Rotational motion for NEET

Many students find rotational motion the hardest chapter of Class 11 mechanics. It usually feels hard because it looks like a new subject. It is not. Every rotational quantity has a linear twin, and every rotational equation copies a linear one. Once you see the pairs, most NEET questions here become one-line substitutions, with the moment of inertia table doing the heavy lifting.

The chapter also carries a few ideas that are tested as concepts rather than numbers: where the centre of mass is, why a skater spins faster, and why a door handle sits far from the hinge.

Centre of mass

For two particles on a line, x_cm = (m₁x₁ + m₂x₂) / (m₁ + m₂). The centre of mass lies on the line joining them, closer to the heavier one. For a uniform symmetric body (rod, ring, disc, sphere), it is at the geometric centre.

Two facts are tested often:

• The centre of mass moves as if the whole mass were there and all external forces acted on it: M a_cm = F_ext.
• Internal forces cannot change its motion. If a shell explodes in mid-air, the centre of mass of the fragments keeps following the original parabola (until a fragment lands).

Linear and rotational pairs

With constant α, the equations of motion carry over: ω = ω₀ + αt, θ = ω₀t + ½αt², ω² = ω₀² + 2αθ.

Torque and angular momentum

Torque τ = r × F. Its magnitude is rF sin θ, which equals the force times its perpendicular distance from the axis. That is why a door handle is placed far from the hinges, and why pushing along the line of the hinge does nothing.

Angular momentum of a particle is L = r × p. For a rigid body turning about a fixed axis, L = Iω. The rotational second law is τ = dL/dt.

Moment of inertia

I = Σmr² measures how hard it is to change a body's rotation. It depends on the mass, on how that mass is spread about the axis, and on which axis you choose. The radius of gyration k is defined by I = Mk²: the distance at which all the mass could be placed to give the same I.

The two axis theorems

• Parallel axes: I = I_cm + Md², where the known axis passes through the centre of mass and the new axis is parallel to it at distance d. Rod about its end: ML²/12 + M(L/2)² = ML²/3.
• Perpendicular axes: I_z = I_x + I_y, for a flat (planar) body only, with x and y in its plane and z perpendicular to it. Ring: MR² = 2 I_diameter, so I_diameter = ½MR².

Conservation of angular momentum

If no external torque acts, L = Iω stays constant. A skater who pulls in the arms reduces I, so ω rises. A diver tucks in to spin faster and opens out to slow down before entering the water. Kepler's second law of planetary motion is the same principle at planetary scale.

Kinetic energy is not conserved in these cases. With L fixed, KE = L²/2I, so it rises when I falls. The extra energy comes from the work the skater's muscles do in pulling the arms in.

Equilibrium of a rigid body

A rigid body is in equilibrium only if both conditions hold:

• Net force is zero (no translation): ΣF = 0.
• Net torque about any point is zero (no rotation): Στ = 0.

Two equal and opposite forces not on the same line form a couple: net force zero, but a turning effect remains. The principle of moments (clockwise moments = anticlockwise moments) is the torque condition applied to levers and see-saws.

Worked numericals

Example 1: centre of mass

Masses of 2 kg at x = 0 and 3 kg at x = 5 m. Find the centre of mass.

• x_cm = (2 × 0 + 3 × 5) / 5 = 3 m, closer to the heavier mass as expected.

Example 2: a flywheel from rest

A torque of 20 N m acts on a wheel with I = 4 kg m², starting from rest, for 4 s.

• α = τ/I = 20/4 = 5 rad s⁻².
• ω = 5 × 4 = 20 rad s⁻¹. Angle turned θ = ½ × 5 × 16 = 40 rad.
• KE = ½ × 4 × 400 = 800 J. Check with work: τθ = 20 × 40 = 800 J ✓.

Example 3: disc about a tangent

A disc of mass 2 kg and radius 0.5 m. Find I about (i) a tangent perpendicular to its plane and (ii) a tangent in its plane.

• (i) Parallel axis from the central perpendicular axis: ½MR² + MR² = 3/2 MR² = 1.5 × 2 × 0.25 = 0.75 kg m².
• (ii) Parallel axis from a diameter: ¼MR² + MR² = 5/4 MR² = 1.25 × 0.5 = 0.625 kg m².

Example 4: skater pulls in the arms

A skater spins at 2 rad s⁻¹ with I = 3 kg m², then pulls in the arms so I = 1 kg m².

• 3 × 2 = 1 × ω, so ω = 6 rad s⁻¹.
• KE before = ½ × 3 × 4 = 6 J. KE after = ½ × 1 × 36 = 18 J, three times larger. The skater's muscles supplied the extra 12 J.

Practice MCQs

1. The moment of inertia of a 2 kg disc of radius 0.5 m about its central perpendicular axis is: (a) 0.125 kg m² (b) 0.25 kg m² (c) 0.5 kg m² (d) 1 kg m²
2. The moment of inertia of a uniform rod (mass M, length L) about a perpendicular axis through one end is: (a) ML²/12 (b) ML²/6 (c) ML²/3 (d) ML²
3. The moment of inertia of a ring (mass M, radius R) about a diameter is: (a) MR² (b) ½MR² (c) ¼MR² (d) 2MR²
4. The radius of gyration of a solid sphere about a diameter is: (a) 0.4R (b) 0.5R (c) about 0.63R (d) R
5. A 10 N force acts on a door 0.5 m from the hinge, at 30° to the door. The torque is: (a) 2.5 N m (b) 4.3 N m (c) 5 N m (d) 10 N m
6. A disc spinning at 10 rad s⁻¹ has its moment of inertia doubled with no external torque. Its new angular speed is: (a) 2.5 rad s⁻¹ (b) 5 rad s⁻¹ (c) 10 rad s⁻¹ (d) 20 rad s⁻¹
7. A 30 kg child sits 2 m from the pivot of a see-saw. Where must a 40 kg child sit on the other side to balance? (a) 1 m (b) 1.5 m (c) 2 m (d) 2.67 m
8. A shell explodes at the top of its parabolic path. The centre of mass of the fragments: (a) falls vertically (b) stops (c) continues along the original parabola (d) moves in a straight line upward

Answers

1. (b) ½ × 2 × 0.25 = 0.25 kg m².
2. (c) ML²/12 + M(L/2)² = ML²/3.
3. (b) Perpendicular axes: MR² = 2 I_d, so I_d = ½MR².
4. (c) k = √(2/5) R ≈ 0.63R.
5. (a) τ = rF sin θ = 0.5 × 10 × 0.5 = 2.5 N m.
6. (b) Iω constant: 10/2 = 5 rad s⁻¹.
7. (b) 30 × 2 = 40 × d, so d = 1.5 m.
8. (c) The explosion's forces are internal; only gravity acts from outside.

What to do next

• Rebuild the moment of inertia table from memory, and derive the ring-diameter and rod-end values with the axis theorems.
• Write the linear–rotational pairs table on your formula sheet and use it to translate each new formula.
• Solve previous-year NEET questions on moment of inertia and angular momentum, timed.
• Go on to gravitation for NEET, where angular momentum returns in Kepler's second law.

For collisions and energy methods that feed into this chapter, see work, energy and power.