"Motion of a Rigid Body"

About

So far we have treated objects as point particles. In reality, bodies have size and shape. This chapter introduces the motion of rigid bodies — objects where the distances between any two constituent particles remain fixed. You will learn about centre of mass, torque (the rotational analogue of force), angular momentum, moment of inertia, and the laws governing rotational motion.

Key Concepts

7.1 Rigid Body

A rigid body is a body in which the separation between any two constituent particles does not change with applied force or during motion. The shape and size of a rigid body remain fixed.

Examples of rigid bodies: A wooden frame with firmly attached rods, a metal sphere, a bicycle wheel.

Not rigid bodies: A heap of sand (particles move relative to each other), water, a rubber band (deforms under force).

7.2 Centre of Mass

The centre of mass (CM) of a system of particles is the point that moves as if the entire mass of the system were concentrated there and all external forces were applied at that point.

For a system of $n$ particles:

$X_{\text{CM}}=\frac{{\sum }_{i=1}^n{m}_i{x}_i}{{\sum }_{i=1}^n{m}_i},Y_{\text{CM}}=\frac{{\sum }_{i=1}^n{m}_i{y}_i}{{\sum }_{i=1}^n{m}_i}$

For two particles: $\frac{r_1}{r_2}=\frac{m_2}{m_1}$

The ratio of distances from the CM is inversely proportional to the ratio of their masses. The heavier particle is closer to the CM.

7.3 Types of Motion of a Rigid Body

7.4 Torque (Moment of Force)

Torque is the rotational analogue of force — it measures the tendency of a force to rotate a body about an axis.

$\vec{\tau }=\vec{r}\times \vec{F}$

Magnitude: $\tau =rF\sin \theta$

SI unit: N⋅m

Torque is a vector quantity. Direction given by the right-hand rule.

• Torque is zero when force acts along the line through the axis ($\theta =0°$ or $180°$)
• Torque is maximum when force is perpendicular to $\vec{r}$ ($\theta =90°$)

7.5 Angular Momentum

Angular momentum is the rotational analogue of linear momentum.

$\vec{L}=\vec{r}\times \vec{p}$

For a particle moving in a circle: $L=mvr=m{r}^2\omega$

SI unit: kg⋅m²/s

Conservation of angular momentum: When no external torque acts on a system, its total angular momentum remains constant.

$I_1{\omega }_1=I_2{\omega }_2$

Example: An ice skater spins faster when pulling arms inward (decreasing $I$ increases $\omega$).

7.6 Moment of Inertia

Moment of inertia ($I$) is the rotational analogue of mass — it measures a body's resistance to angular acceleration.

$I=\sum m_i{r}_i^2$

For continuous bodies: $I=\int r^{2}dm$

SI unit: kg⋅m²

Moment of inertia depends on:

1. Mass of the body
2. Distribution of mass about the axis
3. Position and orientation of the axis of rotation

Radius of gyration ($k$): $I=M{k}^2⟹k=\sqrt{\frac{I}{M}}$

It is the distance from the axis at which the entire mass of the body can be assumed to be concentrated to give the same moment of inertia.

7.7 Theorems of Moment of Inertia

Theorem of Parallel Axes: $I=I_{\text{CM}}+M{d}^2$

Where $d$ is the perpendicular distance between the two parallel axes.

Theorem of Perpendicular Axes (for planar bodies only): $I_z=I_x+I_y$

Where x and y are two perpendicular axes in the plane, and z is perpendicular to the plane.

7.8 Moment of Inertia of Common Bodies

7.9 Rotational Kinetic Energy

$K_{\text{rot}}=\frac{1}{2}I{\omega }^2$

For a body undergoing both translation and rotation (e.g., rolling): $K_{\text{total}}=\frac{1}{2}M{v}_{\text{CM}}^2+\frac{1}{2}{I}_{\text{CM}}{\omega }^2$

7.10 Newton's Second Law for Rotation

$\tau =I\alpha$

Where $\alpha$ is the angular acceleration. This is analogous to $F=ma$ for translation.

INTEXT QUESTIONS 7.1

Q1. A frame is made of six wooden rods. The rods are firmly attached to each other. Can this system be considered a rigid body?

Ans: Yes, this system can be considered a rigid body. A rigid body is one in which the separation between constituent particles does not change with motion. Since the six wooden rods are firmly attached to each other, the distances between all points in the system remain fixed during any motion. The shape and size of the frame are preserved during motion.

Q2. Can a heap of sand be considered a rigid body? Explain your answer.

Ans: No, a heap of sand cannot be considered a rigid body. A rigid body must maintain fixed distances between all its constituent particles during motion. In a heap of sand:

• Individual sand particles can move relative to each other when any force is applied
• Particles can slide and roll, causing the shape and internal structure to change
• The separation between constituent particles does NOT remain constant
• Sand particles are only loosely bound by friction and gravity, not firmly attached

A heap of sand is a deformable body, not a rigid body.

INTEXT QUESTIONS 7.2

Q1. The grid has particles A, B, C, D and E with masses 1.0 kg, 2.0 kg, 3.0 kg, 4.0 kg and 5.0 kg respectively. Calculate the coordinates of the position of the centre of mass of the system.

Given coordinates from the grid:

• A (1, 1), mass = 1 kg
• B (3, 1), mass = 2 kg
• C (3, 3), mass = 3 kg
• D (1, 4), mass = 4 kg
• E (2, 2), mass = 5 kg

Ans:

$X_{\text{CM}}=\frac{\sum m_i{x}_i}{\sum m_i}=\frac{1(1)+2(3)+3(3)+4(1)+5(2)}{1+2+3+4+5}$ $=\frac{1+6+9+4+10}{15}=\frac{30}{15}=2.0$

$Y_{\text{CM}}=\frac{\sum m_i{y}_i}{\sum m_i}=\frac{1(1)+2(1)+3(3)+4(4)+5(2)}{15}$ $=\frac{1+2+9+16+10}{15}=\frac{38}{15}=2.53$

Centre of mass: (2.0, 2.53)

Q2. If three particles of masses m₁ = 1 kg, m₂ = 2 kg, and m₃ = 3 kg are situated at the corners of an equilateral triangle of side 1.0 m, obtain the position coordinates of the centre of mass of the system.

Ans: Place the triangle in a coordinate system:

• m₁ at origin: (0, 0)
• m₂ at: (1.0, 0)
• m₃ at: (0.5, √3/2) = (0.5, 0.866)

$X_{\text{CM}}=\frac{1(0)+2(1.0)+3(0.5)}{1+2+3}=\frac{0+2.0+1.5}{6}=\frac{3.5}{6}=0.583\text{ m}$

$Y_{\text{CM}}=\frac{1(0)+2(0)+3(0.866)}{6}=\frac{0+0+2.598}{6}=0.433\text{ m}$

Centre of mass: (0.583 m, 0.433 m)

Q3. Show that the ratio of the distances of two particles from their common centre of mass is inversely proportional to the ratio of their masses.

Ans: Consider two particles of masses $m_1$ and $m_2$ separated by distance $d$. Let the CM be at distance $r_1$ from $m_1$ and $r_2$ from $m_2$.

$r_1+r_2=d$

For the centre of mass, the moment about CM must be zero: $m_1{r}_1=m_2{r}_2$

$\frac{r_1}{r_2}=\frac{m_2}{m_1}$

Therefore, the ratio of distances ($r_1:r_2$) is inversely proportional to the ratio of masses ($m_1:m_2$). The heavier particle is closer to the centre of mass.

Terminal Exercise

1. Define a rigid body. Give three examples of rigid bodies and three examples of non-rigid bodies.

2. Define centre of mass. How does it differ from centre of gravity?

3. Three particles of masses 1 kg, 2 kg and 3 kg are placed at the vertices of a right-angled triangle with sides 3 m, 4 m and 5 m. Find the position of the centre of mass.

4. Define torque. Show that torque is the rate of change of angular momentum: $\vec{\tau }=\frac{d\vec{L}}{dt}$.

5. State and prove the law of conservation of angular momentum. Explain why an ice skater spins faster when pulling in her arms.

6. Define moment of inertia. On what factors does it depend?

7. State and prove: (a) Theorem of parallel axes, (b) Theorem of perpendicular axes.

8. Derive expressions for the moment of inertia of: (a) a thin rod about an axis through its centre and perpendicular to its length, (b) a solid sphere about its diameter.

9. A solid cylinder of mass 20 kg and radius 0.5 m rotates about its axis at 120 rpm. Calculate: (a) moment of inertia, (b) angular speed in rad/s, (c) rotational kinetic energy.

10. A constant torque of 100 N⋅m acts on a flywheel. If the flywheel starts from rest, find its angular velocity after 10 seconds. (I = 50 kg⋅m²)

11. A uniform rod of length 2 m and mass 10 kg is pivoted at one end. Find its moment of inertia about the pivot.

12. Explain why a tightrope walker carries a long pole to maintain balance.

Worked Examples

Example 1: Centre of Mass

Problem: Find the CM of two particles of masses 2 kg and 3 kg placed 5 m apart.

Solution: Let $r_1$ be distance from the 2 kg mass.

$r_1=\frac{m_2}{m_1+m_2}\times d=\frac{3}{2+3}\times 5=3\text{ m}$

The CM is 3 m from the 2 kg mass (and 2 m from the 3 kg mass).

Example 2: Torque

Problem: A force of 10 N acts at a distance of 0.5 m from the axis of rotation at an angle of 30° to the radius vector. Find the torque.

Solution: $\tau =rF\sin \theta =0.5\times 10\times \sin 30°=0.5\times 10\times 0.5=2.5\text{ N⋅m}$

Example 3: Angular Momentum

Problem: A particle of mass 0.2 kg moves in a circle of radius 2 m with a speed of 5 m/s. Find its angular momentum about the centre.

Solution: $L=mvr=0.2\times 5\times 2=2{\text{ kg⋅m}}^2\text{/s}$

Example 4: Moment of Inertia Using Parallel Axis Theorem

Problem: Find the moment of inertia of a thin rod of mass M and length L about an axis through one end, perpendicular to its length.

Solution: $I_{\text{CM}}=\frac{1}{12}M{L}^2$, $d=\frac{L}{2}$

Using parallel axis theorem: $I=I_{\text{CM}}+M{d}^2=\frac{1}{12}M{L}^2+M{\left(\frac{L}{2}\right)}^2$ $=\frac{M{L}^2}{12}+\frac{M{L}^2}{4}=\frac{M{L}^2+3M{L}^2}{12}=\frac{4M{L}^2}{12}=\frac{1}{3}M{L}^2$

Common Mistakes

1. Confusing centre of mass with centre of gravity: CM depends only on mass distribution; CG also depends on the gravitational field. They coincide in a uniform gravitational field.
2. Forgetting torque is a vector: Direction matters — use the right-hand rule.
3. Applying perpendicular axis theorem to 3D bodies: It works only for planar (2D) bodies.
4. Using the wrong moment of inertia formula: Always note which axis the body rotates about.
5. Forgetting that moment of inertia depends on the axis: The same body has different $I$ for different axes.

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