System of Particles and Rotational Motion — Class 11 Physics

1. Centre of Mass

The centre of mass (COM) is the point whose motion represents the translational motion of a system as a whole. For particles with masses m_i at positions r_i, R = (Σm_ir_i)/(Σm_i).

For two particles on the x-axis, x_COM=(m₁x₁+m₂x₂)/(m₁+m₂). For a uniform symmetric body, the centre of mass lies at its geometric centre.

2. Motion of the Centre of Mass

The total external force determines the acceleration of the centre of mass: F_ext=Ma_COM, where M is the total mass. Internal forces cancel in pairs when considering the system as a whole.

3. Linear Momentum of a System

Total momentum is P=M V_COM. If the net external force is zero, total linear momentum remains constant.

4. Rigid Body and Angular Motion

A rigid body maintains fixed distances between its constituent particles. Angular displacement is θ, angular velocity is ω=dθ/dt, and angular acceleration is α=dω/dt.

For constant angular acceleration: ω=ω₀+αt, θ=ω₀t+½αt², and ω²=ω₀²+2αθ.

5. Torque

Torque measures the turning effect of a force about an axis: τ=r×F, with magnitude τ=rF sinθ. The perpendicular distance from the axis to the line of action of the force is the moment arm.

6. Rotational Equilibrium

A rigid body is in static equilibrium when both the net force and net torque are zero: ΣF=0 and Στ=0. Choosing the torque origin strategically can simplify problems.

7. Moment of Inertia

Moment of inertia measures rotational inertia: I=Σm_ir_i² for discrete particles. It depends on how mass is distributed relative to the chosen axis, not just on total mass.

Common results include: ring about central axis I=MR²; disc about central axis I=½MR²; solid sphere about diameter I=2/5 MR²; rod of length L about its centre I=1/12 ML²; rod about one end I=1/3 ML².

8. Radius of Gyration

If the total mass M is imagined concentrated at distance k from the axis such that the moment of inertia remains unchanged, I=Mk². Thus k=√(I/M).

9. Parallel and Perpendicular Axis Theorems

Parallel axis theorem: I=I_COM+Mh², where h is the separation between parallel axes.

For a plane lamina, perpendicular axis theorem: I_z=I_x+I_y when z is perpendicular to the plane through the same point.

10. Rotational Dynamics

The rotational analogue of Newton’s second law is τ=Iα for a rigid body rotating about a fixed axis when the net torque is considered about that axis.

11. Angular Momentum

For a rigid body rotating about a fixed axis, L=Iω. More generally angular momentum is r×p. If external torque about the relevant point or axis is zero, angular momentum is conserved.

12. Rolling Motion

For pure rolling without slipping, the translational and rotational motions are linked by v_CM=ωR. The condition does not mean that the point of contact has zero acceleration; it means its instantaneous velocity relative to the ground is zero.

Worked Numerical 1 — Centre of Mass

Question: Two particles of masses 2 kg and 3 kg are at x=0 m and x=5 m respectively. Find their centre of mass.

x_COM=(2×0+3×5)/(2+3)=15/5=3 m.

Worked Numerical 2 — Centre of Mass Velocity

Question: A 2 kg object moves at 4 m/s and a 3 kg object moves at 2 m/s in the same direction. Find the velocity of the centre of mass.

V_COM=(2×4+3×2)/5=14/5=2.8 m/s.

Worked Numerical 3 — Torque

Question: A 20 N force acts perpendicular to a 0.5 m lever arm. Find the torque.

τ=rF=0.5×20=10 N m.

Worked Numerical 4 — Angular Kinematics

Question: A wheel starts from rest with angular acceleration 2 rad/s². Find its angular speed after 5 s and angular displacement.

ω=0+2×5=10 rad/s.

θ=½(2)(5²)=25 rad.

Worked Numerical 5 — Moment of Inertia of a Ring

Question: A ring has mass 4 kg and radius 0.5 m. Find its moment of inertia about its central axis.

I=MR²=4(0.5)²=1 kg m².

Worked Numerical 6 — Rotational Dynamics

Question: A net torque of 12 N m acts on a body of moment of inertia 3 kg m². Find angular acceleration.

τ=Iα, so α=12/3=4 rad/s².

Worked Numerical 7 — Angular Momentum

Question: A disc has I=2 kg m² and rotates at 5 rad/s. Find its angular momentum.

L=Iω=2×5=10 kg m²/s.

Worked Numerical 8 — Rolling Without Slipping

Question: A wheel of radius 0.25 m rolls without slipping with centre-of-mass speed 5 m/s. Find its angular speed.

v=ωR, so ω=5/0.25=20 rad/s.

Explained MCQs

1. For a uniform rod, where is its centre of mass?
   Answer: At its geometric centre. Uniform mass distribution and symmetry place the COM at the midpoint.
2. Which quantity measures resistance to angular acceleration?
   Answer: Moment of inertia. It plays the rotational role analogous to mass in translation.
3. If the line of action of a force passes through the rotation axis, its torque about that axis is:
   Answer: Zero. The perpendicular moment arm is zero.
4. If external torque is zero, angular momentum:
   Answer: Remains constant. This follows from dL/dt=τ_ext.
5. For pure rolling, which relation holds?
   Answer: v_CM=ωR. It connects translation and rotation at the no-slip condition.
6. Does a larger mass always mean a larger moment of inertia?
   Answer: No. Distribution of mass relative to the axis is also crucial.

Competency Question

A child sits on a rotating stool while holding two dumbbells. When the child pulls the dumbbells closer to the body, the rotation speeds up. Explain this using conservation of angular momentum and the dependence of moment of inertia on distance from the axis.

HOTS

Two wheels have the same mass and radius, but one is a ring and the other is a solid disc. If the same torque is applied to both, which has the greater angular acceleration? Explain using their moments of inertia rather than simply memorising the result.

Common Mistakes

• Confusing centre of mass with centre of gravity in every situation.
• Using τ=rF without checking the angle or perpendicular moment arm.
• Assuming moment of inertia depends only on mass.
• Applying I=MR² to every circular object.
• Confusing angular velocity with angular acceleration.
• Thinking pure rolling means every point of the wheel has zero velocity.

Quick Revision

• R=(Σmr)/(Σm)
• P=MV_COM
• ω=dθ/dt, α=dω/dt
• τ=r×F
• τ=Iα
• I=Σmr²
• L=Iω
• I=I_COM+Mh²
• I_z=I_x+I_y for a plane lamina
• Pure rolling: v_CM=ωR