System of Particles and Rotational Motion - Class 11 Physics (NCERT)

Understanding System of Particles and Centre of Mass

In our studies so far, we've often treated objects as point masses, ignoring their size and internal structure. While useful, this simplification doesn't always work. For a more accurate description, especially when objects rotate or deform, we consider them as a system of particles. This system can be anything from two colliding billiard balls to a complex rigid body like a wheel or a planet.

At the heart of analyzing such systems is the concept of the Centre of Mass (CM). The Centre of Mass is a unique point within or outside a system of particles where the entire mass of the system appears to be concentrated for the purpose of describing its translational motion. If you apply a force directly through the CM of a rigid body, it will undergo pure translational motion without any rotation. It's the 'balancing point' of the system. For a system of discrete particles with masses $m_1, m_2, ..., m_n$ located at position vectors $\vec{r}_1, \vec{r}_2, ..., \vec{r}_n$, the position vector of the Centre of Mass $\vec{R}_{CM}$ is given by:

$\vec{R}_{CM} = \frac{m_1\vec{r}_1 + m_2\vec{r}_2 + ... + m_n\vec{r}_n}{m_1 + m_2 + ... + m_n} = \frac{\sum_{i=1}^{n} m_i\vec{r}_i}{\sum_{i=1}^{n} m_i} = \frac{1}{M}\sum_{i=1}^{n} m_i\vec{r}_i$

Where $M$ is the total mass of the system. For a continuous body, the summation is replaced by an integration. The velocity and acceleration of the CM are similarly defined, and the total linear momentum of a system of particles is equal to the product of the total mass and the velocity of its Centre of Mass ($P = M V_{CM}$). This simplifies the study of complex interactions, allowing us to separate the translational motion of the entire system from its internal or rotational motion.

Key Concepts in Rotational Dynamics

Rigid Body

A rigid body is an object where the distance between any two constituent particles remains constant, regardless of applied forces. This is an idealization, as all real bodies deform slightly under force, but it's a very useful approximation for rotational motion.

Rotational Motion

Motion where every particle of a body moves in a circle, and the centers of these circles lie on a straight line called the axis of rotation. If the axis of rotation is fixed, all particles move with the same angular velocity.

Torque ($\vec{\tau}$)

The rotational analogue of force. It is the turning effect of a force about an axis or pivot point. Mathematically, it's defined as the vector product of the position vector (from the pivot to the point of force application) and the force: $\vec{\tau} = \vec{r} \times \vec{F}$. Its magnitude is $rF\sin\theta$, where $\theta$ is the angle between $\vec{r}$ and $\vec{F}$. The SI unit is Newton-metre (N m).

Moment of Inertia (I)

The rotational analogue of mass. It is a measure of an object's resistance to changes in its rotational motion (i.e., its rotational inertia). For a single particle, $I = mr^2$. For a system of discrete particles, $I = \sum m_i r_i^2$. For a rigid body, it depends on the mass distribution and the chosen axis of rotation. The SI unit is kilogram-metre squared (kg m²).

Angular Momentum ($\vec{L}$)

The rotational analogue of linear momentum. For a single particle with linear momentum $\vec{p}$ and position vector $\vec{r}$ relative to the origin, $\vec{L} = \vec{r} \times \vec{p}$. For a rigid body rotating about a fixed axis, its magnitude is $L = I\omega$, where $I$ is the moment of inertia and $\omega$ is the angular velocity. The SI unit is kilogram-metre squared per second (kg m²/s).

Worked Examples on Rotational Motion

• Example 1: Centre of Mass of a System Three particles of masses 1 kg, 2 kg, and 3 kg are placed at the corners of an equilateral triangle of side 1 m. Find the coordinates of the centre of mass of the system. Assume the 1 kg mass is at the origin (0,0), and the 2 kg mass is on the x-axis (1,0). Solution: Step 1: Assign coordinates to each mass. $m_1 = 1 \text{ kg}$ at $(x_1, y_1) = (0, 0)$ $m_2 = 2 \text{ kg}$ at $(x_2, y_2) = (1, 0)$ For the 3 kg mass, its x-coordinate will be $1 \times \cos(60^{\circ}) = 1/2 \text{ m}$, and its y-coordinate will be $1 \times \sin(60^{\circ}) = \sqrt{3}/2 \text{ m}$. So, $m_3 = 3 \text{ kg}$ at $(x_3, y_3) = (1/2, \sqrt{3}/2)$. Step 2: Calculate the total mass $M$. $M = m_1 + m_2 + m_3 = 1 + 2 + 3 = 6 \text{ kg}$. Step 3: Calculate the x-coordinate of the Centre of Mass ($X_{CM}$). $X_{CM} = \frac{m_1x_1 + m_2x_2 + m_3x_3}{M} = \frac{(1)(0) + (2)(1) + (3)(1/2)}{6} = \frac{0 + 2 + 1.5}{6} = \frac{3.5}{6} = \frac{7}{12} \text{ m}$. Step 4: Calculate the y-coordinate of the Centre of Mass ($Y_{CM}$). $Y_{CM} = \frac{m_1y_1 + m_2y_2 + m_3y_3}{M} = \frac{(1)(0) + (2)(0) + (3)(\sqrt{3}/2)}{6} = \frac{0 + 0 + 3\sqrt{3}/2}{6} = \frac{3\sqrt{3}}{12} = \frac{\sqrt{3}}{4} \text{ m}$. Final answer: The coordinates of the Centre of Mass are $(\frac{7}{12}, \frac{\sqrt{3}}{4})$.
• Example 2: Calculating Torque A force $\vec{F} = (2\hat{i} + 3\hat{j} - \hat{k})$ N acts on a particle whose position vector relative to the origin is $\vec{r} = (\hat{i} - \hat{j} + 2\hat{k})$ m. Find the torque acting on the particle about the origin. Solution: Step 1: Recall the formula for torque. Torque $\vec{\tau} = \vec{r} \times \vec{F}$. Step 2: Perform the cross product. $\vec{\tau} = (\hat{i} - \hat{j} + 2\hat{k}) \times (2\hat{i} + 3\hat{j} - \hat{k})$ Using the determinant method or distributing terms: $\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 2 \\ 2 & 3 & -1 \end{vmatrix}$ Step 3: Calculate the components. $\hat{i}[(-1)(-1) - (2)(3)] - \hat{j}[(1)(-1) - (2)(2)] + \hat{k}[(1)(3) - (-1)(2)]$ $\hat{i}[1 - 6] - \hat{j}[-1 - 4] + \hat{k}[3 + 2]$ $-5\hat{i} - (-5)\hat{j} + 5\hat{k}$ Final answer: The torque acting on the particle is $\vec{\tau} = (-5\hat{i} + 5\hat{j} + 5\hat{k})$ N m.
• Example 3: Conservation of Angular Momentum An ice skater spins with an angular speed of 10 rad/s with her arms outstretched. Her moment of inertia in this position is 2 kg m². She then pulls her arms in, reducing her moment of inertia to 0.8 kg m². What is her new angular speed? Solution: Step 1: Identify the principle to be used. Since there is no external torque acting on the skater during the process of pulling her arms in, her angular momentum will be conserved. Step 2: Write down the initial conditions. Initial angular speed $\omega_1 = 10 \text{ rad/s}$ Initial moment of inertia $I_1 = 2 \text{ kg m}^2$ Initial angular momentum $L_1 = I_1\omega_1 = (2)(10) = 20 \text{ kg m}^2/s$. Step 3: Write down the final conditions. Final moment of inertia $I_2 = 0.8 \text{ kg m}^2$ Final angular speed $\omega_2 = ?$ (to be found) Final angular momentum $L_2 = I_2\omega_2 = 0.8\omega_2$. Step 4: Apply the conservation of angular momentum. $L_1 = L_2$ $20 = 0.8\omega_2$ Step 5: Solve for $\omega_2$. $\omega_2 = \frac{20}{0.8} = \frac{200}{8} = 25 \text{ rad/s}$. Final answer: Her new angular speed is 25 rad/s.

YoLearn AI Tutor's Exam Tips for Rotational Motion

1. Vector Nature is Crucial: Remember that torque, angular velocity, angular acceleration, and angular momentum are all vector quantities. Pay close attention to their directions, often determined by the right-hand rule. For cross products, ensure the order of vectors is correct (e.g., $\vec{r} \times \vec{F}$, not $\vec{F} \times \vec{r}$). A common mistake is treating them as scalars.

2. Axis of Rotation: Moment of inertia, torque, and angular momentum are always defined with respect to a specific axis of rotation. Clearly identify this axis in each problem. If the axis changes, the moment of inertia will likely change. Make sure to correctly apply the Parallel Axis Theorem ($I = I_{CM} + Md^2$) and Perpendicular Axis Theorem ($I_z = I_x + I_y$) where applicable.

3. Analogy with Translational Motion: Rotational dynamics has strong analogies with linear dynamics. Force becomes torque, mass becomes moment of inertia, linear velocity becomes angular velocity, and linear momentum becomes angular momentum. Using these analogies can help you remember formulas and concepts. For example, $F = ma$ becomes $\tau = I\alpha$, and $p = mv$ becomes $L = I\omega$.

4. Conservation Laws: The conservation of angular momentum is a powerful principle. It applies when the net external torque acting on a system is zero. Common scenarios include ice skaters, diving, and astronomical bodies. Don't forget that if there's external torque, angular momentum is not conserved.

5. Units and Dimensions: Always check your units. Torque is in N m, not Joules (which is N m, but for energy). Moment of inertia is kg m², and angular momentum is kg m²/s. Dimensional analysis can often catch errors.

Practice Questions with Solutions

• Q: A uniform rod of mass $M$ and length $L$ is pivoted at one end. Calculate its moment of inertia about an axis perpendicular to the rod and passing through the pivot. A: Step 1: Recall the formula for the moment of inertia of a uniform rod about an axis passing through its centre and perpendicular to its length, which is $I_{CM} = \frac{1}{12}ML^2$. Step 2: Apply the Parallel Axis Theorem. The distance from the centre of mass (midpoint of the rod) to the pivot (one end) is $d = L/2$. Step 3: Use the Parallel Axis Theorem: $I = I_{CM} + Md^2$. $I = \frac{1}{12}ML^2 + M(\frac{L}{2})^2 = \frac{1}{12}ML^2 + \frac{1}{4}ML^2$ Step 4: Combine the terms. $I = (\frac{1}{12} + \frac{3}{12})ML^2 = \frac{4}{12}ML^2 = \frac{1}{3}ML^2$. Final answer: The moment of inertia of the rod about the pivot is $\frac{1}{3}ML^2$.
• Q: A wheel of moment of inertia 5 kg m² is rotating at 120 rpm. What constant torque is required to stop the wheel in 10 seconds? A: Step 1: Convert initial angular speed to rad/s. $\omega_0 = 120 \text{ rpm} = 120 \times \frac{2\pi}{60} \text{ rad/s} = 4\pi \text{ rad/s}$. Step 2: Identify final angular speed and time. Final angular speed $\omega_f = 0 \text{ rad/s}$ (since it stops). Time $t = 10 \text{ s}$. Step 3: Calculate the angular acceleration $\alpha$. Using rotational kinematic equation: $\omega_f = \omega_0 + \alpha t$ $0 = 4\pi + \alpha(10) \implies 10\alpha = -4\pi \implies \alpha = -0.4\pi \text{ rad/s}^2$. (The negative sign indicates deceleration). Step 4: Calculate the torque $\tau$. Using the rotational equivalent of Newton's second law: $\tau = I\alpha$. $\tau = (5 \text{ kg m}^2)(-0.4\pi \text{ rad/s}^2) = -2\pi \text{ N m}$. Final answer: A constant torque of magnitude $2\pi \text{ N m}$ is required to stop the wheel. The negative sign indicates the torque acts opposite to the direction of rotation.
• Q: Two particles, each of mass $m$, are connected by a massless rod of length $L$. The system rotates about an axis passing through the centre of the rod and perpendicular to its length. Find the moment of inertia of the system about this axis. A: Step 1: Visualize the system. The two particles are at the ends of the rod. The axis of rotation passes through the midpoint of the rod. Step 2: Determine the distance of each particle from the axis of rotation. Since the rod is of length $L$ and the axis is at its centre, each particle is at a distance $r = L/2$ from the axis. Step 3: Apply the formula for moment of inertia for a system of discrete particles: $I = \sum m_i r_i^2$. For particle 1: $I_1 = m(L/2)^2 = m L^2/4$. For particle 2: $I_2 = m(L/2)^2 = m L^2/4$. Step 4: Add the moments of inertia of the individual particles. $I = I_1 + I_2 = mL^2/4 + mL^2/4 = 2(mL^2/4) = mL^2/2$. Final answer: The moment of inertia of the system is $mL^2/2$.
• Q: A satellite orbits the Earth in an elliptical path. Its speed is greater when it is closer to the Earth and smaller when it is farther away. Explain this phenomenon using the principle of conservation of angular momentum. A: Step 1: Identify the system and external forces. The system is the satellite and the Earth. The gravitational force between them acts along the line connecting their centers, passing through the Earth (which is essentially the pivot point). This means the torque due to gravity about the Earth's center is zero ($\vec{r} \times \vec{F}$ will have $\vec{r}$ and $\vec{F}$ parallel or anti-parallel). Step 2: Apply the conservation of angular momentum. Since the net external torque is zero, the angular momentum of the satellite about the Earth's center is conserved: $L = I\omega = \text{constant}$. Also, for an orbiting particle, $L = mvr$, where $m$ is mass, $v$ is tangential speed, and $r$ is the distance from the Earth. Step 3: Relate angular momentum to distance and speed. So, $mvr = \text{constant}$. Since $m$ is constant, this implies $vr = \text{constant}$. Step 4: Explain the observation. When the satellite is closer to Earth ($r$ is smaller), its speed $v$ must be greater to keep the product $vr$ constant. Conversely, when it is farther away from Earth ($r$ is larger), its speed $v$ must be smaller to maintain the constant product $vr$. Final answer: The satellite's angular momentum ($L=mvr$) about the Earth's center is conserved because the gravitational force acts radially, producing zero torque. Therefore, when the radius $r$ decreases (closer to Earth), the tangential speed $v$ must increase, and vice-versa, to maintain constant angular momentum.

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