chapter 4 chuyen dong cua he chất điểm va vat ran revised

Chapter 4: Motion of a System of Particles4.1 Linear Momentum and Its Conservation4.2 Conservation of Energy and Momentum in Collisions4.3 Center of Mass4.4 Rotational Motion4.5 Rotation

Lecturer : Assoc Prof Pham Hong Quang Email : quangph@pvu.edu.vn

TẬP ĐOÀN DẦU KHÍ VIỆT NAM

TRƯỜNG ĐẠI HỌC DẦU KHÍ VIỆT NAM

General Physics I

Chapter 4: Motion of a System of Particles

4.1 Linear Momentum and Its Conservation

4.2 Conservation of Energy and Momentum in Collisions

4.3 Center of Mass

4.4 Rotational Motion

4.5 Rotational Energy

4.6 Rotational Inertia

4.7 Total Mechanical Energy

4.8 Parallel Axis Theorem

4.9 Torque and Momentum

4.10 Conservation of Angular Momentum

4.11 Work in Rotational Motion

Learning outcome

The students should be able to:

•Given the positions of several particles along an axis or a plane,

determine the location of their center of mass

•Locate the center of mass of an extended, symmetric object by

using the symmetry

•For a two-dimensional or three-dimensional extended object

with a uniform distribution of mass, determine the center of mass

by (a) mentally dividing the object into simple geometric figures,

each of which can be replaced by a particle at its center and (b)

finding the center of mass of those particles

•Apply Newton’s second law to a system of particles by relating the net force (of the forces acting on the particles) to the acceleration

of the system’s center of mass

•Calculate the momentum of a system of particles as the product of the system’s total mass and its center-of-mass velocity

•For an isolated system of particles, apply the conservation of

linear momentum to relate the initial momentum of the particles to their momentum at a later instant

•Apply the conservation of momentum for an isolated

one-dimensional collision to relate the initial momentum of the objects

to their momentum after the collision

•For isolated elastic collisions in one dimension, apply the

conservation laws for both the total energy and the net

momentum of the colliding bodies to relate the initial values to

the values after the collision

•Identify that if all parts of a body rotate around a fixed axis locked together, the body is a rigid body

•Identify that the angular position of a rotating rigid body is the

angle that an internal reference line makes with a fixed, external reference line

Learning outcome

•Apply the relationship between average angular velocity,

angular displacement, and the time interval for that

displacement

•Apply the relationship between average angular acceleration,

change in angular velocity, and the time interval for that change

•For constant angular acceleration, apply the relationships

between angular position, angular displacement, angular

velocity, angular acceleration, and elapsed time

•Find the rotational inertia of a particle about a point

•Find the total rotational inertia of many particles moving around the same fixed axis

•Calculate the rotational inertia of a body by integration

over the mass elements of the body

•Apply the parallel-axis theorem for a rotation axis that is

displaced from a parallel axis through the center of mass of

Learning outcome

4.1 Linear Momentum and Its Conservation

Conservation of Momentum

•Law of Conservation of Momentum - The total

momentum of an isolated system of bodies remains constant.

Isolated system - one in which the forces only act between the objects of the system.

Momentum before = momentum after

2 2

1 1

2 2 1

1vi m vi m vf m vf

Momentum conservation works for a rocket as long

as we consider the rocket and its fuel to be one

system, and account for the mass loss of the rocket.

4.1 Linear Momentum and Its Conservation

4.2 Conservation of Energy and Momentum in Collisions (One dimension)

Momentum is conserved

in all collisions.

Collisions in which

kinetic energy is

conserved as well are

called elastic collisions,

and those in which it is

not are called inelastic.

4.2 Conservation of Energy and Momentum in Collisions

Here we have two objects

colliding elastically We

know the masses and the

initial speeds.

Since both momentum and

kinetic energy are

conserved, we can write two

equations This allows us to

solve for the two unknown

final speeds.

4.2 Conservation of Energy and Momentum in Collisions

Equations 9.15 and 9.16 can be solved for the final speeds in

terms of the initial speeds because there are two equations

and two unknowns:

It is important to remember that the appropriate signs for v 1i and

v 2i must be included in Equations 9.20 and 9.21.

4.2 Conservation of Energy and Momentum in Collisions

4.2 Conservation of Energy and Momentum in Collisions

4.3 Center of Mass

In (a), the diver’s motion is pure translation; in (b) it is

translation plus rotation.

There is one point that moves in the same path a particle

would take if subjected to the same force as the diver This point is called the center of mass (CM).

The general motion of an object can be considered

as the sum of the translational motion of the CM, plus rotational, vibration or other forms

of motion about the CM.

h/2

h h/2

h

The center of mass lies at the geometric center for a symmetric, uniform density object.

4.3 Center of Mass

4.3 Center of Mass

Center of mass

N

i i i CM

Consider the following masses and their coordinates

which make up a "discrete mass" rigid body.

What are the coordinates for the center of mass of this system?

4.3 Center of Mass

z

m r

M

y

m r

M

x

m r

N

i i i cm

N

i i i cm

1 ( ) 7 )(

10 ( ) 2 )(

5 (

16

) 17 )(

1 ( ) 2 )(

10 ( ) 0 )(

5 (

16

) 10 )(

1 ( ) 4 )(

10 ( ) 3 )(

5 (

1 1 1

M

z

m r

M

y

m r

M

x

m r

N

i i i cm

N

i i i cm

N

i i i cm

z y x







k j

What if you had a body that had a non-uniform mass

distribution throughout its structure? Since for a rigid body

we used SUMMATION, S, it only make sense that we would use INTEGRATION to sum all of the small individual masses thus allowing us to determine the center of mass of an

4.3 Center of Mass

4.3 Center of Mass

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

The sum of all the forces acting on a system is equal to the total mass of the system multiplied by the acceleration of the center of mass:M aCM F

F a

M CM  



 r s

Angular displacement

Average angular speed

Instantaneous angular speed:

Average angular acceleration:

Instantaneous angular acceleration:

tf i

i f

tf i

i f

Angular velocity is a vector

Right-hand rule for

determining the

direction of this vector

4.4 Rotational Motion

4.4 Rotational Motion

4.4 Rotational Motion

4.4 Rotational Motion

Centripetal Force

4.5 Rotational Energy

The kinetic energy of a rigid rotating body is sum over

kinetic energy all the particles in the body

2

1 2

1 2

1 2

I   m r

Requirement: ωr and ω is the same for all particles in a must be expressed in rad/s.

SI Unit of Rotational Kinetic Energy: joule (J)

4.6 Rotational Inertia

4.6 Rotational Inertia

4.6 Rotational Inertia

4.7 Total mechanical Energy

An object that undergoes combined rotational and

translation motion has two types of kinetic energy:

(1) a rotational kinetic energy due to its rotation about its

center of mass (2) a translational kinetic energy due to translation of its

center of mass The total mechanical energy is:

Example A very common problem is to find the velocity of a

ball rolling down an inclined plane In the past, everything was SLIDING Now the object is rolling and thus has MORE energy than normal So let’s assume the ball is like a thin spherical shell and was released from a position 5 m above the ground

Calculate the velocity at the bottom of the incline.

3

1 2

1

) )(

3

2 ( 2

1 2

1

3 2 2

1 2

1

2 2

2

2 2 2

2

@

2 2

v v

gh

R

v mR mv

mgh

mR I

R

v

I mv

mgh

K K

U

E E

cm sphere

R T

g

after before





s m gh

v

v gh

mv mgh

mv mgh

K U

E E

T g

after before

/ 90 9 ) 5 )(

8 9 ( 2 2

2 1 2 1 2 1

2 2 2

4.8 Parallel Axis Theorem

This theorem will allow us to calculate the moment of

inertia of any rotating body around any axis, provided we

know the moment of inertia about the center of mass

It basically states that the Moment of Inertia ( Ip) around any axis "P" is equal to the known moment of inertia (Icm) about some center of mass plus M ( the total mass of the system) times the square of "d" ( the distance between the two

parallel axes)

2

Md I

I p  cm 

• Consider moment of inertia I of an area A with respect to the axis AA’



 y dA

I 2

• The axis BB’ passes through the

center of mass and is called a

dA y

d dA

y

dA d

y dA

y

I

2 2

2 2

2

2

Ad I

I

4.8 Parallel Axis Theorem

According to the definition of center of mass   y,dA   0

A composite area is made by adding or subtracting a series of

“simple” shaped areas like rectangles, triangles, and circles

For example, the area on the left can be made from a rectangle minus a triangle and circle.

The moment of inertia of these “simpler” shaped areas

about their centroidal axes are found in most engineering

handbooks

Using these data and the parallel-axis theorem, the moment

of inertia for a composite area can easily be calculated.

4.8 Parallel Axis Theorem

The moment of inertia of the shaded area is obtained by

subtracting the moment of inertia of the half-circle from the

moment of inertia of the rectangle

Determine the moment of inertia

of the shaded area with respect

to the x axis.

4.8 Parallel Axis Theorem

4.9 Torque and Angular Momentum

Torque: magnitude and direction

4.9 Torque and Angular Momentum

r

4.9 Torque and Angular Momentum

P dt

r

d dt

L d

P dt

r



 F dt

P d

Here

and

“The net torque acting on a particle is equal to the

time rate of change of its angular momentum”

Relation between Torque and Momentum

4.9 Torque and Angular Momentum

d I dt

4.9 Torque and Angular Momentum

An example

4.9 Torque and Angular Momentum

Solution

T 

' 2

Two masses m1 (5 kg) and m2

(10 kg) are hanging from a

pulley of mass M (3 kg) and

radius R (0.1 m), as shown

There is no slip between

the rope and the pulleys

What will happen when the

masses are released?

Other example

What is the acceleration of a sphere smoothly

rolling down an inclined plane?

4.9 Torque and Angular Momentum

x component Newtons Law

M com  

Assoc.Prof Pham Hong Quang General Physics I

53

Other example (Cont.)

4.9 Torque and Angular Momentum

4.10 Conservation of Angular Momentum

L  tan



“If the net external torque acting on

a system is zero, the total vector angular momentum of the system remains constant”

Then

4.10 Conservation of Angular Momentum

4.11 Work in Rotational Motion

4.12 Summary: Angular and Linear quantities

Linear Momentum; Collisions; Elastic collision; Inelastic

collision; Center of Mass; Rotational Motion; Rotational

Energy; Rotational Inertia; Parallel Axis Theorem; Torque;

Angular Momentum

Key words of the chapter

•Momentum is conserved if the net external force is zero

• Internal forces within a system always sum to zero

• In collision, assume external forces can be ignored

• Inelastic collision: kinetic energy is not conserved

• Completely inelastic collision: the objects stick together

afterward

•Elastic collision: kinetic energy is conserved

• Center of mass:

•Describing rotational motion requires analogs to position,

velocity, and acceleration

•Rolling motion:

•Kinetic energy of rotation:

•Moment of inertia:

•Kinetic energy of an object rolling :

•When solving problems involving conservation of energy, both

the rotational and linear kinetic energy must be taken into

account

•In systems with both rotational and linear motion, Newton’s

second law must be applied separately to each

• Angular momentum:

• For tangential motion,

• Newton’s second law:

• In systems with no external torque, angular momentum is

conserved

•Work done by a torque:

•Rotational quantities are vectors that point along the axis of

rotation, with the direction given by the right-hand rule

Check your understanding 1

A 35 Kg boy jumps (from rest) into a moving trolley of mass 70

Kg and already moving at a velocity of 5 m/s to the right What

is the speed of the trolley after the boy has jumped in ?

A) 2.0 m/s; B) 1.5 m/s; C)- 2.0 m/s; D)3.3 m/s;

Ans D Conservation of momentum 350 = (35 + 70) v

v = 350 / 105 = 3.3 m/s to the right

Check your understanding 2

An object at rest begins to rotate with a constant angular

acceleration If this object rotates through an angle q in the time

t, through what angle did it rotate in the time 1/2 t?

A) 1/2 ; B) 1/4 ; c) 3/4 ; D) 2 

from rest), and there is a quadratic dependence on time

Therefore, in half the time, the object has rotated through

one-quarter the angle

Check your understanding 3

For the area A, we know the centroid’s (C)

location, area, distances between the four

parallel axes, and the MoI about axis 1

We can determine the MoI about axis 2 by

applying the parallel axis theorem

A) directly between the axes 1 and 2

B) between axes 1 and 3 and then

between the axes 3 and 2

C) between axes 1 and 4 and then axes 4

A

•C

Axis

Check your understanding 4

Two spheres have the same radius and equal masses One is

made of solid aluminum, and the other is made from a hollow

Check your understanding 5

A figure skater spins with her arms extended When she pulls in her arms, she reduces her rotational inertia and spins faster so

that her angular momentum is conserved Compared to her initial rotational kinetic energy, her rotational kinetic energy after she

pulls in her arms must be

A) the same; B) larger because

she’s rotating faster; C) smaller

because her rotational inertia is

smaller ;D) It depends on the height

of center of mass of her body

from the work she does on her arms

Thank you!