Board exam focus — System of Particles and Rotational Motion (CBSE & HBSE)

This chapter extends mechanics from point particles to extended rigid bodies. It covers the centre of mass and its motion, torque and angular momentum with their conservation laws, the moment of inertia together with the theorems of parallel and perpendicular axes, and rolling motion. It is a conceptually demanding and high-weightage chapter in both CBSE and HBSE Class 11 Physics, blending vector reasoning with rotational dynamics and energy.

Centre of Mass and its Motion

Centre of Mass

The centre of mass (CM) of a system is the point at which the entire mass of the system may be assumed to be concentrated for describing its translational motion. For a system of particles with masses mᵢ at positions xᵢ, yᵢ:

$$x_{cm} = \frac{\sum m_i x_i}{\sum m_i}, \qquad y_{cm} = \frac{\sum m_i y_i}{\sum m_i}$$

For a two-particle system on a line:

$$x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$$

Motion of the Centre of Mass

The CM moves as if all the mass were concentrated there and all external forces acted there:

$$\vec{F}_{ext} = M\vec{a}_{cm}$$

Key consequences:

• If the net external force is zero, the CM moves with constant velocity (or stays at rest).
• Internal forces (between parts of the system) never affect the motion of the CM.
• The total momentum of a system equals M v_cm.

Tip: The CM may lie outside the material of the body — for a ring or a horseshoe magnet, the CM is in empty space.

CBSE/HBSE trap: When a shell explodes in flight, fragments fly apart but the CM continues on the original parabolic path because gravity (the only external force) still acts as before.

Torque, Angular Momentum and their Conservation

Torque (Moment of Force)

Torque is the rotational analogue of force — it measures the turning effect of a force about an axis:

$$\vec{\tau} = \vec{r}\times\vec{F}, \qquad \tau = rF\sin\theta$$

Its SI unit is N m. Torque is maximum when the force is perpendicular to the position vector (θ = 90°).

Angular Momentum

Angular momentum is the rotational analogue of linear momentum:

$$\vec{L} = \vec{r}\times\vec{p}, \qquad L = I\omega$$

Its SI unit is kg m² s⁻¹ (or J s). The relation between torque and angular momentum is:

$$\vec{\tau} = \frac{d\vec{L}}{dt}$$

which is the rotational form of Newton's second law.

Conservation of Angular Momentum

If the net external torque on a system is zero, its total angular momentum remains constant:

$$\tau_{ext} = 0 \;\Rightarrow\; L = I\omega = \text{constant} \;\Rightarrow\; I_1\omega_1 = I_2\omega_2$$

Applications:

• An ice skater spins faster on pulling in the arms (I decreases, ω increases).
• A diver curls up to rotate faster, then extends to slow down before entering water.

Translational vs Rotational Analogy

CBSE/HBSE trap: Angular momentum is conserved only when external torque is zero; internal forces (like a skater's muscles) change I but not L, so ω adjusts to keep L = Iω constant.

Moment of Inertia, Axis Theorems and Rolling Motion

Moment of Inertia

The moment of inertia (I) is the rotational analogue of mass — it measures a body's resistance to angular acceleration:

$$I = \sum m_i r_i^2$$

It depends on the mass, its distribution, and the axis of rotation. The radius of gyration k is defined by I = Mk².

Moments of Inertia of Common Bodies

Theorem of Parallel Axes

The moment of inertia about any axis equals the moment of inertia about a parallel axis through the CM plus M times the square of the distance between the axes:

$$I = I_{cm} + Md^2$$

Theorem of Perpendicular Axes

For a planar (laminar) body, the moment of inertia about an axis perpendicular to its plane equals the sum of the moments of inertia about two mutually perpendicular axes in its plane intersecting at the same point:

$$I_z = I_x + I_y$$

This theorem applies only to plane laminas, not to three-dimensional bodies.

Rolling Motion

In rolling without slipping, the contact point is instantaneously at rest and v = ωR. The total kinetic energy is the sum of translational and rotational parts:

$$KE_{total} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv^2\left(1 + \frac{k^2}{R^2}\right)$$

For a body rolling down an incline of height h, the speed at the bottom is:

$$v = \sqrt{\frac{2gh}{1 + k^2/R^2}}$$

Since k²/R² differs for different shapes, a solid sphere reaches the bottom fastest and a ring slowest (the body with the smallest k²/R² wins).

Tip: For acceleration down an incline: a = g sin θ / (1 + k²/R²). The smaller the moment of inertia factor, the greater the acceleration.

CBSE/HBSE trap: The perpendicular axis theorem works only for flat (laminar) bodies; the parallel axis theorem applies to any rigid body.

Frequently asked questions

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