Dynamics of Rotational Motion about a Fixed Axis
When an object spins around an axle, every point in it shares the same angular displacement \(\theta\), angular speed \(\omega\), and angular acceleration \(\alpha\). That makes life easy: we can treat the whole rigid body as if it were “one big particle” located at its axle and use neat rotational versions of our usual linear formulas.:contentReference[oaicite:0]{index=0}
1. Side-by-side comparison
| Linear idea | Rotational twin |
|---|---|
| Displacement \(x\) | Angular displacement \(\theta\) |
| Speed \(v=\dfrac{dx}{dt}\) | Angular speed \(\omega=\dfrac{d\theta}{dt}\) |
| Acceleration \(a=\dfrac{dv}{dt}\) | Angular acceleration \(\alpha=\dfrac{d\omega}{dt}\) |
| Mass \(M\) | Moment of inertia \(I\) |
| Force \(F=Ma\) | Torque \(\tau=I\alpha\) |
| Work \(dW=F\,ds\) | Work \(dW=\tau\,d\theta\) |
| Kinetic energy \(K=\dfrac{1}{2}Mv^{2}\) | Kinetic energy \(K=\dfrac{1}{2}I\omega^{2}\) |
| Power \(P=Fv\) | Power \(P=\tau\omega\) |
| Linear momentum \(p=Mv\) | Angular momentum \(L=I\omega\) |
Treat \(I\) exactly like mass—it tells you how stubborn the object is about changing its spin.:contentReference[oaicite:1]{index=1}
2. Torque and angular acceleration
Give a rigid body a net torque \(\tau\) about its fixed axis and you create an angular acceleration
\(\boxed{\tau = I\alpha}\)
This is Newton’s second law in rotation-speak: bigger torque or smaller moment of inertia means a quicker spin-up.:contentReference[oaicite:2]{index=2}
3. Work, power, and energy in rotation
- The tiny work done by a torque during a tiny turn \(d\theta\) is \(dW=\tau\,d\theta\).:contentReference[oaicite:3]{index=3}
- Divide by time to get power: \(P=\tau\omega\).:contentReference[oaicite:4]{index=4}
- Spin a body from rest up to speed \(\omega\) and its rotational kinetic energy becomes \(K=\tfrac12 I\omega^{2}\).:contentReference[oaicite:5]{index=5}
4. Which torques actually matter?
Because the axle stays put, only the torque components pointing along the axle can twist the body. That lets you ignore:
- Forces whose directions run along the axle (their torques point sideways).
- Position-vector parts that happen to line up with the axle (again, sideways torques).
Focus on forces and distances that live in the plane perpendicular to the axle and calculations stay neat.:contentReference[oaicite:6]{index=6}
5. Worked example (flywheel)
A light cord wraps around a 20 kg flywheel of radius 0.20 m. You pull with 25 N.
- Angular acceleration \(\tau = F R = 25\,\text{N}\times0.20\,\text{m}=5.0\,\text{N·m}\), \(I=\tfrac12MR^{2}=0.4\,\text{kg·m}^{2}\), so \(\alpha = \dfrac{\tau}{I}=12.5\,\text{s}^{-2}\).
- Work done while 2 m of cord unwinds: \(W = F\,\ell = 25\,\text{N}\times2\,\text{m}=50\,\text{J}\).
- Final kinetic energy Unwound length ⇒ angle \(\theta =\dfrac{2\,\text{m}}{0.20\,\text{m}} = 10\,\text{rad}\). \(\omega^{2}=2\alpha\theta = 2(12.5)(10)=250\Rightarrow \omega\approx 16\,\text{rad/s}\). \(K=\tfrac12 I\omega^{2}=50\,\text{J}\).
- The work you do equals the energy the wheel gains—no surprises, no friction.
Try the same steps with different pulls or radii—only the numbers change, the method doesn’t.:contentReference[oaicite:7]{index=7}
6. High-yield checkpoints for NEET
- \(\tau = I\alpha\)—linking torque to angular acceleration.
- Energy shortcut \(K=\tfrac12 I\omega^{2}\) for rotating bodies.
- Work-angle relation \(W=\tau\theta\) and power formula \(P=\tau\omega\).
- Moment of inertia playing the same role in rotation as mass does in straight-line motion.
- Strategy: focus only on torque components along the fixed axis to simplify problems.
You’ve got the tools—now spin up those practice questions!
