Physics · System Of Particles And Rotational Motion · NEET
In rotational motion, the work done by a torque is W = τθ (torque times the angle turned), and the power delivered is P = τω (torque times angular speed). These are the exact rotational twins of the linear formulas W = F·s and P = F·v, so if you swap force for torque and distance for angle, you already know them. Memory hook: "τ replaces F, θ replaces s, ω replaces v" — same equations, new letters.
A torque τ turning a wheel through angle θ does work W = τθ and delivers power P = τω. These mirror the linear formulas W = F·s and P = Fv, and the net work equals the change in rotational kinetic energy (1/2)Iω².
Your doubts, answered
Is the work done by a torque equal to the change in rotational kinetic energy?
Yes. This is the work-energy theorem for rotation: W = ΔKE_rot = (1/2)Iω₂² − (1/2)Iω₁². The net work done by all torques on a rotating body equals the change in its rotational kinetic energy, exactly like W = ΔKE in linear motion. NEET loves this: to find the work needed to STOP a spinning body, you just compute its current rotational KE = (1/2)Iω².
Why is rotational power P = τω and not τv?
Power is the rate of doing work. In rotation, work is W = τθ, so P = dW/dt = τ·(dθ/dt) = τω, because ω = dθ/dt is the angular speed. Compare it with linear motion where P = Fv and v = dx/dt. So the linear pair (F, v) simply becomes the rotational pair (τ, ω). Never write τv — v is a linear speed and does not belong here.
How do I calculate the work done by a constant torque?
Use W = τθ, where τ is the torque in newton-metres (N·m) and θ is the angle turned in radians (not degrees). For example, a torque of 5 N·m acting through 2 radians does W = 5 × 2 = 10 J of work. If the torque varies, you integrate: W = ∫τ dθ.
Does a torque do work when the body does not rotate?
No. If θ = 0 (no angular displacement), then W = τθ = 0, even if a large torque is applied. Work in rotation needs an actual turn, just as pushing a wall (large force, zero displacement) does zero work in linear motion. This is a common trap in NEET reasoning questions.
⚠️ The NEET trap ✗ To stop a spinning ring and a spinning sphere of the SAME mass, radius and angular speed, the same work is needed. ✓ The work to stop equals the rotational KE = (1/2)Iω². Since I depends on the shape (ring I = MR², sphere I = (2/5)MR²), the ring needs MORE work than the sphere for the same ω. Different shapes → different I → different work. 🧠 Work to stop = (1/2)Iω². Same ω does NOT mean same work — bigger I always means more work.
Real NEET questions
2018
Three bodies A (solid sphere), B (disc) and C (ring), each of mass M and radius R, spin about their symmetry axes with the same angular speed. The work needed to stop them satisfies:
A · W_B > W_A > W_C
B · W_A > W_B > W_C
C · W_C > W_B > W_A ✓
D · W_A > W_C > W_B
Solution: Step 1: Work to stop a rotating body = its rotational kinetic energy = (1/2)Iω² (work-energy theorem). Step 2: All three have the same ω, so W depends only on I. Step 3: Compare moments of inertia about the symmetry axis: ring I = MR², disc I = (1/2)MR², sphere I = (2/5)MR². Step 4: So I_ring > I_disc > I_sphere, which gives W_C > W_B > W_A. Correct option: C.
2016
A particle of mass 10 g moves on a circle of radius 6.4 cm with constant tangential acceleration. If its kinetic energy becomes 8×10⁻⁴ J by the end of the second revolution, the tangential acceleration is:
A · 0.1 m/s² ✓
B · 0.15 m/s²
C · 0.18 m/s²
D · 0.2 m/s²
Solution: Step 1: Kinetic energy gained equals the work done, and here KE = (1/2)mv². So v = √(2·KE/m) = √(2 × 8×10⁻⁴ / 0.01) = √0.16 = 0.4 m/s. Step 2: Distance covered in 2 revolutions: s = 2 × (2πr) = 2 × 2π × 0.064 = 0.804 m. Step 3: Using v² = u² + 2a_t·s with u = 0: a_t = v²/(2s) = 0.16/(2 × 0.804) = 0.16/1.608 ≈ 0.0995 ≈ 0.1 m/s². Correct option: A.
Solved System Of Particles And Rotational Motion NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
What is the formula for work done in rotational motion?
The work done by a torque τ turning a body through an angle θ (in radians) is W = τθ. If the torque changes with angle, W = ∫τ dθ. This is the rotational version of W = F·s.
What is the formula for power in rotational motion?
Power in rotation is P = τω, where τ is the torque and ω is the angular speed. It is the rotational twin of P = Fv in linear motion.
What is the work-energy theorem for rotation?
The net work done by all torques on a rotating body equals its change in rotational kinetic energy: W = (1/2)Iω₂² − (1/2)Iω₁². To stop a body (ω₂ = 0), the work needed equals (1/2)Iω₁².
How are the rotational formulas related to the linear ones?
They are direct analogues: force F becomes torque τ, displacement s becomes angle θ, and velocity v becomes angular velocity ω. So W = Fs → W = τθ and P = Fv → P = τω. Learn one pair, and you get the other for free.
Why is this topic important for NEET?
NEET regularly asks 'work needed to stop a rotating body' or 'ratio of rotational kinetic energies' questions. These reduce to W = (1/2)Iω², so knowing moment of inertia values for standard shapes lets you solve them in seconds.
