is given by $L = p \times d$,where $p = mv$ is the linear momentum and $d$ is the perpendicular distance from the origin to the line of motion.
Given the line of motion is $Y = X + 4$,which can be written as $X - Y + 4 = 0$.
The perpendicular distance $d$ from the origin $(0, 0)$ to the line $X - Y + 4 = 0$ is calculated as:
$d = \frac{|(1)(0) - (1)(0) + 4|}{\sqrt{1^2 + (-1)^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$ units.
The linear momentum $p = mv = 5 \times 3\sqrt{2} = 15\sqrt{2}$ units.
Therefore,the magnitude of angular momentum is:
$L = p \times d = (15\sqrt{2}) \times (2\sqrt{2}) = 15 \times 2 \times 2 = 60$ units.

(C) The average torque $\bar{\tau}$ acting on a body is given by the rate of change of angular momentum: $\bar{\tau} = \frac{\Delta L}{\Delta t}$.
Given:
Initial angular momentum $L_i = 30\, kg\, m^2/s$
Final angular momentum $L_f = 20\, kg\, m^2/s$
Time interval $\Delta t = 2\, s$
Change in angular momentum $\Delta L = L_f - L_i = 20 - 30 = -10\, kg\, m^2/s$.
The magnitude of the average torque is $|\bar{\tau}| = |\frac{-10}{2}| = 5\, N\, m$.

(D) The angular momentum $\vec{L}$ about the origin is given by $\vec{L} = \vec{r} \times \vec{p} = \vec{r} \times m\vec{v}$.
Given: $\vec{r} = (1\hat{i} + 1\hat{j}) \ m$,$m = 2 \ kg$,and $\vec{v} = 2(\hat{i} - \hat{j} + \hat{k}) \ m/s$.
Therefore,$\vec{p} = m\vec{v} = 2 \times 2(\hat{i} - \hat{j} + \hat{k}) = 4(\hat{i} - \hat{j} + \hat{k}) \ kg \cdot m/s$.
Calculating the cross product $\vec{L} = \vec{r} \times \vec{p}$:
$\vec{L} = (\hat{i} + \hat{j}) \times 4(\hat{i} - \hat{j} + \hat{k})$
$\vec{L} = 4 [(\hat{i} \times \hat{i}) - (\hat{i} \times \hat{j}) + (\hat{i} \times \hat{k}) + (\hat{j} \times \hat{i}) - (\hat{j} \times \hat{j}) + (\hat{j} \times \hat{k})]$
Using cross product rules: $\hat{i} \times \hat{i} = 0$,$\hat{i} \times \hat{j} = \hat{k}$,$\hat{i} \times \hat{k} = -\hat{j}$,$\hat{j} \times \hat{i} = -\hat{k}$,$\hat{j} \times \hat{j} = 0$,$\hat{j} \times \hat{k} = \hat{i}$.
$\vec{L} = 4 [0 - \hat{k} - \hat{j} - \hat{k} - 0 + \hat{i}]$
$\vec{L} = 4(\hat{i} - \hat{j} - 2\hat{k}) = 4\hat{i} - 4\hat{j} - 8\hat{k}$.
The angular momentum about the $z$-axis is the $z$-component of $\vec{L}$,which is $L_z = -8 \ kg \cdot m^2/s$.