Rotational Motion About a Fixed Axis – Quick-Read Notes

When an object spins around a fixed line (the axis), every point on it shares the same angle \( \theta \). Because only that one angle matters, the motion has just one degree of freedom. :contentReference[oaicite:1]{index=1}

1  Big Picture

• One-to-one analogy: \( \theta \) ↔ displacement \( x \), \( \omega \) ↔ speed \( v \), and \( \alpha \) ↔ acceleration \( a \). :contentReference[oaicite:3]{index=3}
• Definitions:
  • Angular speed: \( \displaystyle \omega = \frac{d\theta}{dt} \)
  • Angular acceleration: \( \displaystyle \alpha = \frac{d\omega}{dt} \)
  • Axis is locked in place, so you can treat \( \omega \) and \( \alpha \) as plain numbers. :contentReference[oaicite:5]{index=5}

2  Rotational “SUVAT” Equations (Uniform \( \alpha \))

1. \( \displaystyle \omega = \omega_0 + \alpha t \)  (Eq. 6.36) :contentReference[oaicite:7]{index=7}
2. \( \displaystyle \theta = \theta_0 + \omega_0 t + \tfrac12 \alpha t^{2} \)  (Eq. 6.37) :contentReference[oaicite:9]{index=9}
3. \( \displaystyle \omega^{2} = \omega_0^{2} + 2\alpha(\theta – \theta_0) \)  (Eq. 6.38) :contentReference[oaicite:11]{index=11}

A 30-second proof of Eq. 6.36

Because \( \alpha \) is constant, \( \tfrac{d\omega}{dt} = \alpha \). Integrate once to get \( \omega = \alpha t + C \). At \( t = 0 \), \( \omega = \omega_0 \), so \( C = \omega_0 \). That’s it—done! :contentReference[oaicite:13]{index=13}

3  Worked Example – Speeding-Up Motor Wheel

1. Initial speed: \( 1200 \text{ rpm} = 40\pi \text{ rad s}^{-1} \)
2. Final speed: \( 3120 \text{ rpm} = 104\pi \text{ rad s}^{-1} \)
3. Angular acceleration: \( \alpha = \dfrac{104\pi – 40\pi}{16} = 4\pi \text{ rad s}^{-2} \)
4. Angle turned: \( \theta = \omega_0 t + \tfrac12 \alpha t^{2} = 1152\pi \text{ rad} \)
5. Revolutions: \( N = \dfrac{\theta}{2\pi} = 576 \)

The wheel picks up an angular acceleration of \( 4\pi \text{ rad s}^{-2} \) and completes 576 turns.

4  Moments of Inertia to Memorise

Body	Axis	\( I \)
Thin ring (radius \( R \))	Perpendicular through centre	\( MR^{2} \)
Thin ring	Diameter	\( \tfrac12 MR^{2} \)
Thin rod (length \( L \))	Perpendicular at midpoint	\( \tfrac1{12} ML^{2} \)
Solid disc (radius \( R \))	Perpendicular through centre	\( \tfrac12 MR^{2} \)
Solid disc	Diameter	\( \tfrac14 MR^{2} \)
Hollow cylinder (radius \( R \))	Along axis	\( MR^{2} \)
Solid cylinder (radius \( R \))	Along axis	\( \tfrac12 MR^{2} \)
Solid sphere (radius \( R \))	Diameter	\( \tfrac25 MR^{2} \)

These eight pop up everywhere—flash-card them! :contentReference[oaicite:16]{index=16}

5  Linear ↔ Rotational Cheat-Sheet

• \( x \) ↔ \( \theta \)
• \( v \) ↔ \( \omega \)
• \( a \) ↔ \( \alpha \)

Whenever you solve a straight-line (linear) problem, swap symbols and you’ve got the spinning version. :contentReference[oaicite:18]{index=18}

6  High-Yield Ideas for NEET

1. Mastering the three rotational “SUVAT” equations. :contentReference[oaicite:20]{index=20}
2. Quickly converting rpm to rad s⁻¹ with \( 2\pi (\text{rpm})/60 \). :contentReference[oaicite:22]{index=22}
3. Remembering key moments of inertia (ring, disc, rod, cylinder, sphere). :contentReference[oaicite:24]{index=24}
4. Using linear-to-rotational analogies to build or check solutions. :contentReference[oaicite:26]{index=26}
5. Deriving \( \omega = \omega_0 + \alpha t \) from first principles to cement concepts. :contentReference[oaicite:28]{index=28}

Keep practising—before long, spinning problems will feel as comfortable as straight-line ones!