Chapter 6: System of Particles and Rotational Motion
Unit V – Motion of System of Particles and Rigid Body
6.1 Centre of Mass
The centre of mass (CM) of a system of particles is the point where the whole mass can be assumed concentrated.
For Two-Particle System
\[ x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2} \]
\[ y_{cm} = \frac{m_1 y_1 + m_2 y_2}{m_1 + m_2} \]
For a System of Particles
\[ \vec{r}_{cm} = \frac{\sum m_i \vec{r}_i}{M} \]
For Uniform Rod (length L)
\[ x_{cm} = \frac{L}{2} \]
6.2 Torque (Moment of Force)
The torque (or moment of force) of a force \(\vec{F}\) about a point:
\[ \vec{\tau} = \vec{r} \times \vec{F} \]
\[ |\tau| = rF\sin\theta \]
- SI unit: N·m
- Torque is a vector (direction by right-hand rule)
- Torque = force × perpendicular distance from pivot
6.3 Angular Momentum
The angular momentum \(\vec{L}\) of a particle about a point:
\[ \vec{L} = \vec{r} \times \vec{p} = \vec{r} \times m\vec{v} \]
\[ |\vec{L}| = mvr\sin\theta \]
For a rigid body:
\[ L = I\omega \]
Newton’s 2nd Law for Rotation
\[ \vec{\tau} = \frac{d\vec{L}}{dt} = I\vec{\alpha} \]
6.4 Conservation of Angular Momentum
If net external torque on a system is zero:
\[ \vec{L} = I\omega = \text{constant} \]
Example: A figure skater pulls her arms in → \(I\) decreases → \(\omega\) increases.
\[ I_1\omega_1 = I_2\omega_2 \]
6.5 Moment of Inertia
The moment of inertia (I) is the rotational analog of mass:
\[ I = \sum m_i r_i^2 = \int r^2,dm \]
Moments of Inertia of Common Bodies
| Body | Axis | \(I\) |
|---|---|---|
| Thin rod (length L) | Through centre ⊥ to rod | \(\frac{1}{12}mL^2\) |
| Thin rod (length L) | Through end ⊥ to rod | \(\frac{1}{3}mL^2\) |
| Disk (radius R) | Through centre ⊥ to plane | \(\frac{1}{2}mR^2\) |
| Ring (radius R) | Through centre ⊥ to plane | \(mR^2\) |
| Solid sphere (radius R) | Through diameter | \(\frac{2}{5}mR^2\) |
| Hollow sphere (radius R) | Through diameter | \(\frac{2}{3}mR^2\) |
Radius of Gyration
\[ I = Mk^2 \Rightarrow k = \sqrt{\frac{I}{M}} \]
Parallel Axis Theorem
\[ I = I_{cm} + Md^2 \]
where \(d\) is the distance between the two parallel axes.
Perpendicular Axis Theorem (planar bodies only)
\[ I_z = I_x + I_y \]
6.6 Equations of Rotational Motion
Analogous to linear kinematic equations:
\[ \omega = \omega_0 + \alpha t \]
\[ \theta = \omega_0 t + \frac{1}{2}\alpha t^2 \]
\[ \omega^2 = \omega_0^2 + 2\alpha\theta \]
6.7 Rotational Kinetic Energy
\[ KE_{rot} = \frac{1}{2}I\omega^2 \]
For a rolling body (without slipping):
\[ KE_{total} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 = \frac{1}{2}mv^2\left(1 + \frac{I}{mR^2}\right) \]
6.8 Equilibrium of a Rigid Body
A rigid body is in equilibrium if:
- Translational equilibrium: \(\sum\vec{F} = 0\)
- Rotational equilibrium: \(\sum\vec{\tau} = 0\)
6.9 Comparison: Linear vs. Rotational Motion
Key Formulas Summary
| Formula | Quantity |
|---|---|
| \(x_{cm} = \dfrac{\sum m_i x_i}{M}\) | Centre of mass |
| \(\tau = rF\sin\theta\) | Torque |
| \(L = I\omega\) | Angular momentum |
| \(I = Mk^2\) | Moment of inertia (radius of gyration) |
| \(I = I_{cm} + Md^2\) | Parallel axis theorem |
| \(\tau = I\alpha\) | Rotational Newton’s 2nd law |
Practice Questions
Section A – MCQ (1 mark each)
Q1. The moment of inertia of a solid sphere about its diameter is:
(a) \(\frac{2}{5}mR^2\) (b) \(\frac{2}{3}mR^2\) (c) \(mR^2\) (d) \(\frac{1}{2}mR^2\)
Answer
(a) \(\frac{2}{5}mR^2\)
Q2. A planet moves faster when it is:
(a) Far from the sun (b) Close to the sun (c) Speed is constant (d) At aphelion
Answer
(b) Close to the sun — by conservation of angular momentum, smaller radius → greater velocity.
Section B – Short Answer (2–3 marks)
Q3. A disc (radius 0.5 m, mass 2 kg) rotates at 5 rad/s. Find its angular momentum and rotational kinetic energy.
Answer
\(I = \frac{1}{2}mR^2 = \frac{1}{2} \times 2 \times 0.25 = 0.25 \text{ kg·m}^2\)
\(L = I\omega = 0.25 \times 5 = \mathbf{1.25 \text{ kg·m}^2/\text{s}}\)
\(KE = \frac{1}{2}I\omega^2 = \frac{1}{2} \times 0.25 \times 25 = \mathbf{3.125 \text{ J}}\)
Section D – Competency-Based Questions
Q4. (Case Study) A figure skater has a moment of inertia \(I_1 = 4 \text{ kg·m}^2\) when her arms are spread out, rotating at \(\omega_1 = 2 \text{ rad/s}\). She pulls her arms in, reducing \(I_2 = 1 \text{ kg·m}^2\).
(i) Find her new angular velocity.
(ii) Find the ratio of initial to final KE.
(iii) Does KE increase or decrease? Where does the extra energy come from?
Answer
(i) By conservation of angular momentum: \(I_1\omega_1 = I_2\omega_2\)
\(\omega_2 = \dfrac{4 \times 2}{1} = \mathbf{8 \text{ rad/s}}\)
(ii) \(\dfrac{KE_1}{KE_2} = \dfrac{\frac{1}{2}I_1\omega_1^2}{\frac{1}{2}I_2\omega_2^2} = \dfrac{4 \times 4}{1 \times 64} = \dfrac{16}{64} = \mathbf{1:4}\)
(iii) KE increases 4 times. The extra energy comes from the work done by the skater’s muscles in pulling her arms inward.