Class 11 Physics (Physics Part-I) Chapter 6 Systems Of Particles And Rotational Motion

Key Concept: Formula Application

a) (3, 3)
[Solution Description] For three particles of equal mass, the center of mass $X, Y$ is calculated as $X = \frac{x_1 + x_2 + x_3}{3}$, $Y = \frac{y_1 + y_2 + y_3}{3}$. Substituting the given points: $X = \frac{1 + 3 + 5}{3} = 3$, $Y = \frac{1 + 3 + 5}{3} = 3$. Hence, the center of mass is at (3, 3).

Key Concept: Angular Displacement

b) 6 rad/s
[Solution Description] Angular velocity ($\omega$) is given by the formula: $\omega = \frac{\Delta \theta}{\Delta t}$. Here, $\Delta \theta = 30$ radians and $\Delta t = 5$ seconds.
Therefore, $\omega = \frac{30}{5} = 6$ rad/s.

Key Concept: Fixed Axis Concept

b) They remain stationary

[Solution Description]

For a rigid body rotating about a fixed axis, the particles on the axis of rotation do not experience any linear movement and therefore remain stationary. Thus, their linear velocity is zero (v = 0).

Key Concept: Angular Displacement

c) $30 \, \text{rad}$
[Solution Description]
The relation between angular displacement $\Delta \theta$, angular velocity $\omega$, and time $t$ in uniform rotation is given by:
$\Delta \theta = \omega t.$
Given that $\omega = 6 \, \text{rad/s}$ and $t = 5 \, \text{s}$, we substitute these values into the formula:
$\Delta \theta = 6 \times 5 = 30 \, \text{rad}.$
Therefore, the angular displacement is $30 \, \text{rad}$.

Key Concept: Angular Velocity

c) 24 units/s
[Solution Description] The angular velocity is the derivative of angular displacement with respect to time:
$\omega = \frac{d\theta}{dt} = \frac{d}{dt}(3t^2) = 6t.$
Substituting $t = 4$ gives $\omega = 6 \times 4 = 24\, \text{units/s}$.

Key Concept: Scalar Equations

d) $\sum F_{ix} = 0$
[Solution Description] In equilibrium conditions for forces, the scalar equation along the x-axis is $\sum F_{ix} = 0$, which indicates that the sum of all scalar components of forces in the x-direction is zero.

Key Concept: External Force and Torque Independence, Complex System Dynamics

c) The center of mass accelerates at $\frac{F}{m_1 + m_2}$ independent of $\tau$
[Solution Description]
The acceleration of the center of mass $a_{cm}$ due to external force $F$ is given by:
$a_{cm} = \frac{F}{m_1 + m_2}$
The torque $\tau$ affects rotational motion but does not directly influence the translational motion of the center of mass. Hence, the center of mass moves under the effect of external force $F$ only.
Thus, the center of mass experiences translational motion determined by the net external force independent of the external torque.

Key Concept: Translational Equilibrium

b) The sum of forces is zero
[Solution Description] Translational equilibrium occurs when the vector sum of all forces acting on the body is zero, expressed as $\sum \vec{F} = 0$.

Key Concept: Equation Interpretation

b) $3 \, m/s^2$

[Solution Description] To find the acceleration at time $t = 1 \, s$, substitute $t = 1$ into the given acceleration equation: $a = 4(1)^2 – 3(1) + 2$. Calculate each term: $4 \cdot 1^2 = 4$, $-3 \cdot 1 = -3$, and we have $+2$. Therefore, $a = 4 – 3 + 2 = 3 \, m/s^2$.

Key Concept: Reflection Symmetry

d) Geometric centre

[Solution Description] A homogeneous cube has reflection symmetry along each axis. Therefore, the centre of mass must be at the geometric center where these axes intersect symmetrically.

Key Concept: Conceptual Understanding

c) It remains constant
[Solution Description] When the external torque along the axis of rotation is zero, the angular momentum $L_z$ remains constant due to the conservation principle. The formula used is $L_z = I\omega = \text{constant}$.

Key Concept: Symmetry Consideration

d) At the intersection of diagonals

[Solution Description] Due to the symmetry and uniform distribution of mass in a square plate, the centre of mass is located at the geometric centre. This is because each axis divides the square into symmetrical halves. Hence, the center of mass lies at coordinates $\left(\frac{L}{2}, \frac{L}{2}\right)$.

Key Concept: Angular Acceleration

c) 5 rad/s$^2$
[Solution Description] Angular acceleration ($\alpha$) can be calculated using the formula $\alpha = \frac{\Delta \omega}{\Delta t}$, where $\Delta \omega$ is the change in angular velocity and $\Delta t$ is the time taken for this change. So here, $\Delta \omega = 25 – 10 = 15$ rad/s and $\Delta t = 3$ s. Thus, $\alpha = \frac{15 \, \text{rad/s}}{3 \, \text{s}} = 5 \, \text{rad/s}^2$.

Key Concept: Symmetry Consideration

d) Homogeneous disc

[Solution Description] For symmetric bodies such as a homogeneous disc, the center of mass coincides with the geometric center because the mass is evenly distributed around this point.

Key Concept: Derivation Understanding, Complex Problem Solving

c) Angular Momentum
[Solution Description]
In this scenario, the figure skater uses the principle of conservation of angular momentum. When she pulls her arms in, her moment of inertia decreases, but her angular momentum must remain constant because there is no external torque acting on her.
Angular momentum $L = I\omega$ remains constant since $\tau_{\text{ext}} = 0$. As the skater reduces her moment of inertia $I$, her angular velocity $\omega$ increases to keep $L$ constant. Therefore, even though both $I$ and $\omega$ change, the product $L$ remains unchanged.

Key Concept: Moment of Inertia Formula

b) $I = \frac{1}{2}MR^2$
[Solution Description] For a solid cylinder rotating about its axis, the moment of inertia is calculated using the formula $I = \frac{1}{2}MR^2$. Here, $M$ is the mass of the cylinder and $R$ is its radius.

Key Concept: Vector Product

c) 6
[Solution Description] The formula for the magnitude of the vector product is given by:
$|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin\theta$
Substituting $|\mathbf{a}| = 3$, $|\mathbf{b}| = 4$, and $\theta = 30^\circ$, we have:
$|\mathbf{a} \times \mathbf{b}| = 3 \times 4 \times \sin(30^\circ)$
Since $\sin(30^\circ) = 0.5$, we get:
$|\mathbf{a} \times \mathbf{b}| = 3 \times 4 \times 0.5 = 6$
Therefore, the magnitude of the vector product is $6$.

Key Concept: Basic Definition

d) Mass distribution, axis orientation, and position

[Solution Description] The moment of inertia is dependent on three main factors: mass distribution, orientation of the axis, and position relative to the axis.

Key Concept: Vector Product Application

b) $\langle -20, -15, 0 \rangle \, \text{m/s}$
[Solution Description] The formula for linear velocity is given by the cross product $v = \omega \times r$. Using the vectors:

\[
\mathbf{v} =
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
0 & 0 & 5 \\
3 & 4 & 0
\end{vmatrix}
\]

\[
= (0)(0) – (5)(4)\mathbf{i} – [(0)(0) – (5)(3)]\mathbf{j} + [(0)(4) – (0)(3)]\mathbf{k}
\]

\[
= -20\mathbf{i} – 15\mathbf{j}
\]

\[
= \langle -20, -15, 0 \rangle \, \text{m/s}
\]

Key Concept: Complex Systems, Advanced Reasoning

c) $\omega’ = \frac{2}{3}\omega$
[Solution Description] The moment of inertia of the disk before the mass is added is $I = \frac{1}{2}mR^2$. When the mass $m/4$ is added at the edge, its moment of inertia contribution is $(m/4)R^2$. Therefore, the new moment of inertia is $I’ = \frac{1}{2}mR^2 + (m/4)R^2 = \left(\frac{1}{2}m + \frac{1}{4}m\right)R^2 = \frac{3}{4}mR^2$. By conservation of angular momentum, $I\omega = I’\omega’$, thus $\omega’ = \frac{I\omega}{I’} = \frac{\frac{1}{2}mR^2 \cdot \omega}{\frac{3}{4}mR^2} = \frac{2}{3}\omega$.

Key Concept: Elementary Cross Products

c) $\mathbf{k}$

[Solution Description]

According to the elementary cross products, $\mathbf{i} \times \mathbf{j} = \mathbf{k}$. This follows from the convention of the right-handed coordinate system.

Key Concept: Torque Definition

c) The force applied at a distance from the axis of rotation
[Solution Description] Torque is defined as the cross product of the position vector $\mathbf{r}$ and the force vector $\mathbf{F}$. Mathematically, it can be represented as $\mathbf{\tau} = \mathbf{r} \times \mathbf{F}$. This implies that the torque is a measure of the tendency of a force to rotate an object about an axis.

Key Concept: Translational vs Rotational

d) All particles of the body have the same velocity at any instant.

[Solution Description]

In pure translational motion, every particle in a rigid body has the same velocity at any instant. Unlike rotational motion where particles move in circles, translational motion involves parallel movement without rotation.

Key Concept: Conservation Laws

c) With constant velocity.

[Solution Description] According to the conservation of momentum, in the absence of external forces, the center of mass of a system will continue to move with constant velocity as no change in its state of motion occurs.

Key Concept: Symmetry Concept

d) At the center

[Solution Description]

For a homogeneous solid sphere, due to its symmetry, the center of mass lies at its geometric center. This is because the distribution of mass is uniform throughout the sphere. Thus, the center of mass is at the center point of the sphere.

Key Concept: Magnitude of Cross Product

b) 15
[Solution Description] The formula for the magnitude of the cross product is given by $|a \times b| = ab \sin \theta$. Substituting the given values:
$|a \times b| = 5 \times 6 \times \sin(30^{\circ}) = 30 \times 0.5 = 15$

Key Concept: Application in Simple Systems

c) 1 m/s
[Solution Description] Using the formula $v = r\omega$, where $r = 0.25$ m and $\omega = 4$ rad/s, we have $v = 0.25 \times 4 = 1$ m/s.

Key Concept: Units and Dimensions

b) $kg \, m^2$
[Solution Description] The SI unit for mass is kilograms ($kg$) and for distance is meters ($m$). Therefore, since moment of inertia involves mass times distance squared, its unit is $kg \, m^2$.

Key Concept: Basic Definition

b) $ab \sin \theta$

[Solution Description] The magnitude of the vector product of two vectors $\mathbf{a}$ and $\mathbf{b}$ is given by the formula $c = ab \sin \theta$, where $\theta$ is the angle between the vectors.

Key Concept: Torque and Equilibrium, Real-World Application

d) $57.74 \, \text{N}$
[Solution Description]
Let the tension in the cable be $T$. For equilibrium, the sum of torques around the hinge must be zero.
The torque due to the weight $W = 100 \, \text{N}$ acting at the center of the beam ($2 \, \text{m}$ from the hinge) is:
$\tau_W = W \cdot \frac{L}{2} \cdot \sin(90^\circ) = 100 \cdot 2 \cdot 1 = 200 \, \text{Nm}$
The torque due to the tension in the cable $T$ acting at the full length of the beam ($4 \, \text{m}$) is:
$\tau_T = T \cdot L \cdot \sin(60^\circ) = T \cdot 4 \cdot \frac{\sqrt{3}}{2} = 2\sqrt{3} T$
Setting the net torque to zero for equilibrium:
$200 = 2\sqrt{3} T$
Solving for $T$:
$T = \frac{200}{2\sqrt{3}} = \frac{100}{\sqrt{3}} \approx 57.74 \, \text{N}$

Key Concept: System Momentum

b) $14 \, \text{kg m/s}$
[Solution Description] Total linear momentum $P$ of a system is calculated as $P = m_1v_1 + m_2v_2$. Here, $m_1 = 5 \, \text{kg}, v_1 = 4 \, \text{m/s}$, and $m_2 = 3 \, \text{kg}, v_2 = -2 \, \text{m/s}$. Therefore,
$P = 5 \times 4 + 3 \times (-2)
= 20 – 6
= 14 \, \text{kg m/s}.$
Hence, the total linear momentum of the system is $14 \, \text{kg m/s}$.

Key Concept: Advanced Derivation, Multi-Concept Integration

c) 1000$\pi$ radians
[Solution Description] Convert the angular velocities from rpm to rad/s: Initial $\omega_0 = \frac{500 \times 2\pi}{60} = \frac{50\pi}{3} \, \text{rad/s}$. Final $\omega = \frac{1500 \times 2\pi}{60} = \frac{150\pi}{3} = 50\pi \, \text{rad/s}$. Use the equation $\theta = \theta_0 + \omega_0 t + \frac{1}{2} \alpha t^2$. First, find $\alpha$ using $\omega = \omega_0 + \alpha t$:
\begin{align*}
\alpha &= \frac{\omega – \omega_0}{t} = \frac{50\pi – \frac{50\pi}{3}}{30} = \frac{100\pi}{3 \times 30} = \frac{10\pi}{9} \, \text{rad/s}^2.
\end{align*}
Now, compute $\theta$:
$\theta = 0 + \frac{50\pi}{3} \times 30 + \frac{1}{2} \times \frac{10\pi}{9} \times 30^2 = 500\pi + 500\pi = 1000\pi \, \text{radians}.$

Key Concept: Equation Application

b) $1.25 \, \text{m/s}$
[Solution Description]
To find the velocity of the combined mass after an inelastic collision, we use the conservation of linear momentum. The initial momentum is:
$P_{\text{initial}} = m_1 v_1 + m_2 v_2 = 5 \times 3 + 7 \times 0 = 15 \, \text{kg} \cdot \text{m/s}$
Let $v$ be the velocity of the combined mass after the collision. Then,
$(m_1 + m_2) v = P_{\text{initial}}$
$v = \frac{P_{\text{initial}}}{m_1 + m_2} = \frac{15}{12} = 1.25 \, \text{m/s}$
Therefore, the velocity of the combined mass is $1.25 \, \text{m/s}$.

Key Concept: Mechanical Advantage

c) 4
[Solution Description] Mechanical advantage ($M.A.$) is given by the formula:
$M.A. = \frac{d_2}{d_1}$
Where $d_2$ is the length of the effort arm and $d_1$ is the length of the load arm. Given that $d_2 = 4d_1$, substitute into the formula:
$M.A. = \frac{4d_1}{d_1} = 4$
Thus, the mechanical advantage is 4.

Key Concept: Angular Velocity Basics

d) Linear velocity (v)
[Solution Description] Angular velocity ($\omega$) in rotational motion is analogous to linear velocity ($v$) in translational motion because both describe how fast an object moves, either in rotation or translation.

Key Concept: Basic Concept

b) External forces

[Solution Description] The motion of the centre of mass is determined by external forces, not internal forces. According to Newton’s Second Law for systems of particles, it’s the external forces that affect the overall movement.

Key Concept: Coplanar Forces

c) 3

[Solution Description] When forces are coplanar, three conditions need to be satisfied: two for translational equilibrium and one for rotational equilibrium.

Key Concept: Torque Calculation

c) 0.707 Nm
[Solution Description]
To calculate the torque, we use the formula $\tau = rF \sin \alpha$. Here, $r = 0.25$ m, $F = 8$ N, and $\alpha = 45^\circ$. We find:
$\tau = 0.25 \times 8 \times \sin(45^\circ)$
The sine of $45^\circ$ is $\frac{\sqrt{2}}{2}$. So,
$\tau = 0.25 \times 8 \times \frac{\sqrt{2}}{2} = 1 \times \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2} \approx 0.707 \text{ Nm}$

Key Concept: Angle Consideration

c) $90^\circ$

[Solution Description] For vectors $\mathbf{a}$ and $\mathbf{b}$ that are perpendicular, the angle $\theta$ between them is $90^\circ$. In this case, the sine function reaches its maximum value, i.e., $\sin 90^\circ = 1$.

Key Concept: Basic Calculation

b) At the midpoint

[Solution Description] For a uniform rod, the center of gravity is located at its geometric center because the mass is evenly distributed along its length. Therefore, if it balances at its midpoint, that midpoint corresponds to the center of gravity.

Key Concept: Internal vs External Forces

a) Internal forces

[Solution Description] In a closed system, internal forces do not affect the total linear momentum due to Newton’s third law — for every action, there is an equal and opposite reaction within the system. Thus, they cancel each other out, whereas only external forces can change the total linear momentum.

Key Concept: Complex Systems, Angular Momentum Conservation

b) $x_1 m_1 v_1 – y_2 m_2 v_2$
[Solution Description] The angular momentum $\mathbf{L}$ for each particle is given by $\mathbf{L}_i = \mathbf{r}_i \times \mathbf{p}_i = \mathbf{r}_i \times m_i \mathbf{v}_i$.
First, calculate $\mathbf{L}_1$ for the first particle:
$\mathbf{L}_1 = \mathbf{r}_1 \times m_1 \mathbf{v}_1 = (x_1, 0, 0) \times (0, m_1 v_1, 0) = (0, 0, x_1 m_1 v_1).$
Next, calculate $\mathbf{L}_2$ for the second particle:
$\mathbf{L}_2 = \mathbf{r}_2 \times m_2 \mathbf{v}_2 = (0, y_2, 0) \times (m_2 v_2, 0, 0) = (0, 0, -y_2 m_2 v_2).$
The net angular momentum $\mathbf{L}$ is the sum of $\mathbf{L}_1$ and $\mathbf{L}_2$:
$\mathbf{L} = \mathbf{L}_1 + \mathbf{L}_2 = (0, 0, x_1 m_1 v_1 – y_2 m_2 v_2).$
Thus, the magnitude of the net angular momentum is $|x_1 m_1 v_1 – y_2 m_2 v_2|$ in the z-direction.

Key Concept: Basic Concept

d) The center of mass behaves as if all external forces act upon it.
[Solution Description] When Eq. (6.11), $M \mathbf{A} = \mathbf{F}_{\text{ext}}$, is applied, it implies that the center of mass of the system moves as if all external forces are acting on that point and all the mass is concentrated there.

Key Concept: Cyclic Order

a) $\hat{i} \times \hat{j} = \hat{k}$
[Solution Description]
The cyclic order of unit vectors involves the identity relations:
– $\hat{i} \times \hat{j} = \hat{k}$
– $\hat{j} \times \hat{k} = \hat{i}$
– $\hat{k} \times \hat{i} = \hat{j}$
Checking each option:
A) $\hat{i} \times \hat{j} = \hat{k}$ (True)
B) $\hat{j} \times \hat{k} = -\hat{i}$ (False; should be $\hat{i}$)
C) $\hat{k} \times \hat{i} = \hat{j}$ (True)
D) $\hat{j} \times \hat{i} = -\hat{k}$ (True; opposite order has negative sign)
Thus, option A is true.

Key Concept: Angular Momentum Definition

d) $l = r \times p$

[Solution Description] Angular momentum for a single particle is given by the vector product of its position vector $r$ and linear momentum $p$. Therefore, it is defined as $l = r \times p$.

Key Concept: Application of Equations

c) 45 N
[Solution Description]
The formula to calculate the external force acting on a system based on Newton’s second law for a system of particles is:
$F_{ext} = M \cdot A$
where $M$ is the total mass, and $A$ is the acceleration of the center of mass.
Given: $M = 15 \, \text{kg}$, $A = 3 \, \text{m/s}^2$.
Substitute these values into the equation:
$F_{ext} = 15 \times 3 = 45 \, \text{N}$
So, the external force acting on the system is 45 N.

Key Concept: Angular Momentum

b) $L = \sum r_i \times p_i$

[Solution Description] The angular momentum of a system of particles about a point is calculated using the formula $L = \sum r_i \times p_i$, where $r_i$ is the position vector and $p_i$ is the linear momentum.

Key Concept: Definition Recall

b) $\tau = r \times F$
[Solution Description] Torque is defined as the vector product of the position vector and force, represented by $\tau = r \times F$. This definition shows that torque depends on both the magnitude and direction of the position vector and force.

Key Concept: Fixed Axis Rotation

c) It remains constant
[Solution Description] For rotation about a fixed axis with no external torque, the angular momentum $L_z = I\omega$ remains constant. Thus, the satellite’s angular momentum does not change as long as there’s no external force causing torque.

Key Concept: Vector Product Application

c) Along the positive z-axis
[Solution Description] The linear velocity vector $\mathbf{v}$ is determined by the cross product $\mathbf{v} = \boldsymbol{\omega} \times \mathbf{r}$. The angular velocity vector $\boldsymbol{\omega}$ is directed along the x-axis: $\boldsymbol{\omega} = \langle \omega, 0, 0 \rangle$, and the position vector is $\mathbf{r} = \langle 0, r, 0 \rangle$. Calculating the cross product:
$\boldsymbol{\omega} \times \mathbf{r} =
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k}
\omega & 0 & 0
0 & r & 0
\end{vmatrix}
=
(\omega \cdot 0 – 0 \cdot r)\hat{i} – (\omega \cdot 0 – 0 \cdot 0)\hat{j} + (\omega \cdot r – 0 \cdot 0)\hat{k}
=
\langle 0, 0, \omega r \rangle$
Thus, the direction of $\mathbf{v}$ is along the positive z-axis.

Key Concept: Power in Rotational Motion

b) 200 Watts
[Solution Description] Power generated in rotational motion is given by the expression $P = \tau \cdot \omega$. Substitute the given values:
$P = 20 \, \text{Nm} \times 10 \, \text{rad/s} = 200 \, \text{Watts}.$

Key Concept: Symmetry Consideration

b) 3 m

[Solution Description] For a homogeneous thin rod, the centre of mass is at its geometric center. Therefore, it is at half the length of the rod:

$X = \frac{\text{length of rod}}{2} = \frac{6}{2} = 3 \text{ m}$

Key Concept: Center of Mass Velocity

c) 4 m/s
[Solution Description]
The velocity of the center of mass can be found using the equation for total momentum, $M = P \cdot V$.
Rearranging gives us:
$V = \frac{P}{M}$
Substituting the given values:
$V = \frac{80 \text{ kg}\cdot\text{m/s}}{20 \text{ kg}}$
$V = 4 \text{ m/s}$

Key Concept: Equal Mass Particles

b) 2.5

[Solution Description]

Since the particles have equal mass, the $X$-coordinate of the center of mass is the average of their $x$-coordinates:

$X = \frac{x_1 + x_2 + x_3 + x_4}{4} = \frac{1 + 2 + 3 + 4}{4} = \frac{10}{4} = 2.5$

Therefore, the $X$-coordinate of the center of mass is 2.5.

Key Concept: Conservation Conditions, Advanced Problem Solving

b) -3 m/s
[Solution Description] By conservation of linear momentum, the total initial momentum is zero since both skaters are at rest. Thus, the total momentum after they push off is also zero:
$m \cdot v_1 + m \cdot v_2 = 0$
Solving for the velocity of the second skater $v_2$, we find:
$v_2 = -v_1 = -3\, \text{m/s}$
The negative sign indicates that the second skater moves in the opposite direction.

Key Concept: Elementary Cross Products

b) $\hat{j}$
[Solution Description] According to the properties of elementary cross products, $\hat{k} \times \hat{i} = \hat{j}$. This follows from the cyclic order where $\hat{k}, \hat{i}, \hat{j}$ are in cyclic order and the cross product moves forward in this cycle.

Key Concept: Coordinate Calculation

b) 5.44
[Solution Description]
The y-coordinate of the center of mass is found using the formula:
$Y = \frac{m_1y_1 + m_2y_2 + m_3y_3}{m_1 + m_2 + m_3}$
Substituting the given values:
$Y = \frac{2 \times 3 + 3 \times 5 + 4 \times 7}{2 + 3 + 4}$
Calculating the numerator:
$6 + 15 + 28 = 49$
And the denominator:
$2 + 3 + 4 = 9$
Therefore,
$Y = \frac{49}{9} = 5.44$

Key Concept: Multi-step Problem

c) 250 rad
[Solution Description] The angular displacement is found using the equation $\theta = \omega_0 t + \frac{1}{2} \alpha t^2$. First, calculate angular acceleration $\alpha$ with $\alpha = \frac{\omega – \omega_0}{t} = \frac{45 \, \text{rad/s} – 5 \, \text{rad/s}}{10 \, \text{s}} = 4 \, \text{rad/s}^2$. Substituting into the angular displacement formula: $\theta = 5 \times 10 + \frac{1}{2} \times 4 \times 10^2 = 50 + 200 = 250 \, \text{rad}$.

Key Concept: Angular Velocity

c) $\omega = \frac{d\theta}{dt}$

[Solution Description] Angular velocity $\omega$ is defined as the rate of change of angular displacement $\theta$ with respect to time $t$. Mathematically, this is expressed as $\omega = \frac{d\theta}{dt}$.

Key Concept: Vector Addition Distributive Property

a) $a \times (b + c) = a \times b + a \times c$
[Solution Description] The distributive property of vector products states that $a \times (b + c) = a \times b + a \times c$. This expression directly applies the distributive law to vector cross products.

Key Concept: Definition Recall

c) The point where the mass-weighted mean of positions is located

[Solution Description]

By definition, the centre of mass for a two-particle system is the point where the mass-weighted mean of positions is located.

Key Concept: Translational Equilibrium

b) Their vector sum must be zero

[Solution Description] In translational equilibrium, the vector sum of all forces acting on the body must be zero $sum \vec{F} = 0$. This condition ensures that there is no net force causing the body to accelerate linearly.

Key Concept: Angular Velocity Basics

c) $2\pi \text{ rad/s}$
[Solution Description]
The formula for angular velocity $\omega$ is given by $\omega = \frac{\theta}{t}$, where $\theta$ is the angular displacement and $t$ is the time taken. Here, $\theta = 8\pi \text{ radians}$ and $t = 4 \text{ seconds}$. Therefore,
$\omega = \frac{8\pi}{4} = 2\pi \text{ rad/s}.$

Key Concept: Angular Velocity Role

c) It is analogous to linear velocity in translational motion

[Solution Description] Angular velocity $\omega$ in rotational motion is analogous to linear velocity in translational motion. It represents how fast an object rotates or spins around an axis.

Key Concept: Basic Formula Application

d) $\mathbf{V} = \frac{m_1 \mathbf{v}_1 + m_2 \mathbf{v}_2}{m_1 + m_2}$
[Solution Description] The velocity of the centre of mass is given by:
$M \mathbf{V} = \sum m_i \mathbf{v}_i$
For two particles, this becomes:
$(m_1 + m_2)\mathbf{V} = m_1 \mathbf{v}_1 + m_2 \mathbf{v}_2$
Solving for $\mathbf{V}$, we divide both sides by $(m_1 + m_2)$:
$\mathbf{V} = \frac{m_1 \mathbf{v}_1 + m_2 \mathbf{v}_2}{m_1 + m_2}$

Key Concept: Mass Distribution

b) 2.8
[Solution Description]
To find the x-coordinate of the centre of mass $X$, we use the formula:
$X = \frac{\sum m_i x_i}{M} = \frac{(2 \times 1) + (3 \times 2) + (5 \times 4)}{2 + 3 + 5}$
Calculating step by step:
\begin{align*}
\text{Numerator: } &\quad (2 \times 1) + (3 \times 2) + (5 \times 4) \\
&\quad = 2 + 6 + 20 = 28 \\[8pt]
\text{Denominator: } &\quad 2 + 3 + 5 = 10 \\[8pt]
X &= \frac{28}{10} = 2.8
\end{align*}

Therefore, the x-coordinate of the centre of mass is $2.8$.

Key Concept: Angular Velocity

a) Along the axis of rotation

[Solution Description] The direction of angular velocity in rotational motion about a fixed axis is given by the right-hand rule. It points along the axis of rotation and follows the direction of the thumb if the fingers curl in the direction of rotation.

Key Concept: External Force Impact

d) Total momentum changes over time

[Solution Description]

According to Newton’s second law for a system:

$\frac{d\mathbf{P}}{dt} = \mathbf{F}_{\text{ext}}$

Given $\mathbf{F}_{\text{ext}} = 5\, \text{N}$, this implies:

$\frac{d\mathbf{P}}{dt} = 5\, \text{N}$

Since the derivative of momentum with respect to time is non-zero, the total momentum of the system is not conserved and changes over time.

Key Concept: Angular Displacement Calculation

c) 18 rad
[Solution Description]
The angular displacement $\theta$ can be calculated using the relation $\theta = \omega t$, where $\omega$ is the angular velocity and $t$ is the time.
Given:
\begin{align*}
\omega &= 6 \text{ rad/s}
t &= 3 \text{ s}
\end{align*}
Now calculate $\theta$:
$\theta = \omega t = 6 \times 3 = 18 \text{ rad}$
Thus, the angular displacement is 18 rad.

Key Concept: Definition Recall

a) Increases
[Solution Description] When a figure skater pulls in their arms, they reduce their moment of inertia. Since angular momentum ($I\omega$) is conserved with no external torque, the reduction in $I$ results in an increase in angular speed $\omega$.

Key Concept: Centre of Mass Path

c) The original parabolic path

[Solution Description] In an exploding projectile, the path of the center of mass continues along the original parabolic trajectory despite the explosion.