4.1 The Concepts of Force and Mass

Transcript 4.1 The Concepts of Force and Mass

Review: using F=ma

• Force in

circular motion

– 𝑎 𝑐 and 𝑎 𝑡 in curvilinear motion (e.g. non uniform circular motion)

• Note on other important forces

– Propulsion – Resistive force (Drag in air/liquid), terminal velocicty 𝑣 𝑇 – Conservative force, spring force  Potential energy

Relevant concepts: “inertia” and force a) Inertial reference frame b) accelerated frame of the car - fictitious force

Rolling friction

Conical Pendulum (e.g.

ชิงช ้าหมุน

)

•The object is in equilibrium in the vertical direction.

•It undergoes uniform circular motion in the horizontal direction.

∑𝐹 𝑦 ∑𝐹 𝑥 = 0 → 𝑇 cos 𝜃 = 𝑚𝑔 = 𝑇 sin 𝜃 = 𝑚 𝑎 𝑐 •

is independent of

Car traveling in a banked curve •Design the curve with no friction –in equilibrium in the vertical direction.

–in uniform circular motion in the horizontal direction  force supplies the centripetal force.

a component of the normal •The angle of bank is tan  

•Note: –The banking angle is independent of the mass of the vehicle.

–If the car rounds the curve at less than the design speed, friction is necessary to keep it from sliding down the bank.

–If the car rounds the curve at more than the design speed, friction is necessary to keep it from sliding up the bank.

Car traveling in a horizontal (Flat) Curve •uniform circular motion in the horizontal direction.

•in equilibrium in the vertical direction.

•The force of static friction supplies the centripetal force.

•The maximum speed at which the car can negotiate the curve is:

 

s gr

– Note: this does not depend on the mass of the car.

Section 6.1

Ferris Wheel

•Uniform circular motion with constant speed

(controlled by the motor) 𝑭 𝒏𝒆𝒕 = 𝑚𝑣 2 (− 𝒓) 𝑅 𝑭 𝒏𝒆𝒕 𝑭 𝒏𝒆𝒕 •Under gravity, the child feels apparent weight differently at top and bottom Section 6.1

Ferris Wheel (2)

•At the bottom of the loop, the upward force (the normal) experienced by the object is greater than its weight.

•At the top of the circle, the force exerted on the object is less than its weight.



F n bot



n bot





  1 

  



F n top

 

n top mg

  

v rg



 1  Section 6.1

Non-uniform Circular Motion If the speed also changes in magnitude  there is non-zero

tangential acceleration

Section 6.2

Vertical Circle with Non-Uniform Speed •Launch by shooting (say, from the bottom at initial velocity 𝒗 𝒃𝒐𝒕 ). The string restricts motion to a circle of radius R (1 DOF) •

What is T and acceleration?

•Tension depends on position 𝜃(𝑡) d𝜃 and speed 𝑣 𝑡 = 𝑅 ,both varying 𝑑𝑡 with time •The tension at the bottom is a maximum.

•The tension at the top is a minimum.

•Requires critical speed 𝑣 𝑡𝑜𝑝 𝑅𝑔 to complete the circle ( 𝑇 𝑡𝑜𝑝 ≥ 0 at 𝜃 = 𝜋 ) ≥ Section 6.2



 

 cos   

Resistive force (or drag force in air/fluid) and terminal velocity

𝑣 𝑇 (this topic is not in the test)

Resistive force 𝑭 𝑫 could be of either form(linear or quadratic to 𝑣 ), but always opposite direction to motion ( − ) 1. Linear (Stokes law in liquid) 𝑭 𝑫 ∝ − 𝒗 𝐹 𝐷 = −𝑏𝑣

v T



mg b

2. Quadratic (air resistance) 𝑭 𝑫 ∝ −𝑣 2 𝐹 𝐷 = − 1 2 𝐶 𝐷 𝜌𝐴 𝑣 2 𝐶 𝐷 is the drag coefficient (empirical, no dimension, 1 typically 𝐶 𝐷 ≈ 2 ).

𝜌 is the density of air.

𝐴 is the cross-sectional area of the object 𝑣 𝑇 ≈ 4𝑚𝑔 𝜌𝐴

Lecture 4

Work and Energy

Outline

• Work done by a constant force

–

Projection

and

scalar product

• Force that results in positive work. of vectors – negative work? forces that do no work?

• Work done by varying force with displacement • Work-energy theorem

6.1 Work Done by a Constant Force



1 N



m



1 joule

 

6.1 Work Done by a Constant Force

Work, cont.

•

cos  – A force does no work on the object if the force does not move through a displacement.

– The work done by a force on a moving object is zero when the force applied is perpendicular to the displacement of its point of application.

Section 7.2

6.1 Work Done by a Constant Force

 

cos  

cos 0   1 cos 90   0 cos 180    1

6.1 Work Done by a Constant Force

Example 1 Pulling a Suitcase-on-Wheels

Find the work done if the force is 45.0-N, the angle is 50.0 degrees, and the displacement is 75.0 m.

 

cos     45 .

0 N  cos 50 .

0    75 .

0 m   2170 J

6.1 Work Done by a Constant Force

 

cos 0





Fs W

 

cos 180



 

6.1 Work Done by a Constant Force

Example 3 Accelerating a Crate

The truck is accelerating at a rate of +1.50 m/s 2 . The mass of the crate is 120-kg and it does not slip. The magnitude of the displacement is 65 m.

What is the total work done on the crate by all of the forces acting on it?

f s W



    2   180N    180 N  cos 0   65 m   1 .

2  10 4 J

6.1 Work Done by a Constant Force

The angle between the displacement and the friction force is 0 degrees.

f s



  120 kg   1 .

5 m s 2   180 N

   180 N  cos 0   65 m   1 .

2  10 4 J

Scalar Product of Two Vectors

A B cos



cos    D

Section 7.3

Scalar Product, cont

•The scalar product is commutative.

•The scalar product obeys the distributive law of multiplication.

   Section 7.3

Dot Products of Unit Vectors

1 0 Using component form with vectors:



A x



A y



A z



B x



B y



B z

k A B





 In the special case where



;





2 Section 7.3

6.2 The Work-Energy Theorem and Kinetic Energy

Consider a constant net external force acting on an object.

The object is displaced a distance

, in the same direction as the net force.



F s

The work is simply

  

6.2 The Work-Energy Theorem and Kinetic Energy



  

1 2 



v o

2   1 2

 1 2

o v



v o

2  2    1 2 



v o

2  DEFINITION OF KINETIC ENERGY The kinetic energy KE of and object with mass

and speed

is given by KE  1 2

6.2 The Work-Energy Theorem and Kinetic Energy

THE WORK-ENERGY THEOREM When a net external force does work on and object, the kinetic energy of the object changes according to

 KE f  KE o  1 2

f 2  1 2

mv o

6.2 The Work-Energy Theorem and Kinetic Energy

Example 4 Deep Space 1

The mass of the space probe is 474-kg and its initial velocity is 275 m/s. If the 56.0-mN force acts on the probe through a displacement of 2.42

×10 9 m, what is its final speed?

6.2 The Work-Energy Theorem and Kinetic Energy

W  1 2

f 2  1 2

mv o

2 W     cos  

6.2 The Work-Energy Theorem and Kinetic Energy

   cos  

 1 2

f 2  1 2

mv o

2  5.60

 10 2 N  cos 0   2 .

42  10 9 m   1 2  474 kg 

f 2  1 2  474 kg  275 m s  2

v f

 805 m s

6.2 The Work-Energy Theorem and Kinetic Energy

In this case the net force is 



sin 25  

f k

6.2 The Work-Energy Theorem and Kinetic Energy

Conceptual Example 6 Work and Kinetic Energy

A satellite is moving about the earth in a circular orbit and an elliptical orbit. For these two orbits, determine whether the kinetic energy of the satellite changes during the motion.

W<0 W>0 W>0 W<0

6.3 Gravitational Potential Energy

 

cos  

s W

gravity 



h o



h f



6.3 Gravitational Potential Energy

gravity 



h o



h f

 Work due to gravity is independent of path!

We call this phenomenon “conservative force”  We can define “potential energy” based on this force

6.3 Gravitational Potential Energy

Example 7 A Gymnast on a Trampoline

The gymnast leaves the trampoline at an initial height of 1.20 m and reaches a maximum height of 4.80 m before falling back down. What was the initial speed of the gymnast?

6.3 Gravitational Potential Energy

gravity 

mgh o



mgh f

DEFINITION OF GRAVITATIONAL POTENTIAL ENERGY The gravitational potential energy PE is the energy that an object of mass

has by virtue of its position relative to the surface of the earth. That position is measured by the height

of the object relative to an arbitrary zero level: PE 

mgh

1 N  m  1 joule  

6.4 Conservative Versus Nonconservative Forces

DEFINITION OF A CONSERVATIVE FORCE

Version 1

A force is conservative when the work it does on a moving object is independent of the path between the object ’s initial and final positions.

Version 2

A force is conservative when it does no work on an object moving around a closed path, starting and finishing at the same point.

6.4 Conservative Versus Nonconservative Forces

6.4 Conservative Versus Nonconservative Forces Version 1

A force is conservative when the work it does on a moving object is independent of the path between the object’s initial and final positions.

gravity 



h o



h f



6.4 Conservative Versus Nonconservative Forces Version 2

A force is conservative when it does no work on an object moving around a closed path, starting and finishing at the same point.

gravity 



h o



h f



h o



h f

6.4 Conservative Versus Nonconservative Forces

An example of a nonconservative force is the kinetic frictional force.

 

cos

 



f k

cos 180



 

f k s

The work done by the kinetic frictional force is always negative.

Thus, it is impossible for the work it does on an object that moves around a closed path to be zero. The concept of potential energy is not defined for a nonconservative force.

6.4 Conservative Versus Nonconservative Forces

In normal situations both conservative and nonconservative forces act simultaneously on an object, so the work done by the net external force can be written as



W c



W nc W



KE

f 

KE

o  D

KE

W c



gravity 

mgh o



mgh f



PE

o 

PE

f   D

PE

Spring force and Spring (Elastic) Potential Energy

6.4 Conservative Versus Nonconservative Forces

Work done by conservative force



W c



W nc

Work done by non-conservative force D

KE

  D

PE



W nc

THE WORK-ENERGY THEOREM

W nc

 D

KE

 D

PE

6.5 The Conservation of Mechanical Energy

W nc

 D

KE

 D

PE

 

KE

f 

KE

o  

PE

f 

PE

o 

W nc



E

f 

E

o If the net work on an object by nonconservative forces is zero, then its energy does not change:

E

f 

E

6.5 The Conservation of Mechanical Energy

THE PRINCIPLE OF CONSERVATION OF MECHANICAL ENERGY The total mechanical energy (E = KE + PE) of an object remains constant as the object moves, provided that the net work done by external nononservative forces is zero.

6.5 The Conservation of Mechanical Energy

Example 8 A Daredevil Motorcyclist

A motorcyclist is trying to leap across the canyon by driving horizontally off a cliff 38.0 m/s. Ignoring air resistance, find the speed with which the cycle strikes the ground on the other side.

6.5 The Conservation of Mechanical Energy

E

f 

E

mgh f

 1 2



mgh o

 1 2

o gh f

 1 2



gh o

 1 2

v o

6.5 The Conservation of Mechanical Energy

v f gh f

 1 2



gh o

 1 2

v o

2 

2



h o



h f

 

v o

v f



2



9 .

8 m s

2  

35 .

0 m

 

38 .

0 m s

 2 

46 .

2 m s

6.5 The Conservation of Mechanical Energy

Conceptual Example 9 The Favorite Swimming Hole

The person starts from rest, with the rope held in the horizontal position, swings downward, and then lets go of the rope. Three forces act on him: his weight, the tension in the rope, and the force of air resistance.

Can the principle of conservation of energy be used to calculate his final speed?

6.7 Power

DEFINITION OF AVERAGE POWER Average power is the rate at which work is done, and it is obtained by dividing the work by the time required to perform the work.



Work Time



W t

joule s



watt (W)

6.7 Power



Change in energy Time 1 horsepower



550 foot



pounds second



745.7 watts



F v

6.7 Power

6.8 Other Forms of Energy and the Conservation of Energy

THE PRINCIPLE OF CONSERVATION OF ENERGY Energy can neither be created not destroyed, but can only be converted from one form to another.

• Where does work due to nonconservative force (W nc ) go to, other than kinetic energy? •

Thermal energy (heat)

internal energy of materials

• This means

energy of the universe (isolated system) is always conserved.

– When we say “not energy conserving” we mean just in the sub system.

Lecture 5

Impulse and Momentum

7.1 The Impulse-Momentum Theorem

There are many situations when the force on an object is not constant.

7.1 The Impulse-Momentum Theorem

DEFINITION OF IMPULSE The impulse of a force is the product of the average force and the time interval during which the force acts: 

 

Impulse is a vector quantity and has the same direction as the average force.

newton  seconds (N  s)

7.1 The Impulse-Momentum Theorem



 

7.1 The Impulse-Momentum Theorem

DEFINITION OF LINEAR MOMENTUM The linear momentum of an object is the product of the object’s mass times its velocity: 





Linear momentum is a vector quantity and has the same direction as the velocity.

kilogram  meter/seco nd (kg  m/s)

7.1 The Impulse-Momentum Theorem



 

v f

 D



v o

 





 

 

v f

 D



v o

  D





v f





v o

7.1 The Impulse-Momentum Theorem

IMPULSE-MOMENTUM THEOREM When a net force acts on an object, the impulse of this force is equal to the change in the momentum of the object impulse   D





v f





v o

final momentum initial momentum

7.1 The Impulse-Momentum Theorem

Example 2 A Rain Storm

Rain comes down with a velocity of -15 m/s and hits the roof of a car. The mass of rain per second that strikes the roof of the car is 0.060 kg/s. Assuming that rain comes to rest upon striking the car, find the average force exerted by the rain on the roof.

  D





v f





v o

7.1 The Impulse-Momentum Theorem

Neglecting the weight of the raindrops, the net force on a raindrop is simply the force on the raindrop due to the roof.







v f





v o



  



v o



   0 .

060 kg s   15 m s    0 .

90 N

7.1 The Impulse-Momentum Theorem

Conceptual Example 3 Hailstones Versus Raindrops

Instead of rain, suppose hail is falling. Unlike rain, hail usually bounces off the roof of the car.

If hail fell instead of rain, would the force be smaller than, equal to, or greater than that calculated in Example 2?

7.2 The Principle of Conservation of Linear Momentum

 sum of average external forces  D

 

P f

 

P o

If the sum of the external forces is zero, then 0  

P f

 

P o



P f

 

P o

PRINCIPLE OF CONSERVATION OF LINEAR MOMENTUM The total linear momentum of an isolated system is constant (conserved). An isolated system is one for which the sum of the average external forces acting on the system is zero.

7.2 The Principle of Conservation of Linear Momentum

Conceptual Example 4 Is the Total Momentum Conserved?

Imagine two balls colliding on a billiard table that is friction-free. Use the momentum conservation principle in answering the following questions. (a) Is the total momentum of

the two-ball system

the same before and after the collision? (b) Answer part (a) for

a system that contains only one of the two

colliding balls.

7.2 The Principle of Conservation of Linear Momentum

Example 6 Ice Skaters

Starting from rest, two skaters push off against each other on ice where friction is negligible.

One is a 54-kg woman and one is a 88-kg man. The woman moves away with a speed of +2.5 m/s. Find the recoil velocity of the man.

7.2 The Principle of Conservation of Linear Momentum



P f

 

P o

v f

1 

v f

2  0

v f

2  

v f

2    54 kg   2 .

5 m 88 kg s    1 .

5 m s

7.2 The Principle of Conservation of Linear Momentum

Applying the Principle of Conservation of Linear Momentum

1. Decide which objects are included in the system.

2. Relative to the system, identify the internal and external forces.

3. Verify that the system is isolated.

4. Set the final momentum of the system equal to its initial momentum.

Remember that momentum is a vector.

การชนกัน (collision)

-

ในการชนส่วนใหญ่ เราไม่ทราบแรงชน จึงไม่ สามารถวิเคราะห์ การเคลื่อนที่โดยใช ้กฎของนิวตันได ้

-

แต่หากทราบลักษณะการเคลื่อนที่

(

ความเร็ว

)

การวิเคราะห์การเคลื่อนที่ ส าหรับการชนสามารถใช ้

หลักอนุร ักษ์โมเมนตัม

-

ได้

กรณีที่เป็น

การชนแบบยืดหยุ่น

เป็นการชนแบบ ไม่สูญเสียพลังงานจลน์ สามารถใช ้

หลักอนุร ักษ์พลังงานจลน์

หลังชนมีค่าเท่ากัน ซึ่ง หมายถึง พลังงานจลน์รวมของระบบก่อนและ กรณีที่เป็น

การชนแบบไม่ยืดหยุ่น

พลังงานจลน์ 80

7.3 Collisions in One Dimension

The total linear momentum is conserved when two objects collide, provided they constitute an isolated system.

Elastic collision --

One in which the total kinetic energy of the system after the collision is equal to the total kinetic energy before the collision.

Inelastic collision --

One in which the total kinetic energy of the system after the collision is

not

equal to the total kinetic energy before the collision; if the objects stick together after colliding, the collision is said to be completely inelastic.

Example of elastic collision in 1D

7.3 Collisions in One Dimension

Example 8 A Ballistic Pendulum

The mass of the block of wood is 2.50-kg and the mass of the bullet is 0.0100-kg. The block swings to a maximum height of 0.650 m above the initial position.

Find the initial speed of the bullet.

7.3 Collisions in One Dimension

Apply conservation of momentum to the collision:

v f

1 

v f

2 

v o

1 

v o

2 

1 

2 

v f



v o

1 Inelastic collision

v o

1  

1 

2 

v f m

7.3 Collisions in One Dimension

Applying conservation of energy to the swinging motion:

mgh

 1 2

2 

1 

2 

gh f

 1 2 

1 

2 

f gh f

 1 2

f v f

 2

gh f

 2  9 .

80 m s 2   0 .

650 m 

7.3 Collisions in One Dimension

v f

 2  9 .

80 m s 2   0 .

650 m 

v o

1  

1 

2 

v f m

v o

1    0 .

0100 kg  2 .

50 kg 0.0100

kg   2  9 .

80 m s 2   0 .

650 m    896 m s

7.5 Center of Mass

The center of mass is a point that represents the average location for the total mass of a system.

x cm



x m

1 1  

7.5 Center of Mass

x cm



1 D

x m

1 1  

2 D

v cm



v m

1 1  

7.5 Center of Mass

v cm



v m

1 1  

2 In an isolated system, the total linear momentum does not change, therefore the velocity of the center of mass does not change.

7.5 Center of Mass

BEFORE

v cm



v m

1 1  

2  0 AFTER

v cm

  88 kg   1 .

5 m s 88 kg 54 kg    54 kg 2 .

5 m s   0 .

002  0