Rotational dynamics for JEE Main

Rotation is where many aspirants lose confidence, usually because they try to learn it as a new subject. It is not. Every idea has a linear twin you already know: torque plays the part of force, moment of inertia the part of mass, and angular momentum the part of momentum. Learn the dictionary, then solve problems the way you solved Newton's-law problems.

The rotational motion unit of the syllabus covers torque, angular momentum and its conservation, moment of inertia and radius of gyration, the parallel and perpendicular axis theorems, equilibrium of rigid bodies and the equations of rotational motion.

The linear–rotational dictionary

For constant α, the equations mirror kinematics: ω = ω₀ + αt, θ = ω₀t + ½αt², ω² = ω₀² + 2αθ.

Torque

τ = r × F, with magnitude rF sin θ, which is force × perpendicular distance from the axis to the line of action. A force whose line passes through the axis produces no torque, however large it is.

Moment of inertia

I = Σmr², where r is the perpendicular distance of each particle from the axis. It depends on the mass, how the mass is spread and which axis you choose. The radius of gyration k is defined by I = Mk².

Derivation: rod about one end. A rod of mass M and length L has dm = (M/L) dx.
I = ∫x² (M/L) dx from 0 to L = ML²/3.

The two axis theorems

• Parallel axis (any body): I = I_cm + Md², where d is the distance between the parallel axes and the first axis passes through the centre of mass.
• Perpendicular axis (flat bodies only): I_z = I_x + I_y, with x and y in the plane of the body and z perpendicular to it.

Example: a disc about a tangent in its plane. About a diameter, I = ¼MR²; shift by R: I = ¼MR² + MR² = (5/4)MR².

Angular momentum

For a rigid body about a fixed axis, L = Iω, and τ_ext = dL/dt. For a particle, L = r × p = mvr sin θ about a chosen point.

Conservation: if the external torque is zero, I₁ω₁ = I₂ω₂. A skater pulling their arms in, a diver tucking and a disc dropped onto a spinning disc all follow this.

Kinetic energy is not conserved when I changes. With KE = L²/2I and L fixed, reducing I increases KE; the skater's muscles supply that energy.

Equilibrium of rigid bodies

A rigid body is in equilibrium when ΣF = 0 and Στ = 0. In equilibrium, the torque balance works about any point. Choose a point where unknown forces act, so their torques vanish.

Rolling without slipping

Rolling combines translation and rotation, with v_cm = ωR at the contact point.

• KE = ½Mv² + ½Iω² = ½Mv²(1 + k²/R²).
• Rolling down an incline: a = g sin θ / (1 + k²/R²), and v at the bottom = √(2gh / (1 + k²/R²)).
• The friction needed is static and does no work, so mechanical energy is conserved.

The smallest k²/R² wins a race down an incline, so a solid sphere beats a disc, which beats a ring. Mass and radius do not matter.

Worked problems

Problem 1 (numerical answer): block on a pulley. A 2 kg block hangs from a light string wound on a uniform disc pulley of mass 4 kg and radius 0.2 m, which turns on a fixed frictionless axle. Find the block's acceleration (g = 10 m/s²).

Block: mg − T = ma, so 20 − T = 2a.
Pulley: TR = Iα = ½MR² × (a/R), so T = ½Ma = 2a.
Adding: 20 = 4a, so a = 5 m/s² and T = 10 N.

Problem 2: turntable. A person on a frictionless turntable spins at 2 rev/s with arms outstretched (I = 6 kg m²). They pull their arms in, reducing I to 2 kg m². Find the new rate and the ratio of kinetic energies.

I₁ω₁ = I₂ω₂: 6 × 2 = 2 × ω₂, so ω₂ = 6 rev/s.
KE₂/KE₁ = (2 × 6²)/(6 × 2²) = 72/24 = 3. The kinetic energy triples.

Problem 3: rolling sphere. A solid sphere rolls without slipping from rest down an incline of vertical height 7 m. Find its speed at the bottom (g = 10 m/s²).

v = √(2gh/(1 + 2/5)) = √(10gh/7) = √(10 × 10 × 7/7) = 10 m/s.

Problem 4: parallel axis. Find the moment of inertia of a uniform 3 kg rod, 2 m long, about an axis perpendicular to it and 0.5 m from its centre.

I = ML²/12 + Md² = 3 × 4/12 + 3 × 0.25 = 1 + 0.75 = 1.75 kg m².

Practice set

1. Find the moment of inertia of a rod of mass M and length L about an axis through one end, perpendicular to the rod.
2. For a disc and a ring rolling without slipping, what fraction of the total kinetic energy is rotational?
3. A torque of 1 N m acts on a 2 kg disc of radius 0.5 m, free to turn about its axis, starting from rest. Find α, ω after 5 s, and the angle turned.
4. A disc spins freely at ω. An identical disc, not rotating, is dropped onto it coaxially and they turn together. Find the final ω and the fraction of kinetic energy lost.
5. Find the radius of gyration of a solid sphere of radius 5 cm about a diameter.
6. Find the moment of inertia of a thin ring about a tangent perpendicular to its plane, and about a tangent in its plane.
7. A force F = (4î + 5ĵ) N acts at r = (2î + 3ĵ) m. Find the torque about the origin.
8. If Earth shrank to half its radius with no change in mass, how long would a day be? Treat Earth as a uniform sphere.

Answers

1. ML²/3.
2. Disc: (½)/(1 + ½) = 1/3. Ring: 1/(1 + 1) = 1/2.
3. I = ½ × 2 × 0.25 = 0.25 kg m², so α = 4 rad/s²; ω = 4 × 5 = 20 rad/s; θ = ½ × 4 × 25 = 50 rad.
4. Iω = 2Iω′, so ω′ = ω/2. KE goes from ½Iω² to ½(2I)(ω/2)² = ¼Iω², so half is lost.
5. k = R√(2/5) ≈ 5 × 0.632 ≈ 3.16 cm.
6. Perpendicular to its plane: MR² + MR² = 2MR². In its plane: ½MR² + MR² = (3/2)MR².
7. τ = r × F = (2 × 5 − 3 × 4) k̂ = −2k̂ N m.
8. I ∝ R², so I becomes I/4 and ω becomes 4ω. The day would last 6 hours.

What to do next

• Write the linear–rotational dictionary from memory.
• Derive I for a rod, ring and disc by integration, and practise both axis theorems.
• Solve 25 problems: pulleys with massive discs, angular momentum conservation and rolling.
• Revise centre of mass if the parallel-axis step feels shaky, then move on to gravitation.