To maintain a rotor at a uniform angular speed of 200 rad/s, an engine needs to transmit a torque of 180 Nm. What is the power required by the engine?

Correct Option: B. Solution: Power $$P = \tau \omega = 180 \times 200 = 36000$$ W or 36 kW.

System of Particles and Rotational Motion

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Summary

Chapter 6: Systems of Particles and Rotational Motion

Summary

Introduction to the motion of extended bodies as systems of particles.
Key concepts include:
- Centre of mass
- Motion of centre of mass
- Linear momentum of a system of particles
- Vector product of two vectors
- Angular velocity and its relation with linear velocity
- Torque and angular momentum
- Equilibrium of a rigid body
- Moment of inertia
- Kinematics and dynamics of rotational motion about a fixed axis
- Angular momentum in rotation about a fixed axis

Important Formulas and Definitions

Quantity	Symbols	Dimensions	Units	Remarks
Angular velocity	ω	[T⁻¹]	rad/s	Direction along the axis of rotation.
Angular momentum	L	[ML²T⁻¹]	J·s	L = r x p
Torque	τ	[ML²T⁻²]	Nm	τ = r x F
Moment of inertia	I	[ML²]	kg·m²	I = Σmᵢrᵢ²
Kinetic energy of rotation	K		J	K = 1/2 Iω²
Velocity of centre of mass	V		m/s	V = P/M, where P is linear momentum.

Learning Objectives

Understand the concept of centre of mass and its significance.
Analyze the motion of a system of particles.
Apply the principles of linear momentum and angular momentum.
Solve problems involving torque and moment of inertia.
Differentiate between translational and rotational motion.

Common Mistakes and Exam Tips

Mistake: Confusing linear and angular quantities.
- Tip: Always check units and dimensions when converting between linear and angular motion.
Mistake: Ignoring the effects of external forces on the centre of mass.
- Tip: Remember that the motion of the centre of mass is influenced only by external forces.
Mistake: Misapplying the formulas for torque and moment of inertia.
- Tip: Ensure you understand the geometry of the problem to apply the correct formula.

Important Diagrams

Diagram of Centre of Mass: Illustrates the concept of centre of mass with arrows indicating forces acting on a system of particles.
Diagram of Rotational Motion: Shows the relationship between linear and angular quantities, emphasizing the fixed axis of rotation.

Learning Objectives

Learning Objectives for Chapter Six: Systems of Particles and Rotational Motion
- Understand the concept of the centre of mass and its significance in the motion of extended bodies.
- Analyze the motion of the centre of mass for a system of particles.
- Apply the principles of linear momentum to a system of particles.
- Explore the vector product of two vectors and its applications.
- Relate angular velocity to linear velocity in rotational motion.
- Calculate torque and angular momentum for rigid bodies.
- Assess the conditions for equilibrium in rigid bodies.
- Determine the moment of inertia for various shapes and its implications in rotational dynamics.
- Solve problems involving kinematics and dynamics of rotational motion about a fixed axis.
- Investigate the relationship between angular momentum and rotation about a fixed axis.

Detailed Notes

Chapter 6: Systems of Particles and Rotational Motion

6.1 Introduction

Motion of a single particle was primarily considered in earlier chapters.
Real bodies have finite sizes, requiring a different approach to understand their motion.
The chapter focuses on the motion of extended bodies, treating them as systems of particles.

6.2 Centre of Mass

The centre of mass is a key concept in understanding the motion of a system of particles.

6.3 Motion of Centre of Mass

The motion of the centre of mass can be analyzed without knowledge of internal forces.

6.4 Linear Momentum of a System of Particles

The linear momentum of a system is defined as the product of mass and velocity.

6.5 Vector Product of Two Vectors

The vector product (cross product) of two vectors is defined, with magnitude and direction.

6.6 Angular Velocity and Its Relation with Linear Velocity

Angular velocity is a vector quantity related to linear velocity in rotational motion.

6.7 Torque and Angular Momentum

Torque is the rotational equivalent of force, affecting angular momentum.

6.8 Equilibrium of a Rigid Body

A rigid body is in mechanical equilibrium if:
1. Total external force is zero (ΣFᵢ=0).
2. Total external torque is zero.

6.9 Moment of Inertia

The moment of inertia is defined by the formula: I = Σmᵢrᵢ², where rᵢ is the distance from the axis.

6.10 Kinematics of Rotational Motion About a Fixed Axis

Involves rotational motion with fixed axis and relates to translational motion.

6.11 Dynamics of Rotational Motion About a Fixed Axis

Discusses forces and torques in rotational dynamics.

6.12 Angular Momentum in Case of Rotation About a Fixed Axis

Angular momentum is defined and related to torque.

Points to Ponder

The motion of the centre of mass does not require knowledge of internal forces.
The kinetic energy of a system can be separated into components.
Newton's laws apply to systems of particles as well as single particles.
Total torque and total force conditions are independent.
The centre of gravity coincides with the centre of mass only in uniform gravitational fields.

Important Formulas

Quantity	Symbols	Dimensions	Units	Remarks
Angular velocity	ω	[T⁻¹]	rad/s
Angular momentum	L	[ML²T⁻¹]	J·s	L = r x p
Torque	τ	[ML²T⁻²]	Nm	τ = r x F
Moment of inertia	I	[ML²]	kg·m²	I = Σmᵢrᵢ²
Kinetic energy (rotation)	K			K = ¹/₂ Iω²
Power	P			P = τω
Linear momentum	p	[MLT⁻¹]	kg·m/s	p = mv

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips

Common Pitfalls

Misunderstanding the Center of Mass: Students often confuse the center of mass with the center of gravity. Remember, the center of gravity coincides with the center of mass only in a uniform gravitational field.
Ignoring External Forces: When analyzing the motion of a system of particles, some students forget that only external forces affect the motion of the center of mass.
Neglecting Torque in Equilibrium Problems: In problems involving equilibrium, students may overlook the condition that the total external torque must also be zero, not just the total external force.
Confusing Angular and Linear Quantities: Students sometimes mix up angular velocity and linear velocity, especially in rotational motion problems. Ensure to use the correct formulas for each.

Tips for Success

Visualize the Problem: Draw diagrams to represent forces, torques, and the motion of the center of mass. This can help clarify complex problems.
Use the Right Formulas: Familiarize yourself with key formulas related to angular momentum, torque, and moment of inertia. For example, remember that the moment of inertia for a solid cylinder is given by I = (1/2)MR².
Practice with Real-World Examples: Relate concepts to real-life scenarios, such as the motion of a child on a turntable or the dynamics of a spinning top, to better understand the principles.
Check Units: Always ensure that your units are consistent, especially when calculating quantities like torque (Nm) and moment of inertia (kg m²).
Review Common Problems: Go over typical exam questions related to the motion of rigid bodies, such as finding the center of mass for different shapes or calculating angular momentum.

Practice & Assessment

Multiple Choice Questions

40 rad/s

60 rad/s

90 rad/s

30 rad/s

Correct Answer: A

Solution:

Using the conservation of angular momentum,

I_1 \omega_1 = I_2 \omega_2

, where

I_2 = 1.5 I_1

. Solving for

\omega_2

gives

\omega_2 = \frac{\omega_1}{1.5} = \frac{60}{1.5} = 40

rad/s.

\omega_0

\frac{\omega_0}{2}

2\omega_0

\frac{\omega_0}{4}

Correct Answer: B

Solution:

The conservation of angular momentum states that

I_1 \omega_0 = (I_1 + I_2) \omega_f

, where

I_1 = I_2 = \frac{1}{2}MR^2

. Solving gives

\omega_f = \frac{\omega_0}{2}

At the center of the original disk

At the center of the hole

At a point

R/6

from the center of the original disk towards the hole

At a point

R/4

from the center of the original disk towards the hole

Correct Answer: C

Solution:

The center of gravity of the remaining part can be found using the principle of moments. The mass of the removed section is effectively at its center. The center of gravity of the remaining part shifts towards the hole by

R/6

The center of mass moves as if all the mass of the system is concentrated at this point and all external forces act at it.

The center of mass always moves in a straight line.

The center of mass is always located at the geometric center of the system.

The center of mass is unaffected by external forces.

Correct Answer: A

Solution:

The center of mass of a system of particles moves as if all the mass of the system is concentrated at this point and all external forces act at it.

It always coincides with the center of mass.

It coincides with the center of mass only if the gravitational field is uniform.

It is the point where the total gravitational force acts.

It is the point where the total gravitational torque is maximum.

Correct Answer: B

Solution:

The center of gravity coincides with the center of mass only if the gravitational field is uniform across the body.

625 J

1250 J

2500 J

5000 J

Correct Answer: B

Solution:

The moment of inertia

I

for a solid cylinder rotating about its axis is given by

I = \frac{1}{2} m r^2

. Substituting the given values,

I = \frac{1}{2} \times 20 \times (0.25)^2 = 0.625 \text{ kg m}^2

. The kinetic energy

K

is given by

K = \frac{1}{2} I \omega^2 = \frac{1}{2} \times 0.625 \times (100)^2 = 1250 \text{ J}

Angular momentum is a scalar quantity.

Angular momentum is always conserved in a closed system.

Angular momentum is the cross product of position vector and linear momentum.

Angular momentum is independent of the axis of rotation.

Correct Answer: C

Solution:

Angular momentum is the cross product of the position vector and linear momentum, represented as

L = r \times p

At the center of the original disk

R/6

from the center of the original disk towards the hole

R/4

from the center of the original disk towards the hole

R/3

from the center of the original disk towards the hole

Correct Answer: B

Solution:

The mass of the original disk is proportional to

\pi R^2

and the mass of the removed part is proportional to

\pi (R/2)^2 = \pi R^2/4

. The center of mass of the remaining part is found using the formula for the center of mass of composite bodies:

x_{cm} = \frac{M_1 x_1 - M_2 x_2}{M_1 - M_2}

where

M_1

and

M_2

are the masses of the original and removed parts, respectively. Thus,

x_{cm} = \frac{\pi R^2 \times 0 - \pi R^2/4 \times (R/2)}{\pi R^2 - \pi R^2/4} = \frac{-R/8}{3\pi R^2/4} = -R/6

. The center of gravity is

R/6

towards the hole.

The total external force is zero

The total external torque is zero

Both the total external force and torque are zero

Either the total external force or torque is zero

Correct Answer: C

Solution:

For a rigid body to be in mechanical equilibrium, both the total external force and the total external torque must be zero.

The solid sphere

The hollow cylinder

Both will have the same angular speed

It depends on the material of the objects

Correct Answer: A

Solution:

The solid sphere will have a greater angular speed because it has a smaller moment of inertia compared to the hollow cylinder, allowing it to accelerate more under the same torque.

25 rad/s²

20 rad/s²

15 rad/s²

10 rad/s²

Correct Answer: A

Solution:

The torque

\tau

is given by

\tau = rF = 0.4 \times 30 = 12 \text{ Nm}

. The moment of inertia

I

for a hollow cylinder is

I = m r^2 = 3 \times (0.4)^2 = 0.48 \text{ kg m}^2

. The angular acceleration

\alpha

is given by

\alpha = \frac{\tau}{I} = \frac{12}{0.48} = 25 \text{ rad/s}^2

The center of mass of the fragments follows the original parabolic path.

Both fragments follow the original parabolic path.

The fragments will move in opposite directions along the same path.

The center of mass of the fragments will fall vertically downwards.

Correct Answer: A

Solution:

The center of mass of the fragments continues along the original parabolic path due to the conservation of momentum, as the internal explosion forces do not affect the center of mass motion.

It increases by a factor of 5/2

It decreases by a factor of 5/2

It remains the same

It increases by a factor of 2/5

Correct Answer: A

Solution:

According to the conservation of angular momentum, the product of moment of inertia and angular speed remains constant. Reducing the moment of inertia to 2/5 of its initial value will increase the angular speed by a factor of 5/2.

\frac{\omega_0}{2}

\omega_0

2\omega_0

\frac{3\omega_0}{2}

Correct Answer: A

Solution:

The conservation of angular momentum states that

I_1\omega_0 = (I_1 + I_2)\omega_f

. Since

I_1 = I_2

\omega_f = \frac{\omega_0}{2}

It continues along the original parabolic path.

It moves in a random direction.

It falls straight down due to gravity.

It remains stationary at the explosion point.

Correct Answer: A

Solution:

The center of mass of the fragments continues along the original parabolic path as if the explosion had not occurred.

40 rev/min

80 rev/min

100 rev/min

120 rev/min

Correct Answer: C

Solution:

Using the conservation of angular momentum,

I_1 \omega_1 = I_2 \omega_2

. If

I_2 = \frac{2}{5}I_1

, then

\omega_2 = \frac{I_1}{I_2} \omega_1 = \frac{5}{2} \times 40 = 100

rev/min.

The angular speed increases.

The angular speed decreases.

The angular speed remains the same.

The angular speed becomes zero.

Correct Answer: A

Solution:

Pulling the arms closer to the body reduces the moment of inertia, causing an increase in angular speed due to the conservation of angular momentum.

It increases.

It decreases.

It remains the same.

It becomes zero.

Correct Answer: A

Solution:

Pulling the arms inward reduces the moment of inertia, causing an increase in angular speed due to the conservation of angular momentum.

The angular speed increases

The angular speed decreases

The angular speed remains the same

The angular speed becomes zero

Correct Answer: A

Solution:

According to the conservation of angular momentum, if the moment of inertia decreases, the angular speed increases to keep the angular momentum constant.

0.5 rad/s²

1 rad/s²

2 rad/s²

5 rad/s²

Correct Answer: B

Solution:

The moment of inertia

I

of a solid cylinder about its axis is given by

I = \frac{1}{2}MR^2

. Substituting the given values,

I = \frac{1}{2} \times 10 \times (0.5)^2 = 1.25

kg m². Using the relation

\tau = I\alpha

, where

\tau

is the torque and

\alpha

is the angular acceleration, we have

5 = 1.25 \alpha

. Solving for

\alpha

, we get

\alpha = 4

rad/s².

It increases

It decreases

It remains the same

It becomes zero

Correct Answer: A

Solution:

By conservation of linear momentum, the horizontal momentum before and after the explosion must be equal. Since the heavier fragment falls vertically, the lighter fragment must gain additional horizontal velocity to conserve momentum.

v = r \omega

v = \frac{\omega}{r}

v = \omega^2 r

v = \frac{r}{\omega}

Correct Answer: A

Solution:

In rotational motion about a fixed axis, the linear velocity

v

of a particle is given by

v = r \omega

, where

r

is the radius of the circle.

18 kW

36 kW

54 kW

72 kW

Correct Answer: B

Solution:

Power

P = \tau \omega = 180 \times 200 = 36000

W or 36 kW.

The linear momentum is the product of the total mass and the velocity of the center of mass.

The linear momentum is independent of the velocity of the center of mass.

The linear momentum is half the product of the total mass and the velocity of the center of mass.

The linear momentum is twice the product of the total mass and the velocity of the center of mass.

Correct Answer: A

Solution:

The linear momentum of a system of particles is equal to the product of the total mass of the system and the velocity of its center of mass.

50 g

100 g

150 g

200 g

Correct Answer: B

Solution:

Let the mass of the metre stick be

M

. The torque due to the coins is balanced by the torque due to the stick.

2 \times 5 \times (45 - 12) = M \times (50 - 45)

. Solving,

M = 100

The net external force on the body is zero.

The net external torque on the body is zero.

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

The body must be at rest.

Correct Answer: B

Solution:

For a rigid body to be in rotational equilibrium, the total external torque acting on it must be zero. This does not necessarily mean that the net external force is zero, as translational equilibrium is a separate condition.

It increases.

It decreases.

It remains the same.

It becomes zero.

Correct Answer: A

Solution:

According to the conservation of angular momentum, when the moment of inertia decreases, the angular speed increases.

\frac{1}{2}MR^2

MR^2

\frac{1}{4}MR^2

\frac{2}{5}MR^2

Correct Answer: A

Solution:

The moment of inertia of a solid cylinder about its axis is given by

\frac{1}{2}MR^2

It depends on the mass distribution relative to the axis of rotation.

It is independent of the axis of rotation.

It is the same for all rigid bodies with the same mass.

It is always equal to the mass of the body.

Correct Answer: A

Solution:

The moment of inertia of a rigid body depends on how its mass is distributed relative to the axis of rotation.

It is the point where the total gravitational torque is zero

It is always located at the geometric center of a body

It coincides with the center of mass only in a uniform gravitational field

It is the point where the total mass of the body is concentrated

Correct Answer: A

Solution:

The center of gravity is defined as the point where the total gravitational torque on the body is zero.

The solid sphere

The hollow cylinder

Both will achieve the same angular speed

It depends on the distribution of mass

Correct Answer: A

Solution:

The moment of inertia of a solid sphere is smaller than that of a hollow cylinder for the same mass and radius. Since angular acceleration is inversely proportional to the moment of inertia, the solid sphere will achieve a greater angular speed.

The total linear momentum of the system must be zero.

The total external force and total external torque on the system must be zero.

The total internal forces must be zero.

The total angular momentum of the system must be zero.

Correct Answer: B

Solution:

For a system to be in mechanical equilibrium, both the total external force and the total external torque must be zero.

The hollow sphere

The solid sphere

Both reach at the same time

Cannot be determined without additional information

Correct Answer: B

Solution:

The solid sphere reaches the bottom first because it has a smaller moment of inertia compared to the hollow sphere. This means it accelerates faster down the incline, assuming the only forces acting are gravitational and there is no slipping.

All particles have different angular velocities.

All particles have the same angular velocity.

Angular velocity depends on the distance from the axis.

Angular velocity is zero for particles on the axis.

Correct Answer: B

Solution:

In a rigid body rotating about a fixed axis, every particle has the same angular velocity, as the rotation is uniform throughout the body.

100 g

200 g

300 g

400 g

Correct Answer: D

Solution:

Let the mass of the stick be

M

. The torque due to the coins is

2 \times 5 \times (45 - 12) = 330 \text{ g cm}

. The torque due to the stick's mass is

M \times (50 - 45) = 5M \text{ g cm}

. Setting the torques equal for balance:

330 = 5M \Rightarrow M = 66 \text{ g}

. Converting to kg,

M = 0.4 \text{ kg}

400 \text{ g}

The hollow cylinder will have a greater angular speed.

The solid sphere will have a greater angular speed.

Both will have the same angular speed.

The angular speeds cannot be determined without additional information.

Correct Answer: B

Solution:

The moment of inertia of a hollow cylinder is

I = MR^2

and for a solid sphere is

I = \frac{2}{5}MR^2

. Since torque

\tau = I\alpha

and

\alpha = \frac{\tau}{I}

, the solid sphere with a smaller moment of inertia will have a greater angular acceleration and hence a greater angular speed after the same time.

It increases over time.

It decreases over time.

It remains constant.

It fluctuates randomly.

Correct Answer: C

Solution:

According to the law of conservation of linear momentum, if no external forces act on a system of particles, the total linear momentum of the system remains constant.

The centre of mass of a system of particles is always located at the geometric center.

The centre of mass moves as if all the mass of the system is concentrated at this point and all external forces act at it.

The centre of mass can only move in a straight line.

The centre of mass is always at rest.

Correct Answer: B

Solution:

The centre of mass moves as if all the mass of the system is concentrated at this point and all external forces act at it, regardless of the actual distribution of mass.

\frac{m_1 m_2 L^2}{m_1 + m_2}

\frac{(m_1 + m_2) L^2}{4}

\frac{m_1^2 L^2}{m_1 + m_2}

\frac{m_2^2 L^2}{m_1 + m_2}

Correct Answer: A

Solution:

The moment of inertia of a system about the center of mass is given by

I = \frac{m_1 m_2 L^2}{m_1 + m_2}

, derived from the parallel axis theorem.

25 rad/s²

50 rad/s²

75 rad/s²

100 rad/s²

Correct Answer: A

Solution:

The torque

\tau = rF = 0.4 \times 30 = 12

Nm. The moment of inertia

I = m r^2 = 3 \times (0.4)^2 = 0.48

kg m². Angular acceleration

\alpha = \frac{\tau}{I} = \frac{12}{0.48} = 25

rad/s².

MR^2

\frac{1}{2}MR^2

\frac{1}{4}MR^2

2MR^2

Correct Answer: A

Solution:

The moment of inertia of a thin circular ring about an axis perpendicular to its plane and passing through its center is given by

I = MR^2

40 rev/min

80 rev/min

100 rev/min

120 rev/min

Correct Answer: B

Solution:

Using conservation of angular momentum,

I_1 \omega_1 = I_2 \omega_2

. If

I_2 = \frac{2}{5} I_1

, then

\omega_2 = \frac{I_1}{I_2} \omega_1 = \frac{5}{2} \times 40 = 80

rev/min.

It accelerates

It moves with constant velocity

It remains at rest

It moves in a circular path

Correct Answer: B

Solution:

When the total external force on a system is zero, the center of mass moves with a constant velocity according to the law of conservation of linear momentum.

\frac{1}{2}MR^2

MR^2

\frac{1}{4}MR^2

\frac{2}{5}MR^2

Correct Answer: A

Solution:

The moment of inertia of a solid cylinder about its axis is given by

\frac{1}{2}MR^2

The total external force and total external torque on it are zero.

The total external force on it is zero, but the total external torque is non-zero.

The total external torque on it is zero, but the total external force is non-zero.

Neither the total external force nor the total external torque on it is zero.

Correct Answer: A

Solution:

A rigid body is in mechanical equilibrium when both the total external force and total external torque acting on it are zero.

The solid sphere will reach the bottom first.

The hollow sphere will reach the bottom first.

Both spheres will reach the bottom at the same time.

The hollow sphere will have a greater linear acceleration.

Correct Answer: A

Solution:

The moment of inertia of a solid sphere is less than that of a hollow sphere, meaning the solid sphere will have a greater linear acceleration and reach the bottom first.

36 kW

18 kW

9 kW

72 kW

Correct Answer: A

Solution:

Power

P

is given by

P = \tau \omega = 180 \times 200 = 36000 \text{ W}

36 \text{ kW}

625 J

1250 J

2500 J

5000 J

Correct Answer: B

Solution:

The kinetic energy of rotation is given by

K = \frac{1}{2} I \omega^2

, where

I = \frac{1}{2} m r^2

for a solid cylinder. Substituting the values,

I = \frac{1}{2} \times 20 \times (0.25)^2 = 0.625

kg m². Thus,

K = \frac{1}{2} \times 0.625 \times (100)^2 = 1250

The center of mass moves in a circular path.

The center of mass remains stationary.

The center of mass moves in a straight line with constant velocity.

The center of mass oscillates back and forth.

Correct Answer: C

Solution:

In the absence of external forces, the center of mass of a binary star system moves in a straight line with constant velocity, as per the conservation of linear momentum.

Equal to the initial horizontal velocity of the projectile

Twice the initial horizontal velocity of the projectile

Half the initial horizontal velocity of the projectile

Zero

Correct Answer: B

Solution:

By the conservation of linear momentum, the horizontal momentum before and after the explosion must be equal. Initially, the total horizontal momentum is

mv

, where

m

is the mass of the projectile and

v

is its initial horizontal velocity. After the explosion, one fragment falls vertically, contributing no horizontal momentum. Therefore, the other fragment must have a horizontal momentum of

2mv

to conserve momentum, implying its velocity is

2v

The center of mass follows a straight line path.

The center of mass follows a parabolic path.

The center of mass remains at the explosion point.

The center of mass moves in a circular path.

Correct Answer: B

Solution:

The center of mass of the fragments continues to follow the original parabolic path of the projectile, as the external forces (gravity) remain unchanged.

The total external force on it is zero.

The total external torque on it is zero.

Both the total external force and torque on it are zero.

The total internal force is zero.

Correct Answer: C

Solution:

A rigid body is in mechanical equilibrium if it is in translational equilibrium (total external force is zero) and rotational equilibrium (total external torque is zero).

MR^2

\frac{MR^2}{2}

\frac{MR^2}{4}

2MR^2

Correct Answer: B

Solution:

The moment of inertia of a thin circular ring about its diameter is

\frac{MR^2}{2}

True or False

Correct Answer: True

Solution:

The formula for the kinetic energy of rotation is

K = \frac{1}{2} I \omega^2

, where

I

is the moment of inertia and

\omega

is the angular velocity.

Correct Answer: False

Solution:

The linear velocity of a particle in a rotating rigid body is given by

v = \omega r

, where

r

is the distance from the axis of rotation. Therefore, it depends on

r

Correct Answer: True

Solution:

The excerpt states that if the total external force is zero, the total linear momentum of the system is constant, illustrating the conservation of linear momentum.

Correct Answer: True

Solution:

The moment of inertia of a hollow cylinder about its axis is calculated as

MR^2

, which is consistent with the formula for a hollow cylinder.

Correct Answer: True

Solution:

A rigid body fixed at one point or along a line can only rotate about that point or line, as stated in the excerpt.

Correct Answer: True

Solution:

If there are no external forces, the center of mass of a double star moves like a free particle, in a straight line.

Correct Answer: True

Solution:

The centre of gravity is defined as the point where the total gravitational torque on the body is zero.

Correct Answer: True

Solution:

When the total external force acting on a system of particles is zero, the center of mass moves with a constant velocity, as stated in the excerpt.

Correct Answer: True

Solution:

For a rigid body to be in mechanical equilibrium, it must be in both translational and rotational equilibrium, meaning the total external force and total external torque must be zero.

Correct Answer: True

Solution:

The total torque on a system is independent of the origin when the total external force is zero, as stated in the principles of rotational dynamics.

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 is constant. This implies that the velocity of the center of mass remains constant.

Correct Answer: True

Solution:

In rotation about a fixed axis, every point on the rigid body has the same angular velocity at any instant of time.

Correct Answer: True

Solution:

If there are no external forces, the centre of mass of a binary star system moves like a free particle, i.e., in a straight line.

Correct Answer: True

Solution:

In rotational motion about a fixed axis, each particle of the body moves in a circle in a plane perpendicular to the axis, with the centre of the circle on the axis.

Correct Answer: False

Solution:

Angular velocity is a vector quantity, as it has both magnitude and direction, as described in the excerpt.

Correct Answer: False

Solution:

The centre of gravity of a body coincides with its centre of mass only if the gravitational field does not vary from one part of the body to the other.

Correct Answer: False

Solution:

The center of gravity coincides with the center of mass only if the gravitational field is uniform across the body. Otherwise, they may differ.

Correct Answer: True

Solution:

For a rigid body to be in mechanical equilibrium, it must be in both translational and rotational equilibrium, meaning zero net force and zero net torque.

Correct Answer: True

Solution:

If the total external force acting on a system is zero, the centre of mass moves with a constant velocity, following a straight line.

Correct Answer: True

Solution:

A rigid body not fixed in some way can have either pure translational motion or a combination of translational and rotational motions.

Correct Answer: True

Solution:

The center of gravity coincides with the center of mass only if the gravitational field does not vary from one part of the body to another, ensuring uniform gravitational torque.

Correct Answer: True

Solution:

The centre of gravity coincides with the centre of mass when the gravitational field is uniform, meaning the gravitational force is the same throughout the body.

Correct Answer: True

Solution:

The diagram description explicitly mentions that pulling the arms inward reduces the moment of inertia and increases rotational speed, demonstrating conservation of angular momentum.

Correct Answer: True

Solution:

The moment of inertia of a hollow cylinder about its axis is given by

I = MR^2

, whereas for a solid cylinder it is

I = \frac{1}{2}MR^2

. Thus, the hollow cylinder has a greater moment of inertia.

Correct Answer: True

Solution:

When the total external force acting on a system of particles is zero, the total linear momentum of the system is constant.

Correct Answer: False

Solution:

To determine the motion of the centre of mass of a system, only the external forces on the body are needed, not the internal forces.

Correct Answer: True

Solution:

The angular momentum of a system of particles is affected by external forces and torques, but internal forces, according to Newton's third law, do not contribute to the net angular momentum about an origin.

Correct Answer: False

Solution:

The centre of gravity coincides with the centre of mass only if the gravitational field is uniform across the body. Otherwise, it may not be at the geometric center.

Correct Answer: True

Solution:

For mechanical equilibrium, a rigid body must be in both translational and rotational equilibrium, meaning zero net external force and zero net external torque.

Correct Answer: True

Solution:

According to the conservation of linear momentum, if the total external force acting on a system is zero, the center of mass moves with a constant velocity, i.e., in a straight line.

Correct Answer: True

Solution:

Without external forces, the centre of mass of a binary star system moves like a free particle, as described in the excerpt.

Correct Answer: True

Solution:

When the total external force on a system is zero, the total torque is independent of the choice of origin.

Correct Answer: True

Solution:

The center of gravity of an extended body coincides with its center of mass only if the gravitational field is uniform across the body. If the gravitational field varies, the center of gravity and center of mass do not coincide.

Correct Answer: True

Solution:

The moment of inertia of a solid sphere about its diameter is

\frac{2}{5} MR^2

, where

M

is the mass and

R

is the radius.

Correct Answer: False

Solution:

The moment of inertia of a thin circular ring about its diameter is given by

\frac{MR^2}{2}

, while about an axis perpendicular to its plane at the center, it is

MR^2

. Thus, the moment of inertia is greater about the perpendicular axis.

Correct Answer: True

Solution:

The total angular momentum of a system of particles is conserved if the total external torque acting on the system is zero, as per the conservation of angular momentum.

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.

Correct Answer: True

Solution:

Pulling the arms inward reduces the moment of inertia, causing an increase in rotational speed, which demonstrates the conservation of angular momentum.

Correct Answer: True

Solution:

The moment of inertia of a thin circular ring about its diameter is

\frac{MR^2}{2}

, as listed in the table of moments of inertia.

Correct Answer: False

Solution:

The vector product (or cross product) of two vectors is a vector, not a scalar.

Correct Answer: True

Solution:

According to the law of conservation of linear momentum, if the total external force on a system is zero, the centre of mass moves with constant velocity.

Correct Answer: False

Solution:

The moment of inertia of a solid sphere about its diameter is

\frac{2}{5} MR^2

, while for a hollow cylinder about its axis, it is

MR^2

. Thus, the moment of inertia of the hollow cylinder is greater.

Correct Answer: True

Solution:

A rigid body fixed at one point or along a line can only undergo rotational motion, as its translational motion is restricted.

Correct Answer: True

Solution:

The center of gravity of an extended body is defined as the point where the total gravitational torque on the body is zero.

Correct Answer: True

Solution:

Stretching the arms increases the moment of inertia about the axis of rotation, resulting in a decrease in angular speed due to the conservation of angular momentum.

Correct Answer: False

Solution:

The total torque on a system is independent of the origin if the total external force is zero, not the angular momentum.

Correct Answer: True

Solution:

According to the provided table, the moment of inertia of a solid sphere about its diameter is indeed

\frac{2}{5} M R^2

Correct Answer: True

Solution:

The linear momentum of a system of particles is defined as the vector sum of all individual particles, which is equal to the product of the total mass and the velocity of the center of mass.

Correct Answer: False

Solution:

In a binary star system with no external forces, the center of mass moves like a free particle, meaning it can move with constant velocity, not necessarily stationary.

Correct Answer: True

Solution:

According to the law of conservation of linear momentum, if the total external force acting on a system is zero, the center of mass moves with a constant velocity, i.e., in a straight line.

Correct Answer: False

Solution:

A rigid body not fixed in some way can have either pure translational motion or a combination of translational and rotational motions.

Correct Answer: True

Solution:

The linear momentum of a system of particles is defined as the vector sum of the linear momenta of the individual particles, as stated in the excerpt.

Correct Answer: False

Solution:

The moment of inertia of a solid sphere about its diameter is

\frac{2}{5}MR^2

, while that of a hollow cylinder about its axis is

MR^2

. The moment of inertia of the hollow cylinder is greater.

Correct Answer: True

Solution:

In rotational motion about a fixed axis, all particles of the rigid body move in circles around the axis and have the same angular velocity.

Correct Answer: True

Solution:

A rigid body not fixed in some way can exhibit both translational and rotational motions simultaneously.

System of Particles and Rotational Motion

AI Learning Assistant

Summary

Chapter 6: Systems of Particles and Rotational Motion

Summary

Important Formulas and Definitions

Learning Objectives

Common Mistakes and Exam Tips

Important Diagrams

Learning Objectives

Detailed Notes

Chapter 6: Systems of Particles and Rotational Motion

6.1 Introduction

6.2 Centre of Mass

6.3 Motion of Centre of Mass

6.4 Linear Momentum of a System of Particles

6.5 Vector Product of Two Vectors

6.6 Angular Velocity and Its Relation with Linear Velocity

6.7 Torque and Angular Momentum

6.8 Equilibrium of a Rigid Body

6.9 Moment of Inertia

6.10 Kinematics of Rotational Motion About a Fixed Axis

6.11 Dynamics of Rotational Motion About a Fixed Axis

6.12 Angular Momentum in Case of Rotation About a Fixed Axis

Points to Ponder

Important Formulas

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips

Common Pitfalls

Tips for Success

Practice & Assessment

Multiple Choice Questions

Q1.A turntable is rotating with an angular speed of 60 rad/s. If a child sitting at the center extends his arms, increasing the moment of inertia by 50%, what will be the new angular speed of the turntable?

Correct Answer: A

Q2.A disk of mass MMM and radius RRR is spinning freely with angular velocity ω0\omega_0ω0​. A second identical disk, initially at rest, is dropped onto the first disk. Assuming no external torques and perfect inelastic collision, what is the final angular velocity of the system?

Correct Answer: B

Q3.From a uniform disk of radius RRR, a circular hole of radius R/2R/2R/2 is cut out. The center of the hole is at R/2R/2R/2 from the center of the original disc. Locate the center of gravity of the resulting flat body.

Correct Answer: C

Q4.Which of the following statements is true about the center of mass of a system of particles?

Correct Answer: A

Q5.Which of the following statements about the center of gravity of a body is true?

Correct Answer: B

Q6.A solid cylinder of mass 20 kg rotates about its axis with an angular speed of 100 rad/s. The radius of the cylinder is 0.25 m. What is the kinetic energy associated with the rotation of the cylinder?

Correct Answer: B

Q7.Which of the following is a correct statement about angular momentum?

Correct Answer: C

Q8.From a uniform disk of radius RRR, a circular hole of radius R/2R/2R/2 is cut out. The center of the hole is at R/2R/2R/2 from the center of the original disk. Locate the center of gravity of the resulting flat body.

Correct Answer: B

Q9.A rigid body is in mechanical equilibrium. Which of the following conditions must be satisfied?

Correct Answer: C

Q10.A solid sphere and a hollow cylinder, both with the same mass and radius, are subjected to equal torques. Which will have a greater angular speed after a given time?

Correct Answer: A

Q11.A rope of negligible mass is wound around a hollow cylinder of mass 3 kg and radius 40 cm. If the rope is pulled with a force of 30 N, what is the angular acceleration of the cylinder?

Correct Answer: A

Q12.A projectile explodes at the top of its parabolic trajectory into two fragments of equal mass. Which of the following statements about the motion of the fragments is true?

Correct Answer: A

Q13.A child stands at the center of a rotating turntable with arms extended. If the child folds their arms, reducing the moment of inertia to 2/5 of its initial value, what happens to the angular speed of the child?

Correct Answer: A

Q14.A disk of mass MMM and radius RRR is spinning with an angular velocity ω0\omega_0ω0​. If a second identical disk is placed on top of the first disk without slipping, what will be the final angular velocity of the system?

Correct Answer: A

Q15.If a projectile explodes mid-air, what happens to the center of mass of the fragments?

Correct Answer: A

Q16.A child stands at the center of a frictionless turntable with arms outstretched. The turntable is rotating with an angular speed of 40 rev/min. If the child folds his arms back, reducing his moment of inertia to 2/5 times the initial value, what will be the new angular speed of the turntable?

Correct Answer: C

Q17.What happens to the angular speed of a skater when they pull their arms closer to their body while spinning?

Correct Answer: A

Q18.When a child on a rotating turntable pulls his arms inward, what happens to his angular speed?

Correct Answer: A

Q19.What happens to the angular speed of a rotating body if its moment of inertia decreases?

Correct Answer: A

Q20.A solid cylinder of mass 10 kg and radius 0.5 m is rotating about its axis with an angular velocity of 50 rad/s. A torque of 5 Nm is applied to the cylinder. What will be the angular acceleration of the cylinder?

Correct Answer: B

Q21.A projectile explodes at the top of its trajectory into two fragments of unequal mass. If the heavier fragment falls vertically downwards, what happens to the horizontal velocity of the lighter fragment immediately after the explosion?

Correct Answer: A

Q22.In a rotational motion about a fixed axis, every particle of the rigid body moves in a circle. What is the relation between the linear velocity vvv of a particle and its angular velocity ω\omegaω?

Correct Answer: A

Q23.To maintain a rotor at a uniform angular speed of 200 rad/s, an engine needs to transmit a torque of 180 Nm. What is the power required by the engine?

Correct Answer: B

Q24.What is the relation between the linear momentum of a system of particles and the velocity of its center of mass?

Correct Answer: A