Angular Momentum and Angular Impulse Questions in English

(C) The angular momentum of a particle about a point is given by $\vec{L} = \vec{r} \times \vec{p} = m(\vec{r} \times \vec{v})$.
For each mass,the perpendicular distance $r_{\perp}$ from the centroid of the equilateral triangle to the velocity vector (which lies along the side) is the distance from the centroid to the side.
In an equilateral triangle of side $a$,the distance from the centroid to any side is $r_{\perp} = \frac{a}{2\sqrt{3}}$.
The velocity of each mass is $V_0$ directed along the sides of the triangle.
The angular momentum of one mass about the centroid is $L_1 = m V_0 r_{\perp} = m V_0 \left( \frac{a}{2\sqrt{3}} \right)$.
Since the velocities are directed such that all three masses contribute to the angular momentum in the same rotational sense (clockwise or counter-clockwise),the total angular momentum is $L = 3 \times L_1$.
$L = 3 \times m V_0 \left( \frac{a}{2\sqrt{3}} \right) = \frac{3}{2\sqrt{3}} m V_0 a = \frac{\sqrt{3}}{2} m V_0 a$.

(A) The position vector of the particle is $\vec{r} = (1 \hat{i} + 1 \hat{j} + 0 \hat{k}) \ m$.
The velocity vector is $\vec{v} = (4 \hat{i} + 2 \hat{j}) \ m/s$ and mass $m = 4 \ kg$.
The angular momentum $\vec{L}$ is given by the cross product $\vec{L} = \vec{r} \times \vec{p} = \vec{r} \times (m \vec{v})$.
Substituting the values: $\vec{L} = (\hat{i} + \hat{j}) \times [4 \times (4 \hat{i} + 2 \hat{j})]$.
$\vec{L} = (\hat{i} + \hat{j}) \times (16 \hat{i} + 8 \hat{j})$.
Calculating the cross product: $\vec{L} = 16(\hat{i} \times \hat{i}) + 8(\hat{i} \times \hat{j}) + 16(\hat{j} \times \hat{i}) + 8(\hat{j} \times \hat{j})$.
Since $\hat{i} \times \hat{i} = 0$,$\hat{j} \times \hat{j} = 0$,$\hat{i} \times \hat{j} = \hat{k}$,and $\hat{j} \times \hat{i} = -\hat{k}$.
$\vec{L} = 0 + 8(\hat{k}) + 16(-\hat{k}) + 0 = 8 \hat{k} - 16 \hat{k} = -8 \hat{k} \ kg \cdot m^2/s$.

(C) For a particle moving in a circular path,the angular momentum is defined as $\vec{L} = \vec{r} \times \vec{p}$.
The position vector $\vec{r}$ and the linear momentum vector $\vec{p}$ both lie in the plane of the circular motion.
According to the right-hand rule,the direction of the angular momentum vector $\vec{L}$ is perpendicular to the plane of motion.
Since the particle is constrained to move in a fixed circular path,the plane of motion does not change.
Therefore,the direction of the angular momentum vector remains constant throughout the motion,regardless of the change in speed.

(B) The angular momentum $L$ is given by the formula $L = I \omega$,where $I$ is the moment of inertia and $\omega$ is the angular velocity.
For a solid sphere,the moment of inertia about its axis of rotation is $I = \frac{2}{5} MR^2$.
The angular velocity $\omega$ in terms of the period of rotation $T$ is given by $\omega = \frac{2 \pi}{T}$.
Substituting these values into the angular momentum formula:
$L = I \omega = \left( \frac{2}{5} MR^2 \right) \left( \frac{2 \pi}{T} \right)$.
Therefore,$L = \frac{4 \pi MR^2}{5 T}$.

(A) The angular momentum $L$ of a particle with respect to the origin is given by the cross product of its position vector $r$ and its linear momentum $p = Mv$.
$L = r \times (Mv)$
Since the mass $M$ is moving with a constant velocity $v$ parallel to the $X$-axis, we can write its position as $r = xi + yj$, where $y$ is the constant perpendicular distance from the $X$-axis.
The velocity vector is $v = vi$.
The angular momentum is $L = (xi + yj) \times (Mvi) = Mv(xi \times i) + Mv(yj \times i)$.
Since $i \times i = 0$ and $j \times i = -k$, we get $L = -Mvyk$.
As $M$, $v$, and $y$ are all constants, the magnitude and direction of the angular momentum $L$ remain constant over time.

(B) The correct option is $B$.
Concept: Angular momentum $L$ of a particle of mass $m$ moving in a circle of radius $r$ with angular velocity $\omega$ is given by $L = I\omega = mr^2\omega$.
Given that the mass $m$ and angular velocity $\omega$ remain constant.
Initially,$L = m\omega r^2$.
When the length of the string is halved,the new radius becomes $r' = \frac{r}{2}$.
The new angular momentum $L'$ is given by:
$L' = m\omega(r')^2 = m\omega\left(\frac{r}{2}\right)^2$
$L' = m\omega\left(\frac{r^2}{4}\right) = \frac{1}{4}(mr^2\omega)$
$L' = \frac{L}{4}$.

(A) Concept: The angular impulse applied to the rod is equal to the change in its angular momentum.
Angular impulse about the center of mass $O$ is given by $J_{\theta} = P \cdot \frac{l}{2}$.
The angular momentum of the rod about its center of mass is $L = I \omega$,where $I = \frac{m l^2}{12}$ is the moment of inertia of the rod about its center.
Equating angular impulse to change in angular momentum:
$P \cdot \frac{l}{2} = I \omega$
$P \cdot \frac{l}{2} = \left( \frac{m l^2}{12} \right) \omega$
Solving for angular velocity $\omega$:
$\omega = \frac{P \cdot l}{2} \cdot \frac{12}{m l^2} = \frac{6 P}{m l}$
The rod rotates with a constant angular velocity $\omega$. The time $\Delta t$ taken to turn through an angle $\Delta \theta = \frac{\pi}{2}$ is:
$\Delta t = \frac{\Delta \theta}{\omega} = \frac{\pi / 2}{6 P / (m l)} = \frac{\pi}{2} \cdot \frac{m l}{6 P} = \frac{\pi m l}{12 P}$

(B) 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.
From this,we can write $L = \sqrt{2IE}$.
Given $I_1 = \frac{I_2}{3} \implies I_2 = 3I_1$ and $E_1 = 27E_2$.
The ratio of angular momenta is $\frac{L_1}{L_2} = \frac{\sqrt{2I_1E_1}}{\sqrt{2I_2E_2}}$.
Substituting the given values:
$\frac{L_1}{L_2} = \sqrt{\frac{I_1}{I_2} \cdot \frac{E_1}{E_2}} = \sqrt{\frac{I_1}{3I_1} \cdot \frac{27E_2}{E_2}}$.
$\frac{L_1}{L_2} = \sqrt{\frac{1}{3} \cdot 27} = \sqrt{9} = 3$.
Thus,the ratio $L_1 : L_2$ is $3: 1$.

(B) The angular momentum of a particle is defined by the cross product of its position vector and linear momentum: $\overrightarrow{L} = \overrightarrow{r} \times \overrightarrow{p}$.
By the properties of the cross product,the vector $\overrightarrow{L}$ is perpendicular to both $\overrightarrow{r}$ and $\overrightarrow{p}$.
The magnitude of the angular momentum is given by $L = rp \sin \theta$,where $\theta$ is the angle between $\overrightarrow{r}$ and $\overrightarrow{p}$.
For the magnitude $L$ to be maximum,$\sin \theta$ must be maximum,which occurs when $\theta = 90^{\circ}$.
Therefore,$\overrightarrow{L}$ is maximum when $\overrightarrow{p}$ is perpendicular to $\overrightarrow{r}$.

(A) The relationship between torque $(\tau)$ and the change in angular momentum $(\Delta L)$ is given by the angular impulse-momentum theorem: $\Delta L = \int \tau dt$.
Since the torque is constant, the change in angular momentum is simply the product of torque and time: $\Delta L = \tau \times \Delta t$.
Given:
Torque $(\tau)$ = $200 \, N-m$
Time $(\Delta t)$ = $4 \, s$
Calculation:
$\Delta L = 200 \, N-m \times 4 \, s = 800 \, kg-m^{2}/s$.
Therefore, the change in angular momentum is $800 \, kg-m^{2}/s$.

(A) The relationship between torque $\tau$ and the change in angular momentum $\Delta L$ is given by the impulse-momentum theorem for rotation: $\Delta L = \int \tau \ dt$.
Since the torque $\tau = 50 \ Nm$ is constant and acts for a time interval $\Delta t = 8 \ s$,the change in angular momentum is:
$\Delta L = \tau \times \Delta t$
$\Delta L = 50 \ Nm \times 8 \ s$
$\Delta L = 400 \ kg \cdot m^2/s$.
Therefore,the change in angular momentum is $400 \ kg \cdot m^2/s$.

(A) The angular momentum $L$ of a particle about a point $O$ is given by the cross product $L = \vec{r} \times \vec{p} = \vec{r} \times (m\vec{v})$.
The magnitude of angular momentum is $L = mvr \sin(\theta)$,where $r$ is the position vector of the particle from the origin $O$,$v$ is the velocity,and $\theta$ is the angle between $\vec{r}$ and $\vec{v}$.
This can also be written as $L = mv d$,where $d = r \sin(\theta)$ 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,the perpendicular distance $d$ from the origin $O$ to the line of motion remains constant throughout the motion.
Since the mass $m$ and the speed $v$ are also constant (uniform motion),the product $mvd$ remains constant.
Therefore,the angular momentum of the particle about $O$ remains constant during the motion from $A$ to $B$.

(C) The equation of the line is $y = x + a$. Comparing this with the slope-intercept form $y = mx + c$,we get the slope $m_{slope} = 1$. Since $\tan \theta = m_{slope} = 1$,the angle $\theta = 45^{\circ}$.
The velocity vector $\vec{v}$ makes an angle of $45^{\circ}$ with the positive $x$-axis. Thus,$\vec{v} = v \cos 45^{\circ} \hat{i} + v \sin 45^{\circ} \hat{j} = \frac{v}{\sqrt{2}} (\hat{i} + \hat{j})$.
The angular momentum $\vec{L}$ about the origin is given by $\vec{L} = \vec{r} \times \vec{p} = m (\vec{r} \times \vec{v})$.
The perpendicular distance $d$ from the origin $(0, 0)$ to the line $x - y + a = 0$ is given by $d = \frac{|(1)(0) - (1)(0) + a|}{\sqrt{1^2 + (-1)^2}} = \frac{a}{\sqrt{2}}$.
The magnitude of angular momentum is $L = mvd = m v \left( \frac{a}{\sqrt{2}} \right) = \frac{mva}{\sqrt{2}}$.

(B) Given: Diameter $D = 1 \,m$,so radius $r = 0.5 \,m$. Mass $m = 20 \,kg$. Frequency $n = 120$ rotations per minute $= 2$ rotations per second.
The moment of inertia $I$ of a flywheel (disk) about its axis is $I = \frac{1}{2}mr^2$.
The angular velocity $\omega = 2\pi n = 2 \times 3.14 \times 2 = 12.56 \,rad/s$.
Angular momentum $L = I\omega = (\frac{1}{2}mr^2) \times (2\pi n)$.
Substituting the values: $L = \frac{20 \times (0.5)^2}{2} \times 2 \times 3.14 \times 2$.
$L = 5 \times 3.14 \times 2 = 31.4 \,kg-m^2/s$.

(C) The angular momentum $\vec{L}$ of a particle about the origin is given by $\vec{L} = \vec{r} \times \vec{p}$.
At the maximum height,the velocity of the particle is purely horizontal,given by $v_x = u \cos \theta$.
The position vector of the particle at maximum height is $\vec{r} = x \hat{i} + h_{\max} \hat{j}$,where $h_{\max} = \frac{u^2 \sin^2 \theta}{2g}$.
The momentum is $\vec{p} = m v_x \hat{i} = m u \cos \theta \hat{i}$.
The angular momentum is $\vec{L} = (x \hat{i} + h_{\max} \hat{j}) \times (m u \cos \theta \hat{i}) = -m u \cos \theta h_{\max} \hat{k}$.
The magnitude is $L = m u \cos \theta \left( \frac{u^2 \sin^2 \theta}{2g} \right) = \frac{m u^3 \sin^2 \theta \cos \theta}{2g}$.
Thus,$L \propto u^3$.

(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 $\vec{r} = x(t) \hat{i} + b \hat{j}$ and $\vec{v} = v \hat{i}$.
$\vec{L} = (x(t) \hat{i} + b \hat{j}) \times (m v \hat{i}) = x(t) m v (\hat{i} \times \hat{i}) + b m v (\hat{j} \times \hat{i})$.
Since $\hat{i} \times \hat{i} = 0$ and $\hat{j} \times \hat{i} = -\hat{k}$,we get $\vec{L} = -b m v \hat{k}$.
The magnitude of the angular momentum is $|\vec{L}| = | -b m v | = b m v$.
Since $b$,$m$,and $v$ are all constants,the magnitude of the angular momentum $|\vec{L}|$ is constant and independent of the angle $\theta$.
Therefore,the graph of $|\vec{L}|$ versus $\theta$ is a horizontal straight line.

(A) The angular momentum $L$ of a particle about a point $O$ is given by $L = \vec{r} \times \vec{p} = m(\vec{r} \times \vec{v})$. The magnitude is $L = mvr \sin \theta$,where $r \sin \theta$ is the perpendicular distance from $O$ to the line of motion.
For particle $A$: $m_A = 6.5 \ kg$,$v_A = 2.2 \ m/s$,and the perpendicular distance $r_A = 1.5 \ m$. The motion is clockwise,so $L_A = -m_A v_A r_A = -(6.5 \times 2.2 \times 1.5) = -21.45 \ kg \ m^2/s$.
For particle $B$: $m_B = 3.1 \ kg$,$v_B = 3.6 \ m/s$,and the perpendicular distance $r_B = 2.8 \ m$. The motion is counter-clockwise,so $L_B = +m_B v_B r_B = +(3.1 \times 3.6 \times 2.8) = +31.248 \ kg \ m^2/s$.
The total angular momentum is $L = L_A + L_B = -21.45 + 31.248 = 9.798 \ kg \ m^2/s \approx 9.8 \ kg \ m^2/s$.

(D) The position vector is given by $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$.
When $\vec{r} \rightarrow -\vec{r}$,the velocity vector $\vec{v} = \frac{d\vec{r}}{dt}$ also changes sign to $-\vec{v}$.
Linear momentum $\vec{p} = m\vec{v}$ changes sign to $-m\vec{v} = -\vec{p}$.
Acceleration $\vec{a} = \frac{d\vec{v}}{dt}$ changes sign to $-\vec{a}$.
Angular momentum is defined as $\vec{L} = \vec{r} \times \vec{p} = \vec{r} \times (m\vec{v})$.
When $\vec{r} \rightarrow -\vec{r}$ and $\vec{v} \rightarrow -\vec{v}$,the new angular momentum $\vec{L}' = (-\vec{r}) \times (-m\vec{v}) = (-1)(-1)(\vec{r} \times m\vec{v}) = \vec{L}$.
Thus,the angular momentum does not flip its sign.

(B) The angular momentum of a particle relative to another particle is given by $L = m \cdot v_{\text{rel}} \cdot r_{\perp}$.
Here,the mass of car $A$ is $m = 10^3 \text{ kg}$.
The relative velocity of car $A$ with respect to car $B$ is $v_{\text{rel}} = v_A - v_B = 72 \text{ km/h} - 36 \text{ km/h} = 36 \text{ km/h}$.
Converting the relative velocity to $SI$ units: $v_{\text{rel}} = 36 \times \frac{5}{18} \text{ m/s} = 10 \text{ m/s}$.
The perpendicular distance between the tracks is $r_{\perp} = 10 \text{ m}$.
Substituting these values into the formula: $L = 1000 \times 10 \times 10 = 10^5 \text{ kg} \cdot \text{m}^2/\text{s}$ (or $\text{J} \cdot \text{s}$).