Angular Momentum and Angular Impulse Questions in English

(C) Given: Mass $m = 0.5\, kg$,Radius $r = 2\, m$,Centripetal force $F_c = 9\, N$.
We know that centripetal force $F_c = \frac{mv^2}{r}$.
Substituting the values: $9 = \frac{0.5 \times v^2}{2}$.
$9 = 0.25 \times v^2 \implies v^2 = \frac{9}{0.25} = 36$.
So,velocity $v = 6\, m/s$.
Angular momentum $L = mvr$.
$L = 0.5 \times 6 \times 2 = 6\, J\cdot s$.

(C) The rate of change of angular momentum $(L)$ is equal to the net torque $(\tau)$ acting on the system about the point of rotation.
$\frac{dL}{dt} = \tau$
$\Rightarrow dL = \tau dt$
Taking torque about the bottommost point (point of contact with the ground),the perpendicular distance of the line of action of the force $F$ from the bottommost point is $(R + r)$.
$\tau = F \times (R + r) = 2t(R + r)$
Substituting this into the angular momentum equation:
$dL = 2t(R + r) dt$
Integrating from $t = 0$ to $t$ (assuming initial angular momentum is zero):
$L = \int_{0}^{t} 2t(R + r) dt$
$L = 2(R + r) \int_{0}^{t} t dt$
$L = 2(R + r) \left[ \frac{t^2}{2} \right]_{0}^{t}$
$L = (R + r)t^2$

(B) The angular momentum $L$ of a particle about a point $P$ is given by $L = r \times p$,where $r$ is the position vector and $p$ is the linear momentum.
For a particle moving with velocity $v$ at an angle $\theta$ with the horizontal,the perpendicular distance from point $P$ to the line of motion is $d \cos \theta$.
Initial angular momentum $L_i = m v (d \cos \theta)$ (in the clockwise direction).
Since the collision is perfectly elastic,the ball rebounds with the same speed $v$ at the same angle $\theta$ with the horizontal.
Final angular momentum $L_f = -m v (d \cos \theta)$ (in the counter-clockwise direction).
The change in angular momentum $\Delta L = L_f - L_i = -m v d \cos \theta - (m v d \cos \theta) = -2m v d \cos \theta$.
The magnitude of the change in angular momentum is $2mvd \cos \theta$.

(B) Let the impulse $J$ be applied at one end of the rod. The impulse causes a change in linear momentum and angular momentum.
Linear impulse: $J = M v_{cm}$,where $v_{cm}$ is the velocity of the center of mass.
Angular impulse about the center of mass: $J \times \frac{L}{2} = I_{cm} \omega = \left(\frac{ML^2}{12}\right) \omega$.
From this,$\omega = \frac{6J}{ML}$.
The velocity of the other end is $V = v_{cm} - \omega \frac{L}{2}$ (since the rotation opposes the linear motion at the other end).
Substituting the values: $V = \frac{J}{M} - \left(\frac{6J}{ML}\right) \frac{L}{2} = \frac{J}{M} - \frac{3J}{M} = -\frac{2J}{M}$.
Taking the magnitude: $V = \frac{2J}{M} \Rightarrow J = \frac{MV}{2}$.

(C) The angular momentum of the system about the midpoint $A$ is conserved during the collision because there are no external torques acting on the system about point $A$.
Before the collision,the rod is at rest,so its angular momentum is zero.
The angular momentum of the clay ball of mass $m$ moving with speed $v$ at a distance $L/2$ from the midpoint $A$ is given by $L = r \times p = (L/2) \times (mv) = \frac{mvL}{2}$.
Since the external torque about $A$ is zero,the angular momentum of the system remains constant.
Therefore,the angular momentum of the clay-rod system about $A$ after the collision is equal to the angular momentum of the clay ball before the collision,which is $\frac{mvL}{2}$.

(B) The forces acting on the bob are the tension $T$ in the string and the gravitational force $mg$ acting downwards.
Taking the point of suspension as the origin,the torque $\vec{\tau}$ due to the gravitational force is $\vec{\tau} = \vec{r} \times \vec{F} = \vec{r} \times m\vec{g}$.
The magnitude of this torque is $\tau = mg \ell \sin \theta$,where $\theta$ is the angle the string makes with the vertical.
This torque is always directed horizontally,perpendicular to the plane containing the string and the vertical axis.
Since the rate of change of angular momentum $\frac{d\vec{L}}{dt} = \vec{\tau}$,the torque causes the direction of the angular momentum vector $\vec{L}$ to change continuously as it precesses around the vertical axis.
However,because the torque is always perpendicular to the angular momentum vector $\vec{L}$,the magnitude of the angular momentum remains constant.
Therefore,the angular momentum changes in direction but not in magnitude.

(C) Let $V$ be the final velocity of the center of mass and $\omega$ be the angular velocity acquired by the rod.
From the impulse-momentum theorem for linear motion: $J = m(V - u) \quad ...(1)$
From the angular impulse-momentum theorem about the center of mass: $J \cdot \frac{l}{4} = I \omega = \left(\frac{ml^2}{12}\right) \omega \quad ...(2)$
From equations $(1)$ and $(2)$,we get: $m(V - u) \cdot \frac{l}{4} = \frac{ml^2}{12} \omega \implies V - u = \frac{l}{3} \omega \implies V = u + \frac{l}{3} \omega \quad ...(3)$
The velocity of point $P$ at distance $l/4$ from the center is given by $V_P = V + \omega \cdot \frac{l}{4}$.
Given $V_P = 2u$,so $2u = V + \frac{l}{4} \omega \quad ...(4)$
Substitute $V$ from $(3)$ into $(4)$: $2u = (u + \frac{l}{3} \omega) + \frac{l}{4} \omega$
$u = \omega l \left(\frac{1}{3} + \frac{1}{4}\right) = \omega l \left(\frac{7}{12}\right) \implies \omega l = \frac{12}{7} u$
Now,substitute $\omega l$ back into the expression for $V$: $V = u + \frac{1}{3} \left(\frac{12}{7} u\right) = u + \frac{4}{7} u = \frac{11}{7} u$.

(A) The momentum of the particle is given by $p = m \times V = 5 \times 3 \sqrt{2} = 15 \sqrt{2} \text{ units}$.
The particle moves along the line $Y = X + 4$,which can be rewritten as $X - Y + 4 = 0$.
The perpendicular distance $r_{\perp}$ from the origin $(0, 0)$ to the line $AX + BY + C = 0$ is given by $r_{\perp} = \frac{|C|}{\sqrt{A^2 + B^2}}$.
Substituting the values $A = 1, B = -1, C = 4$,we get $r_{\perp} = \frac{|4|}{\sqrt{1^2 + (-1)^2}} = \frac{4}{\sqrt{2}} = 2 \sqrt{2} \text{ units}$.
The magnitude of the angular momentum $L$ about the origin is $L = p \times r_{\perp}$.
$L = (15 \sqrt{2}) \times (2 \sqrt{2}) = 15 \times 2 \times 2 = 60 \text{ units}$.

(B) The equation of the line is $Y = X + 4$,which can be written as $X - Y + 4 = 0$.
The perpendicular distance $d$ from the origin $(0, 0)$ to this line is given by the formula $d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$.
Substituting the values,$d = \frac{|1(0) - 1(0) + 4|}{\sqrt{1^2 + (-1)^2}} = \frac{4}{\sqrt{2}}$.
The angular momentum $L$ about the origin is given by $L = mvd$.
Substituting the given values $m = 5$,$v = 3\sqrt{2}$,and $d = \frac{4}{\sqrt{2}}$:
$L = 5 \times 3\sqrt{2} \times \frac{4}{\sqrt{2}} = 5 \times 3 \times 4 = 60$ units.

(D) The angular momentum $\vec{L}$ of a particle about a point is given by the cross product $\vec{L} = \vec{r} \times \vec{p} = \vec{r} \times (m\vec{v})$.
Since the particle is moving along a straight line and the fixed point lies on the same line,the position vector $\vec{r}$ and the velocity vector $\vec{v}$ are always collinear (parallel or anti-parallel).
The angle $\theta$ between $\vec{r}$ and $\vec{v}$ is either $0^\circ$ or $180^\circ$.
Therefore,the magnitude of angular momentum is $L = |\vec{r}| |m\vec{v}| \sin(\theta) = 0$,because $\sin(0^\circ) = 0$ and $\sin(180^\circ) = 0$.
Thus,the angular momentum remains zero throughout the motion.

(C) The angular momentum $\vec{L}$ of a particle about the origin is given by $\vec{L} = \vec{r} \times \vec{p} = \vec{r} \times (m\vec{v})$.
In magnitude,$L = mvr \sin \theta$,where $\theta$ is the angle between the position vector $\vec{r}$ and the velocity vector $\vec{v}$.
From the geometry of the motion,$r \sin \theta = h$,where $h$ is the perpendicular distance of the line of motion from the origin.
Since the particle moves in a straight line parallel to the $x-$ axis with constant velocity $v$,both $v$ and $h$ remain constant throughout the motion.
Therefore,$L = mvh = \text{constant}$.
Thus,the angular momentum of the particle about the origin remains constant.

(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.
Given:
$I = 0.6 \,kg \cdot m^2$
Frequency $f = 0.5 \,rev/s$
The angular velocity $\omega$ is related to frequency by $\omega = 2\pi f$.
Substituting the values:
$\omega = 2 \times \pi \times 0.5 = \pi \,rad/s$.
Now,calculating the angular momentum:
$L = 0.6 \times \pi = 0.6\pi \,kg \cdot m^2/s$.

(B) The angular momentum $\vec{L}$ of a particle about a point $O$ is given by $\vec{L} = \vec{r} \times \vec{p} = \vec{r} \times (m\vec{v})$.
Here,the particle moves in a horizontal circle of radius $r = 0.6\, m$ at a height $h = 0.8\, m$ above the ground.
Let the point on the ground directly below the center of the circle be $O'$. The position vector $\vec{r}$ of the particle relative to $O'$ has a horizontal component $r = 0.6\, m$ and a vertical component $h = 0.8\, m$.
The velocity of the particle is $\vec{v}$,which is tangential to the circular path,so its magnitude is $v = r\omega = 0.6 \times 12 = 7.2\, m/s$.
The angular momentum about $O'$ is $L = |\vec{r} \times m\vec{v}| = mvr_{\perp}$,where $r_{\perp}$ is the perpendicular distance from the axis of rotation passing through $O'$ to the velocity vector. Since the velocity is horizontal and the point $O'$ is on the vertical axis,the perpendicular distance from the line of velocity to the point $O'$ is the radius $r = 0.6\, m$.
Thus,$L = mvr = (2\, kg) \times (7.2\, m/s) \times (0.6\, m) = 8.64\, kg\, m^2\, s^{-1}$.

(A) The angular momentum $L$ of a particle about a point $O$ is given by $L = m(\vec{r} \times \vec{v})$. For a particle of mass $m$ moving with speed $v$,the magnitude is $L = mvr_{\perp}$,where $r_{\perp}$ is the perpendicular distance from the origin $O$ to the line of velocity.
From the geometry of the circle,the position vector $\vec{r}$ makes an angle $\theta$ with the $x$-axis. The velocity vector $\vec{v}$ is tangent to the circle at the particle's position. The angle between the position vector $\vec{r}$ and the velocity vector $\vec{v}$ is $(90^\circ + \theta)$.
The distance of the particle from the origin $O$ is $r = 2a \cos \theta$.
The angular momentum $L = mvr \sin(90^\circ + \theta) = mvr \cos \theta$.
Substituting $r = 2a \cos \theta$,we get $L = mv(2a \cos \theta) \cos \theta = 2mva \cos^2 \theta$.
Using the identity $2 \cos^2 \theta = 1 + \cos 2\theta$,we get $L = mva(1 + \cos 2\theta)$.
Assuming unit mass $(m=1)$,the angular momentum is $va(1 + \cos 2\theta)$.

(A) The angular momentum $\vec{L}$ of a particle about the origin is given by $\vec{L} = \vec{r} \times \vec{p} = \vec{r} \times (m\vec{v})$.
Given mass $m = 2\, kg$ and position vector $\vec{r}(t) = 5\hat{i} - 2t^2\hat{j}$.
The velocity vector $\vec{v}(t) = \frac{d\vec{r}}{dt} = \frac{d}{dt}(5\hat{i} - 2t^2\hat{j}) = -4t\hat{j}$.
At $t = 2\, s$:
Position $\vec{r}(2) = 5\hat{i} - 2(2)^2\hat{j} = 5\hat{i} - 8\hat{j}$.
Velocity $\vec{v}(2) = -4(2)\hat{j} = -8\hat{j}$.
Angular momentum $\vec{L} = m(\vec{r} \times \vec{v}) = 2 \times [(5\hat{i} - 8\hat{j}) \times (-8\hat{j})]$.
Since $\hat{i} \times \hat{j} = \hat{k}$ and $\hat{j} \times \hat{j} = 0$:
$\vec{L} = 2 \times [5\hat{i} \times (-8\hat{j}) - 8\hat{j} \times (-8\hat{j})] = 2 \times [-40\hat{k} - 0] = -80\hat{k}\, kg\, m^2\, s^{-1}$.

(A) Given: $\vec r(t) = 2t \hat i - 3t^2 \hat j$ and $m = 2 \ kg$.
Velocity $\vec v = \frac{d\vec r}{dt} = 2 \hat i - 6t \hat j$.
At $t = 2 \ s$,position $\vec r = 2(2) \hat i - 3(2)^2 \hat j = 4 \hat i - 12 \hat j$.
At $t = 2 \ s$,velocity $\vec v = 2 \hat i - 6(2) \hat j = 2 \hat i - 12 \hat j$.
Momentum $\vec p = m \vec v = 2(2 \hat i - 12 \hat j) = 4 \hat i - 24 \hat j$.
Angular momentum $\vec L = \vec r \times \vec p = (4 \hat i - 12 \hat j) \times (4 \hat i - 24 \hat j)$.
Using the cross product: $\vec L = [4(-24) - (-12)(4)] \hat k = (-96 + 48) \hat k = -48 \hat k$.

(B) vector quantity is a physical quantity that has both magnitude and direction.
Distance,heat,and energy are scalar quantities because they only have magnitude.
Angular momentum is defined as the cross product of position vector $\vec{r}$ and linear momentum $\vec{p}$,given by $\vec{L} = \vec{r} \times \vec{p}$.
Since it is defined by a cross product and has a specific direction (determined by the right-hand rule),angular momentum is a vector quantity.

(D) The angular momentum $\vec{L}$ of a particle with respect to the origin $O$ is given by $\vec{L} = \vec{r} \times \vec{p} = \vec{r} \times (m\vec{v})$.
As shown in the figure,the particle moves parallel to the $X$-axis at a constant perpendicular distance $a$ from the origin $O$.
The magnitude of the angular momentum is $|L| = m v a \sin(\theta)$,where $\theta$ is the angle between the position vector $\vec{r}$ and the velocity vector $\vec{v}$.
Since $a = r \sin(\theta)$,we have $|L| = m v a$.
Since $m$,$v$,and $a$ are all constants,the magnitude of the angular momentum remains constant.
Furthermore,the direction of $\vec{L}$ (given by the right-hand rule) is perpendicular to the $X-Y$ plane (along the $Z$-axis) and does not change as the particle moves.
Therefore,the angular momentum is constant.

(A) Since the belt does not slip,the linear speed at the rims of both wheels must be equal:
$v_A = v_C$
Using the relation $v = r\omega$,we have:
$r_A \omega_A = r_C \omega_C$ --- $(i)$
Given that the radius of the larger wheel $(C)$ is three times that of the smaller wheel $(A)$:
$r_C = 3r_A$
Substituting this into equation $(i)$:
$r_A \omega_A = (3r_A) \omega_C$
$\omega_A = 3\omega_C$
If both wheels have the same angular momentum $(L)$:
$L_A = L_C$
$I_A \omega_A = I_C \omega_C$
$\frac{I_C}{I_A} = \frac{\omega_A}{\omega_C}$
Substituting $\omega_A = 3\omega_C$:
$\frac{I_C}{I_A} = \frac{3\omega_C}{\omega_C} = 3$
Thus,the ratio of the rotational inertia of the larger wheel to the smaller wheel is $3$.

(D) The angular momentum $\vec{L}$ of a particle in orbital motion is defined by the cross product of its position vector $\vec{r}$ and its linear momentum $\vec{p}$,given by $\vec{L} = \vec{r} \times \vec{p}$.
By the definition of the cross product,the resulting vector $\vec{L}$ is always perpendicular to the plane containing both $\vec{r}$ and $\vec{p}$.
Since both the position vector and the linear momentum vector lie within the orbital plane,the angular momentum vector $\vec{L}$ must be perpendicular to the orbital plane. This is consistent with the right-hand rule,where the direction of $\vec{L}$ points along the axis of rotation.

(C) The position vector of the particle at time $t$ is given by:
$\vec{r} = (v_0 \cos \theta) t \hat{i} + ((v_0 \sin \theta) t - \frac{1}{2} g t^2) \hat{j}$
The velocity vector of the particle at time $t$ is given by:
$\vec{v} = (v_0 \cos \theta) \hat{i} + (v_0 \sin \theta - gt) \hat{j}$
The angular momentum $\vec{L}$ is defined as $\vec{L} = m(\vec{r} \times \vec{v})$:
$\vec{L} = m [((v_0 \cos \theta) t \hat{i} + ((v_0 \sin \theta) t - \frac{1}{2} g t^2) \hat{j}) \times ((v_0 \cos \theta) \hat{i} + (v_0 \sin \theta - gt) \hat{j})]$
Performing the cross product:
$\vec{L} = m [((v_0 \cos \theta) t) (v_0 \sin \theta - gt) (\hat{i} \times \hat{j}) + ((v_0 \sin \theta) t - \frac{1}{2} g t^2) (v_0 \cos \theta) (\hat{j} \times \hat{i})]$
Since $\hat{i} \times \hat{j} = \hat{k}$ and $\hat{j} \times \hat{i} = -\hat{k}$:
$\vec{L} = m [((v_0^2 \sin \theta \cos \theta) t - v_0 g t^2 \cos \theta) \hat{k} - ((v_0^2 \sin \theta \cos \theta) t - \frac{1}{2} v_0 g t^2 \cos \theta) \hat{k}]$
$\vec{L} = m [v_0^2 \sin \theta \cos \theta t - v_0 g t^2 \cos \theta - v_0^2 \sin \theta \cos \theta t + \frac{1}{2} v_0 g t^2 \cos \theta] \hat{k}$
$\vec{L} = m [-\frac{1}{2} v_0 g t^2 \cos \theta] \hat{k} = -\frac{1}{2} mg v_0 t^2 \cos \theta \hat{k}$

(N/A) Let the particle with mass $m$ and velocity $v$ be at point $P$ at some instant $t$. We want to calculate the angular momentum of the particle about an arbitrary point $O$.
The angular momentum is given by $l = r \times mv$. Its magnitude is $mvr \sin \theta$,where $\theta$ is the angle between $r$ and $v$ as shown in the figure.
Although the particle changes position with time,the line of direction of $v$ remains the same. The perpendicular distance from $O$ to the line of motion of the particle is $d = r \sin \theta$,which is a constant.
Further,the direction of $l$ is perpendicular to the plane containing $r$ and $v$. This direction does not change with time. Thus,$l$ remains the same in magnitude and direction and is therefore conserved.

(A) The angular momentum $\vec{l}$ is defined as the cross product of the position vector $\vec{r}$ and the linear momentum vector $\vec{p}$:
$\vec{l} = \vec{r} \times \vec{p} = (x \hat{i} + y \hat{j} + z \hat{k}) \times (p_x \hat{i} + p_y \hat{j} + p_z \hat{k})$
Using the determinant form:
$\vec{l} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ x & y & z \\ p_x & p_y & p_z \end{vmatrix} = \hat{i}(y p_z - z p_y) - \hat{j}(x p_z - z p_x) + \hat{k}(x p_y - y p_x)$
Comparing components,we get:
$l_x = y p_z - z p_y$
$l_y = z p_x - x p_z$
$l_z = x p_y - y p_x$
If the particle moves only in the $x-y$ plane,then $z = 0$ and $p_z = 0$. Substituting these into the expressions:
$l_x = y(0) - (0)p_y = 0$
$l_y = (0)p_x - x(0) = 0$
$l_z = x p_y - y p_x$
Since $l_x = 0$ and $l_y = 0$,the angular momentum has only a $z$-component.

(N/A) Let at a certain instant two particles be at points $P$ and $Q$,as shown in the figure.
Angular momentum of the system about point $P$:
$\vec{L}_{P} = m v \times 0 + m v \times d = m v d$ ... $(i)$
Angular momentum of the system about point $Q$:
$\vec{L}_{Q} = m v \times d + m v \times 0 = m v d$ ... $(ii)$
Consider a point $R$,which is at a distance $y$ from point $Q$,i.e.,$QR = y$ and $PR = d - y$.
Angular momentum of the system about point $R$:
$\vec{L}_{R} = m v \times (d - y) + m v \times y$
$= m v d - m v y + m v y$
$= m v d$ ... $(iii)$
Comparing equations $(i)$,$(ii)$,and $(iii)$,we get:
$\vec{L}_{P} = \vec{L}_{Q} = \vec{L}_{R}$ ... $(iv)$
We infer from equation $(iv)$ that the angular momentum of a system of two particles moving in opposite directions along parallel lines does not depend on the point about which it is calculated.

(N/A) Just as the moment of a force is the rotational analogue of force,the quantity angular momentum is the rotational analogue of linear momentum.
In the figure,$Q$ is a particle of mass $m$,having position vector $\vec{OQ} = \vec{r}$ in a Cartesian coordinate system.
$\vec{v}$ is the linear velocity of the particle. So,its linear momentum is $\vec{p} = m\vec{v}$.
It is not necessary that the particle $Q$ belongs to a rigid body or that it moves along a curved path.
Let the angle between $\vec{r}$ and $\vec{p}$ be $\theta$.
The vector product of $\vec{r}$ and $\vec{p}$ is defined as the angular momentum $\vec{l}$ of the particle with respect to point $O$.
$\therefore \vec{l} = \vec{r} \times \vec{p}$
The unit of angular momentum is $kg \cdot m^2 \cdot s^{-1}$ or $J \cdot s$,and its dimensional formula is $[M^1 L^2 T^{-1}]$.
The magnitude of $\vec{l}$ depends on the selection of the reference point; therefore,while defining the angular momentum of a particle,it is necessary to mention the reference point.
The direction of $\vec{l}$ can be obtained with the help of the right-handed screw rule. Here,$\vec{l}$ is in the $OZ$ direction.
Now,$\vec{l} = \vec{r} \times \vec{p}$.
$\therefore |\vec{l}| = r p \sin \theta = p(r \sin \theta) = p(OR)$.
$\therefore$ Angular momentum of a particle = (magnitude of linear momentum) $\times$ (the perpendicular distance of the line of action of linear momentum from the reference point).

(N/A) The angular momentum $\vec{l}$ of a particle is defined as the cross product of its position vector $\vec{r}$ and linear momentum $\vec{p}$:
$\vec{l} = \vec{r} \times \vec{p}$
In Cartesian coordinates,$\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ and $\vec{p} = p_x\hat{i} + p_y\hat{j} + p_z\hat{k}$.
Calculating the cross product using the determinant form:
$\vec{l} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ x & y & z \\ p_x & p_y & p_z \end{vmatrix}$
Expanding the determinant:
$\vec{l} = \hat{i}(yp_z - zp_y) + \hat{j}(zp_x - xp_z) + \hat{k}(xp_y - yp_x)$
Comparing this with $\vec{l} = l_x\hat{i} + l_y\hat{j} + l_z\hat{k}$,we identify the Cartesian components:
$l_x = yp_z - zp_y$
$l_y = zp_x - xp_z$
$l_z = xp_y - yp_x$
These represent the components of angular momentum along the $X$,$Y$,and $Z$ axes,respectively.

The total angular momentum of a system of particles is the vector sum of the angular momenta of individual particles. For a system of $n$ particles,
$\overrightarrow{L} = \overrightarrow{l_{1}} + \overrightarrow{l_{2}} + \overrightarrow{l_{3}} + \ldots + \overrightarrow{l_{n}} = \sum_{i=1}^{n} \overrightarrow{l_{i}}$
where $\overrightarrow{l_{i}} = \overrightarrow{r_{i}} \times \overrightarrow{p_{i}}$ is the angular momentum of the $i^{\text{th}}$ particle,$\overrightarrow{r_{i}}$ is its position vector,and $\overrightarrow{p_{i}}$ is its linear momentum.
Differentiating the total angular momentum with respect to time $t$:
$\frac{d\overrightarrow{L}}{dt} = \sum_{i=1}^{n} \frac{d\overrightarrow{l_{i}}}{dt} = \sum_{i=1}^{n} \left( \frac{d\overrightarrow{r_{i}}}{dt} \times \overrightarrow{p_{i}} + \overrightarrow{r_{i}} \times \frac{d\overrightarrow{p_{i}}}{dt} \right)$
Since $\frac{d\overrightarrow{r_{i}}}{dt} = \overrightarrow{v_{i}}$ and $\overrightarrow{v_{i}} \times \overrightarrow{p_{i}} = \overrightarrow{v_{i}} \times (m\overrightarrow{v_{i}}) = 0$,the expression simplifies to:
$\frac{d\overrightarrow{L}}{dt} = \sum_{i=1}^{n} (\overrightarrow{r_{i}} \times \overrightarrow{F_{i}}) = \sum_{i=1}^{n} \overrightarrow{\tau_{i}} = \overrightarrow{\tau}_{ext} + \overrightarrow{\tau}_{int}$
Since internal forces occur in equal and opposite pairs along the same line of action (Newton's third law),their net torque $\overrightarrow{\tau}_{int} = 0$. Thus,the relation is:
$\frac{d\overrightarrow{L}}{dt} = \overrightarrow{\tau}_{ext}$

(N/A) Angular momentum $(L)$ of a particle with respect to a point is defined as the cross product of its position vector $(r)$ and its linear momentum $(p)$.
Mathematically,it is expressed as: $L = r \times p$.
Since $p = mv$,where $m$ is the mass and $v$ is the velocity of the particle,the angular momentum can also be written as: $L = r \times (mv) = m(r \times v)$.
The $SI$ unit of angular momentum is $\text{kg} \cdot \text{m}^2/\text{s}$ or $\text{J} \cdot \text{s}$.

(N/A) Angular momentum $(L)$ is defined as the product of the position vector $(r)$ and linear momentum $(p)$,given by the formula $L = r \times p$.
Since $p = mv$,the dimensions of angular momentum are $[M^1 L^1 T^{-1}] \times [L^1] = [M^1 L^2 T^{-1}]$.
The $SI$ unit of mass is $kg$,velocity is $m/s$,and distance is $m$.
Therefore,the $SI$ unit of angular momentum is $kg \cdot m^2/s$ or $J \cdot s$.

(N/A) For a rigid body rotating about a fixed $Z$-axis,the angular momentum of an individual particle $i$ is given by $\vec{l}_i = \vec{r}_i \times \vec{p}_i$.
The total angular momentum $\vec{L}$ of the rigid body is the vector sum of the angular momenta of all its particles:
$\vec{L} = \sum_{i=1}^{n} \vec{l}_i = \sum_{i=1}^{n} (\vec{r}_i \times \vec{p}_i)$.
Since $\vec{p}_i = m_i \vec{v}_i$ and for rotational motion $\vec{v}_i = \vec{\omega} \times \vec{r}_i$,we have:
$\vec{L} = \sum_{i=1}^{n} [\vec{r}_i \times (m_i (\vec{\omega} \times \vec{r}_i))]$.
For rotation about a fixed axis,the angular velocity $\vec{\omega}$ is constant for all particles and directed along the axis of rotation (let it be the $Z$-axis,$\vec{\omega} = \omega \hat{k}$).
The component of angular momentum along the $Z$-axis is:
$L_Z = \sum_{i=1}^{n} m_i r_i^2 \omega = (\sum_{i=1}^{n} m_i r_i^2) \omega$.
Since the moment of inertia $I = \sum_{i=1}^{n} m_i r_i^2$,we get:
$L_Z = I \omega$.
In vector form,$\vec{L} = I \vec{\omega}$,which is analogous to $\vec{p} = m \vec{v}$ in linear motion.

(N/A) For a rigid body rotating about a fixed axis,the total angular momentum $L$ is given by the product of the moment of inertia $I$ of the body about that axis and its angular velocity $\omega$.
The formula is: $L = I \omega$
Where:
$L$ is the angular momentum,
$I$ is the moment of inertia about the axis of rotation,
$\omega$ is the angular velocity of the body.

(A) In rotational motion about a fixed axis,every particle of the rigid body moves in a circular path in a plane perpendicular to the axis of rotation.
Let the axis of rotation be the $z$-axis. The position vector $\vec{r}$ of a particle can be decomposed into two components: $\vec{r}_{\parallel}$ (along the $z$-axis) and $\vec{r}_{\perp}$ (in the $xy$-plane).
The velocity vector $\vec{v}$ of any particle is given by $\vec{v} = \vec{\omega} \times \vec{r}$. Since $\vec{\omega}$ is along the $z$-axis,$\vec{v}$ lies entirely in the $xy$-plane.
The angular momentum is $\vec{L} = \sum (\vec{r} \times \vec{p}) = \sum m(\vec{r} \times \vec{v})$.
Substituting $\vec{r} = \vec{r}_{\parallel} + \vec{r}_{\perp}$,we get $\vec{L} = \sum m((\vec{r}_{\parallel} + \vec{r}_{\perp}) \times \vec{v})$.
Since $\vec{r}_{\parallel}$ is parallel to the $z$-axis and $\vec{v}$ is in the $xy$-plane,their cross product $\vec{r}_{\parallel} \times \vec{v}$ results in a vector perpendicular to the $z$-axis.
However,for a symmetric body or when summing over the entire rigid body,the components of angular momentum perpendicular to the axis of rotation cancel out due to the circular symmetry of the motion,leaving only the component along the axis of rotation $(L_z)$.
Thus,the net angular momentum perpendicular to the axis,${L_ \bot }$,is zero.

(A) Yes,an object moving in a straight line can have angular momentum relative to a point that is not on the line of motion. The formula for angular momentum is $\vec{L} = \vec{r} \times \vec{p}$,where $\vec{r}$ is the position vector from the reference point to the object and $\vec{p}$ is the linear momentum. As long as the line of motion does not pass through the reference point,the cross product $\vec{r} \times \vec{p}$ will be non-zero.

(A) Yes,the angular momentum of a body changes when its axis of rotation changes.
This is because angular momentum is defined as $\vec{L} = \vec{r} \times \vec{p}$.
When the axis of rotation changes,the position vector $\vec{r}$ relative to the new axis changes,and the direction of the angular velocity vector $\vec{\omega}$ also changes.
Since $\vec{L} = I\vec{\omega}$,a change in the axis of rotation alters the moment of inertia $I$ and the direction of $\vec{\omega}$,thereby changing the angular momentum $\vec{L}$.

(B) The angular momentum $\vec{L}$ of a particle is defined as $\vec{L} = \vec{r} \times \vec{p}$,where $\vec{r}$ is the position vector and $\vec{p}$ is the linear momentum.
For the magnitude of angular momentum to be zero,the cross product $\vec{r} \times \vec{p}$ must be zero.
This occurs when the position vector $\vec{r}$ is parallel or anti-parallel to the linear momentum vector $\vec{p}$,or when either $\vec{r}$ or $\vec{p}$ is zero.
Geometrically,this means the line of action of the linear momentum passes through the origin (the reference point for the position vector).

(D) The angular momentum is defined as $\vec{L} = \vec{r} \times \vec{p} = m(\vec{r} \times \vec{v})$.
For point $A$ (the center of the circular path): The position vector $\vec{r}_A$ is in the plane of the circle,and the velocity vector $\vec{v}$ is tangential to the circle. The cross product $\vec{r}_A \times \vec{v}$ always points in the positive $z$-direction (perpendicular to the plane of the circle). Since the speed $v$ and radius $r$ are constant,the magnitude $L_A = mvr$ is constant. Thus,$L_A$ is constant in both magnitude and direction.
For point $B$ (a point on the axis of rotation above $A$): The position vector $\vec{r}_B$ from $B$ to $M$ changes its direction as the mass $M$ rotates. Therefore,the angular momentum $\vec{L}_B = \vec{r}_B \times \vec{p}$ will have a constant magnitude,but its direction will change continuously as it precesses around the $z$-axis.

(B) The angular momentum $\vec{L}$ of a particle is given by $\vec{L} = \vec{r} \times \vec{p} = \vec{r} \times (m\vec{v})$.
For a particle moving with constant speed $v$ in a circular path of radius $r$,the magnitude of angular momentum is $L = mvr \sin(90^{\circ}) = mvr$,which is constant.
The direction of angular momentum is given by the right-hand rule,which is perpendicular to the plane of the circular path (along the axis of rotation).
Since the speed and radius are constant and the plane of motion does not change,both the magnitude and the direction of the angular momentum vector remain constant throughout the motion.

(B) The position vector is $\overrightarrow{r} = 10 \alpha t^2 \hat{i} + 5 \beta (t - 5) \hat{j}$.
The velocity vector is $\overrightarrow{v} = \frac{d\overrightarrow{r}}{dt} = 20 \alpha t \hat{i} + 5 \beta \hat{j}$.
The angular momentum is given by $\overrightarrow{L} = m (\overrightarrow{r} \times \overrightarrow{v})$.
$\overrightarrow{L} = m [10 \alpha t^2 \hat{i} + 5 \beta (t - 5) \hat{j}] \times [20 \alpha t \hat{i} + 5 \beta \hat{j}]$.
Calculating the cross product: $\overrightarrow{L} = m [ (10 \alpha t^2)(5 \beta) \hat{k} - (5 \beta (t - 5))(20 \alpha t) \hat{k} ]$.
$\overrightarrow{L} = m [ 50 \alpha \beta t^2 - 100 \alpha \beta t (t - 5) ] \hat{k}$.
At $t = 0$,$\overrightarrow{L} = m [ 0 - 0 ] \hat{k} = 0$.
We want $\overrightarrow{L} = 0$ at time $t > 0$:
$50 \alpha \beta t^2 - 100 \alpha \beta t (t - 5) = 0$.
Dividing by $50 \alpha \beta t$ (since $t \neq 0$):
$t - 2(t - 5) = 0$.
$t - 2t + 10 = 0 \implies -t = -10 \implies t = 10 \text{ seconds}$.

(B) The angular momentum $\overrightarrow{L}$ is given by the cross product of the position vector $\overrightarrow{r}$ and the linear momentum $\overrightarrow{p} = m\overrightarrow{v}$.
Given $m = 1\,kg$,$\overrightarrow{r} = (3\hat{i} - \hat{j})\,m$,and $\overrightarrow{v} = (3\hat{j} + \hat{k})\,m/s$.
$\overrightarrow{L} = \overrightarrow{r} \times (m\overrightarrow{v}) = 1 \cdot [(3\hat{i} - \hat{j}) \times (3\hat{j} + \hat{k})]$.
Calculating the cross product:
$\overrightarrow{L} = 3\hat{i} \times 3\hat{j} + 3\hat{i} \times \hat{k} - \hat{j} \times 3\hat{j} - \hat{j} \times \hat{k}$.
Using unit vector properties ($\hat{i} \times \hat{j} = \hat{k}$,$\hat{i} \times \hat{k} = -\hat{j}$,$\hat{j} \times \hat{j} = 0$,$\hat{j} \times \hat{k} = \hat{i}$):
$\overrightarrow{L} = 9\hat{k} - 3\hat{j} - 0 - \hat{i} = -\hat{i} - 3\hat{j} + 9\hat{k}$.
The magnitude is $|\overrightarrow{L}| = \sqrt{(-1)^2 + (-3)^2 + (9)^2} = \sqrt{1 + 9 + 81} = \sqrt{91}$.
Thus,$x = 91$.

(A) The angular momentum $\vec{L}$ of a particle with respect to the origin is given by $\vec{L} = \vec{r} \times \vec{p} = \vec{r} \times (m\vec{v})$.
For a particle moving in a circle with uniform speed $v$,the magnitude of angular momentum is $L = mvr_{\perp}$,where $r_{\perp}$ is the perpendicular distance from the origin to the line of velocity. Since the circle is not centered at the origin,the position vector $\vec{r}$ relative to the origin changes continuously as the particle moves along the circle.
Consequently,both the magnitude and the direction of the cross product $\vec{r} \times \vec{v}$ change as the particle rotates. Therefore,the angular momentum is variable in both magnitude and direction.

(B) The angular momentum $\vec{L}$ of a particle about the origin is given by the cross product of its position vector $\vec{r}$ and its linear momentum $\vec{p} = m\vec{v}$.
$\vec{L} = \vec{r} \times \vec{p} = \vec{r} \times (m\vec{v})$.
Let the position of the particle be $\vec{r} = x \hat{i} + b \hat{j}$,where $x$ is the horizontal distance from the $y$-axis and $b$ is the constant vertical distance from the $x$-axis.
The velocity of the particle is $\vec{v} = v \hat{i}$ (since it moves parallel to the $x$-axis).
Therefore,$\vec{L} = (x \hat{i} + b \hat{j}) \times (m v \hat{i})$.
Using the distributive property of the cross product:
$\vec{L} = (x \hat{i} \times m v \hat{i}) + (b \hat{j} \times m v \hat{i})$.
Since $\hat{i} \times \hat{i} = 0$ and $\hat{j} \times \hat{i} = -\hat{k}$,we get:
$\vec{L} = 0 + m v b (-\hat{k}) = -m v b \hat{k}$.
Thus,the angular momentum is $-m v b \hat{k}$.

(C) The angular impulse applied to a body is equal to the change in its angular momentum,given by $\tau \Delta t = \Delta L = I \Delta \omega$.
For a hollow cylinder,the moment of inertia $I$ is $MR^2$.
Given: Mass $M = 2 \,kg$,Radius $R = 20 \,cm = 0.2 \,m$,Angular impulse $= 20 \,Nms$.
The moment of inertia $I = MR^2 = 2 \times (0.2)^2 = 2 \times 0.04 = 0.08 \,kg \cdot m^2$.
Using the relation: $20 = I \Delta \omega = 0.08 \times \Delta \omega$.
$\Delta \omega = \frac{20}{0.08} = \frac{2000}{8} = 250 \,rad/s$.

(A) The angular momentum $L$ of a particle about a point is given by the cross product of its position vector $r$ and its linear momentum $p = mv$.
Mathematically,$L = r \times p$.
The magnitude of angular momentum is given by $L = m v r_{\perp}$,where $r_{\perp}$ is the perpendicular distance of the line of motion of the particle from the origin $O$.
From the given figure,the particle is moving parallel to the $x$-axis at a constant perpendicular distance $b$ from the $x$-axis (and thus from the origin $O$ along the $y$-direction).
Therefore,the perpendicular distance $r_{\perp} = b$.
Substituting this into the formula,we get $L = m v b$.
Thus,the correct option is $A$.

(B) The angular momentum $\vec{L}$ of a particle about the origin is given by $\vec{L} = \vec{r} \times \vec{p} = \vec{r} \times (m\vec{v})$.
Given the initial position $\vec{r}_0 = 8 \hat{j} \, m$ and velocity $\vec{v} = 3 \hat{i} \, m/s$.
The position of the particle at time $t = 5 \, s$ is $\vec{r} = \vec{r}_0 + \vec{v}t = 8 \hat{j} + (3 \hat{i})(5) = 15 \hat{i} + 8 \hat{j} \, m$.
The momentum of the particle is $\vec{p} = m\vec{v} = (1)(3 \hat{i}) = 3 \hat{i} \, kg \cdot m/s$.
Now,$\vec{L} = (15 \hat{i} + 8 \hat{j}) \times (3 \hat{i}) = 15 \times 3 (\hat{i} \times \hat{i}) + 8 \times 3 (\hat{j} \times \hat{i})$.
Since $\hat{i} \times \hat{i} = 0$ and $\hat{j} \times \hat{i} = -\hat{k}$,we get $\vec{L} = 0 + 24(-\hat{k}) = -24 \hat{k} \, kg \cdot m^2/s$.

(A) The initial velocity components are $u_x = u \cos 45^{\circ} = 20 \sqrt{2} \times \frac{1}{\sqrt{2}} = 20 \,m/s$ and $u_y = u \sin 45^{\circ} = 20 \sqrt{2} \times \frac{1}{\sqrt{2}} = 20 \,m/s$.
After $t = 2 \,s$,the position coordinates are:
$x = u_x t = 20 \times 2 = 40 \,m$
$y = u_y t - \frac{1}{2} g t^2 = 20 \times 2 - \frac{1}{2} \times 10 \times 2^2 = 40 - 20 = 20 \,m$.
The velocity components at $t = 2 \,s$ are:
$v_x = u_x = 20 \,m/s$
$v_y = u_y - gt = 20 - 10 \times 2 = 0 \,m/s$.
The angular momentum $\vec{L}$ about the origin is given by $\vec{L} = \vec{r} \times \vec{p} = m(\vec{r} \times \vec{v})$.
$\vec{r} = 40 \hat{i} + 20 \hat{j}$ and $\vec{v} = 20 \hat{i} + 0 \hat{j}$.
$\vec{L} = 1 \times [(40 \hat{i} + 20 \hat{j}) \times (20 \hat{i})] = 1 \times [40 \times 20 (\hat{i} \times \hat{i}) + 20 \times 20 (\hat{j} \times \hat{i})]$.
Since $\hat{i} \times \hat{i} = 0$ and $\hat{j} \times \hat{i} = -\hat{k}$,we get:
$\vec{L} = 400(-\hat{k}) = -400 \hat{k} \,kg \cdot m^2/s$.

(B) The angular momentum $L$ of a particle about the origin is given by $L = \vec{r} \times \vec{p} = m(\vec{r} \times \vec{v})$.
Alternatively,the torque about the origin is $\vec{\tau} = \vec{r} \times \vec{F} = \vec{r} \times (m\vec{g})$.
Since $\vec{g}$ acts downwards ($-y$ direction),$\vec{\tau} = (x\hat{i} + y\hat{j}) \times (-mg\hat{j}) = -xmg\hat{k}$.
The magnitude of torque is $\tau = xmg = (v_x t)mg$.
Integrating torque to find angular momentum: $L = \int_0^t \tau dt = \int_0^t (v_x t)mg dt = mg v_x \frac{t^2}{2}$.
Given: $m = 100\,g = 0.1\,kg$,$v = 20\,ms^{-1}$,$\theta = 45^{\circ}$,$t = 2\,s$,$g = 10\,ms^{-2}$.
$v_x = v \cos 45^{\circ} = 20 \times \frac{1}{\sqrt{2}} = 10\sqrt{2}\,ms^{-1}$.
$L = (0.1)(10)(10\sqrt{2}) \frac{2^2}{2} = 10 \times 10\sqrt{2} \times 2 = 200\sqrt{2} = \sqrt{40000 \times 2} = \sqrt{80000}$.
Wait,re-evaluating: $L = (0.1)(10)(10\sqrt{2}) \times 2 = 20\sqrt{2} = \sqrt{400 \times 2} = \sqrt{800}$.
Thus,$K = 800$.

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

(C) The angular momentum about a point is given by $\vec{L} = \vec{r} \times \vec{p} = \vec{r} \times (m\vec{v})$.
For point $O$ (the center of the circle),the position vector $\vec{r}$ is in the $x-y$ plane and the velocity $\vec{v}$ is tangential to the circle. The angular momentum $\vec{L}_O = \vec{r} \times m\vec{v}$ is directed along the $z$-axis (perpendicular to the plane of motion). Since the speed and radius are constant,the magnitude and direction of $\vec{L}_O$ remain constant.
For point $P$ (a point on the $z$-axis),the position vector $\vec{r}'$ from $P$ to the mass $m$ changes direction as the mass moves in a circle. Consequently,the cross product $\vec{L}_P = \vec{r}' \times m\vec{v}$ changes its direction continuously as the mass rotates,even though its magnitude remains constant. Therefore,$\vec{L}_P$ varies with time.
Thus,$\vec{L}_O$ remains constant while $\vec{L}_P$ varies with time.

(C) Let the mass of the rod be $m$. The distance of the center of mass of the rod from the pivot $O$ is $r_{cm} = L + L/2 = 3L/2$.
Linear impulse $P = \Delta p = m v_{cm} = m (\omega r_{cm}) = m \omega (3L/2)$.
Thus, $P = \frac{3}{2} m \omega L$ --- $(i)$
Angular impulse about pivot $O$ is $J_{\theta} = \int \tau dt = \Delta L_{ang}$.
The impulse $P$ is applied at a distance $r = L + L/2 - x = 3L/2 - x$ from the pivot $O$ (based on the figure, the impulse is at $x$ from the midpoint towards the pivot end).
$P (3L/2 - x) = I_O \omega$, where $I_O$ is the moment of inertia about $O$.
$I_O = I_{cm} + m(r_{cm})^2 = \frac{mL^2}{12} + m(3L/2)^2 = mL^2 (\frac{1}{12} + \frac{9}{4}) = mL^2 (\frac{1+27}{12}) = \frac{28}{12} mL^2 = \frac{7}{3} mL^2$.
So, $P (3L/2 - x) = (\frac{7}{3} mL^2) \omega$ --- $(ii)$
Dividing $(ii)$ by $(i)$:
$\frac{P(3L/2 - x)}{P} = \frac{(7/3) mL^2 \omega}{(3/2) m \omega L}$
$3L/2 - x = \frac{7}{3} \cdot \frac{2}{3} L = \frac{14}{9} L$
$x = \frac{3}{2} L - \frac{14}{9} L = \frac{27-28}{18} L$. Since $x$ is a distance, we take the magnitude: $x = L/18$.
Comparing $x = L/n$ with $x = L/18$, we get $n = 18$.

(D) The initial velocity components are $v_x = v_0 \cos 45^{\circ} = \frac{v_0}{\sqrt{2}}$ and $v_y = v_0 \sin 45^{\circ} = \frac{v_0}{\sqrt{2}}$.
At maximum height $H$,the vertical velocity component is zero,and the horizontal velocity is $v_x = \frac{v_0}{\sqrt{2}}$.
The maximum height $H$ is given by $H = \frac{v_y^2}{2g} = \frac{(v_0/\sqrt{2})^2}{2g} = \frac{v_0^2}{4g}$.
The angular momentum $\vec{L}$ with respect to the origin is $\vec{L} = \vec{r} \times \vec{p} = m(\vec{r} \times \vec{v})$.
At maximum height,the position vector is $\vec{r} = x\hat{i} + H\hat{j}$ and the velocity vector is $\vec{v} = v_x\hat{i}$.
Thus,$\vec{L} = m(x\hat{i} + H\hat{j}) \times (v_x\hat{i}) = mH v_x (\hat{j} \times \hat{i}) = -mH v_x \hat{k}$.
Substituting the values: $L = m \left( \frac{v_0^2}{4g} \right) \left( \frac{v_0}{\sqrt{2}} \right) = \frac{m v_0^3}{4 \sqrt{2} g}$.
The direction is along the negative $z$-axis $(- \hat{k})$.