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Class 11 Science (PCM)

Physics Part 1

System of Particles and Rotational Motion

System of Particles and Rotational Motion

Study Action Plan (Sorted by Difficulty)
Angular Momentum
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Centre of Mass
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Conservation of Linear Momentum
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Dynamics of Rotation About Fixed Axis
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Equilibrium of Rigid Bodies
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Moment of Inertia
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Motion of Centre of Mass
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Rolling Motion and Combined Rotation-Translation
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Rotational Motion Kinematics
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Theorems of Moment of Inertia
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Torque and Angular Momentum Relation
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Types of Motion of Rigid Bodies
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Vector Product of Two Vectors
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Learning Objectives

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Learning Objectives

  • Understand the mathematical determination of the center of mass for systems of particles and rigid bodies, including derivations for linear and spatial distributions.
  • Calculate angular momentum for particles and systems, and comprehend its conservation and relation to torque.
  • Derive the relationship between torque and the rate of change of angular momentum, analogous to force and linear momentum.
  • Define and explore the properties of the vector (cross) product, including its application in defining torque and angular momentum.
  • Analyze the kinematics of rotational motion, including angular velocity and its relation to linear velocity, and angular acceleration.
  • Formulate the conditions for mechanical equilibrium involving both translational and rotational components, with vector and scalar equation formulations.
  • Apply Newton's laws to systems of particles, emphasizing the conservation of linear momentum in isolated systems.
  • Differentiate between types of motion of rigid bodies, including pure translation, pure rotation about a fixed axis, and combined translation plus rotation such as rolling motion.
  • Examine the motion of the center of mass under external forces, using the equation MA=FextMA = F_{\text{ext}}, with examples like projectile explosion/fragments and binary systems.
  • Understand the concept of moment of inertia as the rotational analogue of mass, its dependence on mass distribution and axis, and standard results for simple bodies.
  • Utilize the perpendicular-axis theorem and parallel-axis theorem with applications to lamina, rod, disc, ring, and composite bodies.
  • Explore the dynamics of rotation about a fixed axis, including the relation τ=Iα\tau = I\alpha, rotational kinetic energy, work-energy relation for rotation, and comparison with linear dynamics.
  • Investigate rolling motion and combined rotation-translation, including rolling without slipping, the relation v=Rωv = R\omega, kinetic energy of a rolling body, and rolling on an inclined plane.

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Revision Notes & Summary

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Chapter Notes

Centre of Mass

  • Definition: The center of mass (COM) of a system of particles is the point where the total mass of the system can be considered to be concentrated.
  • Mathematical Determination:
    • For a system of particles, the position of the COM is given by: X=∑mixi∑mi,Y=∑miyi∑mi,Z=∑mizi∑miX = \frac{\sum m_i x_i}{\sum m_i}, \quad Y = \frac{\sum m_i y_i}{\sum m_i}, \quad Z = \frac{\sum m_i z_i}{\sum m_i}
    • For continuous mass distributions, integrals replace summations.
  • Properties: The COM moves as if all external forces act at this point.

Angular Momentum

  • Definition: Angular momentum (LL) of a particle is defined as the vector product of the position vector (rr) and linear momentum (pp): L=r×pL = r \times p
  • Conservation: If the net external torque on a system is zero, the total angular momentum is conserved.

Torque and Angular Momentum Relation

  • Relation: The rate of change of angular momentum of a particle is equal to the torque (τ\tau) acting on it: dLdt=τ\frac{dL}{dt} = \tau
  • Torque: Defined as the vector product of the position vector and force (FF): τ=r×F\tau = r \times F

Vector Product of Two Vectors

  • Definition: The vector product (cross product) of two vectors a\mathbf{a} and b\mathbf{b} is a vector c\mathbf{c} given by: c=a×b=absin⁡θn\mathbf{c} = \mathbf{a} \times \mathbf{b} = ab \sin \theta \mathbf{n} where θ\theta is the angle between a\mathbf{a} and b\mathbf{b}, and n\mathbf{n} is a unit vector perpendicular to the plane containing a\mathbf{a} and b\mathbf{b}.

Rotational Motion Kinematics

  • Angular Velocity (ω\omega): The rate of change of angular displacement.
  • Relation to Linear Velocity: For a particle moving in a circle of radius rr, the linear velocity vv is related to angular velocity by v=ωrv = \omega r.

Equilibrium of Rigid Bodies

  • Conditions for Equilibrium:
    • Translational Equilibrium: The vector sum of all forces is zero.
    • Rotational Equilibrium: The vector sum of all torques is zero.

Conservation of Linear Momentum

  • Principle: In an isolated system, the total linear momentum remains constant if no external forces act on it.

Types of Motion of Rigid Bodies

  • Pure Translation: All points of the body move with the same velocity.
  • Pure Rotation: All points move in circles about a common axis.
  • Combined Motion: Includes rolling motion, where both translation and rotation occur.

Motion of Centre of Mass

  • Equation: MA=FextM \mathbf{A} = \mathbf{F}_{\text{ext}}, where MM is the total mass and A\mathbf{A} is the acceleration of the COM.

Moment of Inertia

  • Definition: A measure of an object's resistance to changes in its rotation rate.
  • Dependence: Depends on the mass distribution relative to the axis of rotation.

Theorems of Moment of Inertia

  • Parallel-Axis Theorem: I=Icm+Md2I = I_{\text{cm}} + Md^2
  • Perpendicular-Axis Theorem: For a planar body, Iz=Ix+IyI_z = I_x + I_y

Dynamics of Rotation About Fixed Axis

  • Equation: τ=Iα\tau = I\alpha, where α\alpha is the angular acceleration.
  • Rotational Kinetic Energy: K=12Iω2K = \frac{1}{2} I \omega^2

Rolling Motion and Combined Rotation-Translation

  • Rolling Without Slipping: v=Rωv = R\omega
  • Kinetic Energy: Combination of translational and rotational kinetic energy.
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Practice Test – MCQs, True/False

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Multiple Choice Questions

A.

dLdt=τ\frac{dL}{dt} = \tau

B.

L=τ×rL = \tau \times r

C.

τ=dLdr\tau = \frac{dL}{dr}

D.

τ=L×ω\tau = L \times \omega
Correct Answer: A

Solution:

The relationship between torque τ\tau and angular momentum LL for a single particle is given by dLdt=τ\frac{dL}{dt} = \tau, which is the rotational analogue of Newton's second law.

Chapter Concept:

Vector Product of Two Vectors

A.

ω\omega is perpendicular to the axis of rotation.

B.

ω\omega is parallel to the linear velocity vector.

C.

ω\omega lies along the axis of rotation.

D.

ω\omega changes direction with time.
Correct Answer: C

Solution:

For rotation about a fixed axis, the angular velocity vector ω\omega lies along the axis of rotation.

Chapter Concept:

Rotational Motion Kinematics

A.

v=ωrv = \omega r

B.

v=ω/rv = \omega / r

C.

v=r/ωv = r / \omega

D.

v=ω2rv = \omega^2 r
Correct Answer: A

Solution:

The linear velocity vv of a particle moving in a circle of radius rr is related to the angular velocity ω\omega by the equation v=ωrv = \omega r.

Chapter Concept:

Angular Momentum

A.

v=ωrv = \omega r

B.

v=ωrv = \frac{\omega}{r}

C.

v=ω2rv = \omega^2 r

D.

v=rωv = \frac{r}{\omega}
Correct Answer: A

Solution:

The linear velocity vv of a point on a rotating body is given by the relation v=ωrv = \omega r, where ω\omega is the angular velocity and rr is the distance from the axis of rotation.

Chapter Concept:

Torque and Angular Momentum Relation

A.

The point of contact has the maximum velocity.

B.

The point of contact has zero velocity.

C.

The point of contact has the same velocity as the center of mass.

D.

The point of contact has the maximum acceleration.
Correct Answer: B

Solution:

In rolling motion without slipping, the point of contact between the rolling object and the surface has zero velocity at any instant.

Chapter Concept:

Rolling Motion and Combined Rotation-Translation

A.

v=ωrv = \omega r

B.

v=ω/rv = \omega / r

C.

v=r/ωv = r / \omega

D.

v=ω2rv = \omega^2 r
Correct Answer: A

Solution:

The linear velocity vv of a particle moving in a circle is related to the angular velocity ω\omega by the equation v=ωrv = \omega r, where rr is the radius of the circle.

Chapter Concept:

Rotational Motion Kinematics

A.

The net force acting on the body is zero.

B.

The net torque acting on the body is zero.

C.

Both the net force and net torque acting on the body are zero.

D.

The net angular momentum of the body is zero.
Correct Answer: C

Solution:

For a rigid body to be in mechanical equilibrium, both the net force and the net torque acting on it must be zero.

Chapter Concept:

Centre of Mass

A.

2Ma23\frac{2Ma^2}{3}

B.

Ma23\frac{Ma^2}{3}

C.

4Ma23\frac{4Ma^2}{3}

D.

5Ma23\frac{5Ma^2}{3}
Correct Answer: A

Solution:

The moment of inertia of each square about its center is Ma26\frac{Ma^2}{6}. Using the parallel-axis theorem and symmetry, the total moment of inertia is 2Ma23\frac{2Ma^2}{3}.

Chapter Concept:

Theorems of Moment of Inertia

A.

Pure translation

B.

Pure rotation about a fixed axis

C.

Combined translation and rotation

D.

Vibrational motion
Correct Answer: B

Solution:

Pure rotation about a fixed axis is characterized by all parts of the body having the same angular velocity at any instant of time.

Chapter Concept:

Types of Motion of Rigid Bodies

A.

I+ML2I + ML^2

B.

I+ML24I + \frac{ML^2}{4}

C.

I+ML22I + \frac{ML^2}{2}

D.

I+3ML24I + \frac{3ML^2}{4}
Correct Answer: C

Solution:

Using the parallel-axis theorem, the moment of inertia about an axis through one end is I+ML22I + \frac{ML^2}{2}.

Chapter Concept:

Theorems of Moment of Inertia

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System of Particles and Rotational Motion

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True or False

Correct Answer: False

Solution:

The vector product, also known as the cross product, is not commutative. This means a×b≠b×a\mathbf{a} \times \mathbf{b} \neq \mathbf{b} \times \mathbf{a}. In fact, a×b=−(b×a)\mathbf{a} \times \mathbf{b} = - (\mathbf{b} \times \mathbf{a}).

Chapter Concept :

Vector Product of Two Vectors

Correct Answer: True

Solution:

According to the law of conservation of linear momentum, if the total external force acting on a system of particles is zero, the total linear momentum of the system remains constant.

Chapter Concept :

Conservation of Linear Momentum

Correct Answer: False

Solution:

The perpendicular-axis theorem is applicable only to planar laminae, not to any rigid body regardless of its shape.

Chapter Concept :

Theorems of Moment of Inertia

Correct Answer: True

Solution:

In pure translational motion, as described, all particles of the body move together with the same velocity at any instant of time.

Chapter Concept :

Types of Motion of Rigid Bodies

Correct Answer: False

Solution:

The vector product, also known as the cross product, is not commutative. The direction of a×b\mathbf{a} \times \mathbf{b} is opposite to that of b×a\mathbf{b} \times \mathbf{a}, so a×b=−(b×a)\mathbf{a} \times \mathbf{b} = - (\mathbf{b} \times \mathbf{a}).

Chapter Concept :

Vector Product of Two Vectors

Correct Answer: True

Solution:

The moment of inertia is a measure of an object's resistance to changes in its rotational motion and depends on how the mass is distributed relative to the axis of rotation.

Chapter Concept :

Moment of Inertia

Correct Answer: False

Solution:

The time rate of change of angular momentum of a particle is equal to the torque acting on it, not the force.

Chapter Concept :

Angular Momentum

Correct Answer: False

Solution:

If no external forces act on a system of particles, the center of mass moves with a constant velocity, as per the conservation of linear momentum. Therefore, it cannot have an accelerating motion.

Chapter Concept :

Conservation of Linear Momentum

Correct Answer: True

Solution:

The moment of a force vanishes if the line of action of the force passes through the origin, as the perpendicular distance is zero.

Chapter Concept :

Torque and Angular Momentum Relation

Correct Answer: True

Solution:

The motion of the center of mass is determined by external forces only, as internal forces cancel each other out due to Newton's third law.

Chapter Concept :

Motion of Centre of Mass

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