System of Particles and Rotational Motion: Class 11 NCERT Guide

The Concept of Centre of Mass (CM)

Imagine trying to balance a cricket bat on your finger. You'll find there's a unique point where it balances perfectly. That point is its Centre of Mass. For any collection of particles or a solid body, the Centre of Mass (CM) is a hypothetical point where the entire mass of the system can be assumed to be concentrated for the purpose of describing its translational motion.

Why is this useful? It simplifies complex problems. Instead of tracking the motion of billions of individual particles in a thrown cricket ball, we can just track the parabolic path of its CM. The ball might be spinning and wobbling, but its CM follows a predictable path as if it were a single particle thrown with the same initial velocity. For a system of 'n' particles, the position vector of the CM, R_CM, is the mass-weighted average of the position vectors r_i of the individual particles:

R_CM = (m₁r₁ + m₂r₂ + ... + mₙrₙ) / (m₁ + m₂ + ... + mₙ) = (Σ mᵢrᵢ) / M

Where M is the total mass of the system. This single equation allows us to pinpoint the 'average' location of the system's mass.

Key Quantities in Rotational Motion

Torque (τ)

The rotational analogue of force. It is the turning effect of a force about an axis. Mathematically, τ = r × F, where r is the position vector from the axis to the point of application of force F. Its SI unit is Newton-meter (Nm).

Moment of Inertia (I)

The rotational analogue of mass. It measures an object's resistance to changes in its rotational motion. It depends not just on the mass, but also on how that mass is distributed around the axis of rotation. For a single particle, I = mr². For a system, it's the sum Σmᵢrᵢ². Its SI unit is kg m².

Angular Momentum (L)

The rotational analogue of linear momentum. It is the 'quantity of rotation' of a body. For a particle, L = r × p, where p is the linear momentum. For a rigid body rotating about a fixed axis, L = Iω. Its SI unit is kg m²/s.

Angular Velocity (ω)

The rate of change of angular displacement (dθ/dt). It describes how fast an object is rotating. Its SI unit is radians per second (rad/s).

Worked Examples: Moment of Inertia & Torque

• Example 1: Calculating Moment of Inertia for a System of Particles Find the moment of inertia of a system of three masses: 1 kg, 2 kg, and 3 kg, placed at coordinates (1, 0), (0, 2), and (1, 1) respectively, about the y-axis. Step 1: Understand the formula for Moment of Inertia. The moment of inertia (I) is the sum of the products of each mass (m) and the square of its perpendicular distance (r) from the axis of rotation. I = Σmᵢrᵢ² Step 2: Identify the perpendicular distance for each mass from the y-axis. The perpendicular distance of a point (x, y) from the y-axis is simply its x-coordinate. For m₁ = 1 kg at (1, 0), r₁ = 1 m. For m₂ = 2 kg at (0, 2), r₂ = 0 m (it's on the y-axis). For m₃ = 3 kg at (1, 1), r₃ = 1 m. Step 3: Calculate the total moment of inertia. I = m₁r₁² + m₂r₂² + m₃r₃² I = (1 kg)(1 m)² + (2 kg)(0 m)² + (3 kg)(1 m)² I = 1 + 0 + 3 = 4 kg m² Final Answer: The moment of inertia of the system about the y-axis is 4 kg m².
• Example 2: Calculating Torque A force F = (2î + 3ĵ - k̂) N is applied at a point with position vector r = (î - ĵ + 2k̂) m relative to the origin. Calculate the torque about the origin. Step 1: Recall the formula for torque. Torque (τ) is the cross product of the position vector (r) and the force vector (F). τ = r × F Step 2: Set up the determinant for the cross product. τ = | î ĵ k̂ | | 1 -1 2 | | 2 3 -1 | Step 3: Evaluate the determinant. τ = î [(-1)(-1) - (2)(3)] - ĵ [(1)(-1) - (2)(2)] + k̂ [(1)(3) - (-1)(2)] τ = î [1 - 6] - ĵ [-1 - 4] + k̂ [3 + 2] τ = î [-5] - ĵ [-5] + k̂ [5] τ = -5î + 5ĵ + 5k̂ Final Answer: The torque about the origin is (-5î + 5ĵ + 5k̂) Nm.

Exam Focus: Parallel and Perpendicular Axis Theorems

These two theorems are extremely important for your exams! They are powerful shortcuts for calculating the moment of inertia without complex integration.

• Theorem of Perpendicular Axes: Use this only for 2D, planar objects (like a thin disc or a lamina). It states that the moment of inertia about an axis perpendicular to its plane (say, the z-axis) is the sum of the moments of inertia about two perpendicular axes in its plane (x and y axes) that intersect at the same point.

I_z = I_x + I_y
Common Trap: Do not apply this to 3D objects like spheres or cylinders.

• Theorem of Parallel Axes: Use this to find the moment of inertia about any axis, provided you know the moment of inertia about a parallel axis passing through the centre of mass. It works for both 2D and 3D objects.

I = I_cm + Md²
Where I_cm is the moment of inertia about the CM, M is the total mass, and d is the perpendicular distance between the two parallel axes.
Common Trap: Forgetting to square the distance 'd' or using an axis that is not parallel.

Practice Questions with Solutions

• Q: Three point masses m₁=1kg, m₂=2kg and m₃=3kg are placed at the vertices of an equilateral triangle of side 1m. Find the coordinates of the centre of mass of the system, assuming m₁ is at the origin and m₂ is on the x-axis. A: Step 1: Determine the coordinates of the three masses. m₁ is at (0, 0). m₂ is at (1, 0). For an equilateral triangle, the third vertex m₃ will be at (x, y) where x = 1/2 and y = (√3)/2. So, m₃ is at (0.5, 0.866). Step 2: Use the formulas for the coordinates of the Centre of Mass (X_cm, Y_cm). X_cm = (m₁x₁ + m₂x₂ + m₃x₃) / (m₁ + m₂ + m₃) Y_cm = (m₁y₁ + m₂y₂ + m₃y₃) / (m₁ + m₂ + m₃) Step 3: Substitute the values. X_cm = (10 + 21 + 30.5) / (1 + 2 + 3) = (0 + 2 + 1.5) / 6 = 3.5 / 6 ≈ 0.583 m. Y_cm = (10 + 20 + 30.866) / (1 + 2 + 3) = (0 + 0 + 2.598) / 6 = 2.598 / 6 ≈ 0.433 m. Final answer: The centre of mass is located at approximately (0.583 m, 0.433 m).
• Q: A solid sphere of mass 2 kg and radius 0.1 m is rotating about its diameter with an angular velocity of 10 rad/s. What is its rotational kinetic energy? A: Step 1: Recall the formula for rotational kinetic energy (KE_rot). KE_rot = (1/2) I ω² Step 2: Find the moment of inertia (I) for a solid sphere about its diameter. The standard formula is I = (2/5)MR². I = (2/5) (2 kg) (0.1 m)² = (2/5) 2 0.01 = 0.008 kg m². Step 3: Substitute I and ω into the kinetic energy formula. ω = 10 rad/s. KE_rot = (1/2) (0.008 kg m²) (10 rad/s)² KE_rot = (1/2) 0.008 100 = 0.4 J. Final answer: The rotational kinetic energy of the sphere is 0.4 Joules.
• Q: A figure skater is spinning at an angular speed of 2 rad/s with her arms outstretched. Her moment of inertia in this position is 4.5 kg m². When she pulls her arms in, her moment of inertia reduces to 1.5 kg m². What is her new angular speed? A: Step 1: Identify the principle to be used. Since there are no external torques acting on the skater, her angular momentum is conserved. The principle is the Conservation of Angular Momentum: L_initial = L_final. Step 2: Write the equation for conservation of angular momentum. L = Iω, so I₁ω₁ = I₂ω₂. Step 3: Substitute the known values and solve for the unknown ω₂. I₁ = 4.5 kg m², ω₁ = 2 rad/s. I₂ = 1.5 kg m². (4.5) (2) = (1.5) ω₂ 9.0 = 1.5 * ω₂ ω₂ = 9.0 / 1.5 = 6 rad/s. Final answer: Her new angular speed is 6 rad/s.
• Q: A wheel of radius 0.5 m rolls without slipping on a horizontal surface. If the speed of its centre of mass is 10 m/s, what is the linear velocity of a point on the top of the wheel? A: Step 1: Understand the condition of rolling without slipping. For rolling without slipping, the velocity of the centre of mass (v_cm) is related to the angular velocity (ω) by v_cm = Rω. The motion of any point on the wheel is a superposition of the translational motion of the CM and the rotational motion about the CM. Step 2: Determine the velocity of the top point due to rotation. The point at the top of the wheel is moving forward due to rotation with a speed v_rot = Rω. Since v_cm = Rω, this means v_rot = v_cm = 10 m/s. Step 3: Combine the translational and rotational velocities. The velocity of the CM (v_cm = 10 m/s) is the translational velocity of the entire wheel. The velocity of the top point relative to the center is also 10 m/s in the same direction. So, the total velocity of the top point is v_top = v_cm + v_rot = 10 m/s + 10 m/s = 20 m/s. Final answer: The linear velocity of a point on the top of the wheel is 20 m/s.

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