Rotational Motion – Study Notes (VITEEE Physics)
Goal: Understand the mechanics of rotating bodies, the equations that govern them, and how to solve typical problems that appear in the VITEEE exam.
1. Simple Explanation
Rotational motion is the motion of a rigid body around a fixed axis.
Just as linear motion is described by displacement, velocity, and acceleration, rotational motion is described by
| Linear Quantity | Rotational Counterpart |
|---|---|
| Displacement (s) | Angular displacement (θ) |
| Velocity (v) | Angular velocity (ω) |
| Acceleration (a) | Angular acceleration (α) |
| Force (F) | Torque (τ) |
| Mass (m) | Moment of inertia (I) |
The fundamental law linking torque and angular acceleration is Newton’s 2nd law for rotation:
[ \boxed{\tau = I , \alpha} ]
2. Key Points
| Concept | Symbol | Units | Typical Formula |
|---|---|---|---|
| Angular displacement | θ | rad | θ = s / r |
| Angular velocity | ω | rad s⁻¹ | ω = dθ/dt |
| Angular acceleration | α | rad s⁻² | α = dω/dt |
| Linear speed (at radius r) | v | m s⁻¹ | v = r ω |
| Centripetal acceleration | a_c | m s⁻² | a_c = r ω² = v²/r |
| Torque | τ | N m | τ = r F sin φ |
| Moment of inertia | I | kg m² | I = Σ m_i r_i² |
| Rotational kinetic energy | K_rot | J | K_rot = ½ I ω² |
| Angular momentum | L | N m s | L = I ω |
| Conservation of angular momentum | If no external torque → I₁ω₁ = I₂ω₂ | ||
| Rotational analogues of Newton’s laws | τ = Iα (2nd law), Στ = 0 (3rd law) |
Common Rigid Bodies (Moment of Inertia)
| Body | I (about axis through centre) | Formula |
|---|---|---|
| Solid cylinder (axis through centre) | ½ m r² | I = ½ m r² |
| Thin cylindrical shell | m r² | I = m r² |
| Solid sphere | ⅔ m r² | I = ⅔ m r² |
| Thin spherical shell | 2/3 m r² | I = 2/3 m r² |
| Point mass at distance r | m r² | I = m r² |
3. Worked Example
Problem (VITEEE‑style)
A solid disc of radius 0.3 m and mass 2 kg is initially at rest. A constant torque of 1.5 N·m is applied about its central axis. Find:
- Angular acceleration α.
- Angular speed ω after 4 s.
- Rotational kinetic energy at that instant.
Solution
Step 1: Find Moment of Inertia (I)
For a solid disc: ( I = \frac{1}{2} m r^2 )
[ I = \frac{1}{2} (2,\text{kg}) (0.3,\text{m})^2 = \frac{1}{2} \times 2 \times 0.09 = 0.09,\text{kg·m}^2 ]
Step 2: Angular Acceleration (α)
Using Newton’s 2nd law for rotation:
[ α = \frac{τ}{I} = \frac{1.5,\text{N·m}}{0.09,\text{kg·m}^2} = 16.67,\text{rad s}^{-2} ]
Step 3: Angular Speed after 4 s (ω)
With constant α and initial ω₀ = 0:
[ ω = ω_0 + α t = 0 + (16.67)(4) = 66.68,\text{rad s}^{-1} ]
Step 4: Rotational Kinetic Energy (K_rot)
[ K_{\text{rot}} = \frac{1}{2} I ω^2 = \frac{1}{2} (0.09)(66.68)^2 = 0.045 \times 4444.22 \approx 200,\text{J} ]
Answers
| Quantity | Value |
|---|---|
| α | 16.67 rad s⁻² |
| ω after 4 s | 66.68 rad s⁻¹ |
| K_rot | ≈ 200 J |
4. Quick Revision Summary
| Concept | Formula | Typical Use |
|---|---|---|
| θ | θ = s / r | Convert linear to angular displacement |
| ω | ω = Δθ / Δt | Angular speed |
| α | α = Δω / Δt | Angular acceleration |
| τ | τ = rF sinφ | Torque from a force |
| I | I = Σ m r² | Find for any body |
| τ = Iα | Compute α or τ | |
| K_rot | ½ I ω² | Rotational kinetic energy |
| L | I ω | Angular momentum |
| Conservation (no external τ) | I₁ω₁ = I₂ω₂ | Predict speed after shape change |