Dynamics Chapter 2. Kinetics of Particles ENGG1010 2014 2015

Chapter

KINETICS of Particles

Introduction

Observational Experiments: Newton’s Laws of Motion Kinematics’ goal: Describing object’s motion

• Via answering the following questions:

• Without asking the

When ?

question: Why is

Where ?

object moving in a

How fast ?

certain way ?

How far ? How long ?

Observational Experiments: Newton’s Laws of Motion Kinetics’ goal: Describing object’s motion • Via answering the main questions:  Why is the object moving in a certain way?  What causes the object to change its velocity ?

 How the interaction between objects influence their motion ?

Observational Experiments: Newton’s Laws of Motion Kinetics’ goal: Describing object’s motion • Dynamics studies motion on a deeper level than kinematics: it studies the causes of changes in objects’ motion !

The motion of an object depends on the forces acting on it. APPLICATIONS A freight elevator is lifted using a motor attached to a cable and pulley system as shown.

Typical Problem: How can we determine the tension force in the cable required to lift the elevator at a given acceleration? Is the tension force in the cable greater than the weight of the elevator and its load?

Bridging Kinematics and Dynamics Kinematics

Dynamics

Question

- Quantity

Question

Where?

- Position

How much matter? - Mass

When?

- Clock reading

How strong is the interaction?

- Force

What is the effect of the interaction?

- Acceleration

For how long? - Time interval How fast? - Velocity, - speed, - Acceleration

Acceleration

- Quantity

Lecture

1

Newton’s Laws of Motion Isaac Newton’s work represents one of the greatest contributions to science ever made by an individual. Most notably, Newton derived the law of universal gravitation, invented the branch of mathematics called calculus, and performed experiments investigating the nature of light and color.

Newton’s Three Laws of Motion First Law: A particle originally at rest, or moving in a straight line at constant velocity, will remain in this state if the resultant force acting on the particle is zero.

Newton’s Three Laws of Motion Third Law: The mutual forces of action and reaction between two particles are equal, opposite and collinear.

Newton’s Three Laws of Motion Second Law: A particle acted upon by an unbalanced force F experiences an acceleration a that has the same direction as the force and a magnitude that is directly proportional to the force. If F is applied to a particle of mass m then:

12

Newton’s Second Law of Motion Newton’s second law forms the basis of the study of dynamics of particles:

    Fext  Fnet  m a

Where: F is the resultant force acting on the particle,

a is the acceleration of the particle. m is the mass of the particle.

Note: for rigid bodies we will need another equation:

Language and Questions of Dynamics Force: The measure of interaction between two objects (pull or push). It is a vector quantity – it has a magnitude and direction

Direction

𝑨

MASS AND WEIGHT Mass: The measure of how difficult it is to change object’s velocity (inertia of the object) Weight: The force of the Earth is pulling the object with. Weight is a vector quantity, it has a magnitude and direction

It is important to understand the difference between the mass and weight of a body!

MASS AND WEIGHT Mass is an absolute property of a body. It is independent of the gravitational field in which it is measured. The mass provides a measure of the resistance of a body to a change in velocity, (m = F/a).

The weight of a body is not absolute, since it depends on the gravitational field in which it is measured.

Weight is : where g is the acceleration due to gravity.

UNITS

F=ma units of force are ( Mass Length / Time 2 )

Metric Absolute (SI):

Length

Time

Mass

Force

meter

sec

kg

N = kg·m/s2

Method of Solving…

Σ𝐹𝑒𝑥𝑡 = 𝑚 𝑎 Free-Body Diagram

=

Kinetic Diagram

EXAMPLE

Given: A crate of mass m is pulled by a cable attached to a truck. The coefficient of kinetic friction between the crate and road is mk.

Find: Draw the free-body and kinetic diagrams of the crate. Plan: 1) Define an inertial coordinate system. 2) Draw the crate’s free-body diagram, showing all external forces applied to the crate in the proper directions. 3) Draw the crate’s kinetic diagram, showing the inertial force vector ma in the proper direction.

EXAMPLE (continued) Solution: 1) An inertial x-y frame can be defined as fixed to the ground. 2) Draw the free-body diagram of the crate:

W = mg

y

T 30°

x

N

F = mk N

The weight force (W) acts through the crate’s center of mass. T is the tension force in the cable. The normal force (N) is perpendicular to the surface. The friction force (F = mKN) acts in a direction opposite to the motion of the crate.

3) Draw the kinetic diagram of the crate:

ma

The crate will be pulled to the right. The acceleration vector can be directed to the right if the truck is speeding up or to the left if it is slowing down.

System

Free-Body Diagram

Kinetic Diagram

W = mg

y

T 30°

x

N

F = mkN

ma

Free-Body Diagram

Kinetic Diagram

Note The second law only provides solutions for forces and accelerations. If velocity or position have to be found, kinematics equations are used once the acceleration is found from the equation of motion.

Any of the tools learned in the previous chapter may be needed to solve a problem.

Example 1 : Breaking Force (1-D case) A twenty-ton-train cart (m = 20T) is moving at 27 m/s. What breaking force is needed to stop the cart in 50m? v0= 27 m/s

vf = 0 m/s

x Dx = 50 m

Solution

F

ext

Dx 

v 2f  v02

We know the mass of the cart but how can we find its acceleration?

 ma a

To find acceleration we have to use our knowledge of kinematics. If we know the stopping distance and the initial velocity, we can find acceleration!

v 2f  v02

2a 2Dx (0 2 )  (27m / s ) 2 a  7.3m / s 2 2(50m)

F  ma F  (20,000kg)( 7.3m / s 2 ) F  146,000 N

What does this answer mean? How can the force be negative?

Example 2 : Pushing Three Boxes You are pushing three boxes by applying a 7.5 N force as shown in the picture on a frictionless surface. (a) Find the acceleration of the boxes and (b) all the contact forces between them.

Notice, we have to draw a FBD to solve the problem!

Solution     Fext  Fnet  ma F  (m1  m2  m3 )a 7.5 N a  0.8m / s 2 9.4kg

x

  Box1 : Fnet  ma

7.5 N

7.5N  F2 on1  (1.3kg)(0.8m / s ) 2

F2 on1  7.5N  1.04 N  6.46 N

1.3kg

F2 on 1

a

To solve the problem, we first look at three boxes as one system to find a, and then we looked at each box separately.

Solution

   Box 2 :  Fext  Fnet  ma

F1 on 2  F3 on 2  (3.2kg)(0.8m / s )

x F1 on 2

2

F2 on 1   F1 on 2 6.46 N  F3 on 2  2.56 N F3 on 2  6.46 N  2.56 N  3.9 N F3 on 2   F2 on 3 

3.2kg

F3 on 2

a F2 on 3

4.9kg a

Box 3 : F2 on 3  3.9 N Notice, that the negative answer for the force means that the force is directed in the opposite direction to the chosen axis.

Making Sense of the Answers F3 on 2   F2 on 3  F2 on 3  3.9 N 3.9 N a3   0.8m / s 2 4.9kg a3  a  0.8m / s 2

x F2 on 3

4.9kg

a

We checked our answer: the acceleration of the third box equals to 0.8m/s2 as we expected. If we didn’t use the third law of Newton in conjunction with the second one, we wouldn’t be able to solve the problem! Also please notice how proper indices for forces helped us to solve the problem!

Quiz : Two Cars Train A light train made up of two cars is traveling at 90 km/h, when the brakes are applied to both cars. Knowing that car A has a mass of 25 Mg and car B a mass of 20 Mg, and the braking force is 30 kN on each car, Determine: (a) the distance traveled by the train before it comes to a stop, (b) the force in the coupling between the cars while the train is slowing down. 90 km/h

A

B

Quiz : Dependent Motion Knowing that the system shown starts from rest, find the velocity at t = 1.2s of block A. Neglect the masses of the pulleys and the effect of friction. 10 kg A 25 N B 15 kg

EXAMPLE

Given: WA = 10 kg

WB = 20 kg voA = 2 m/s mk = 0.2 Find: vA when A has moved 4 m.

Plan: Since both forces and velocity are involved, this problem requires both the equation of motion and kinematics.

• First, draw free body diagrams of A and B. • Apply the equation of motion. • Using dependent motion equations, derive a relationship between aA and aB and use with the equation of motion formulas.

EXAMPLE (continued) Solution: Free-body and kinetic diagrams of B:

2T

= WB

mB aB

   Fy  m a y Apply the equation of motion to B:

WB  2T  mB aB

mB g  2 T  mB aB

(1)

EXAMPLE (continued) Free-body and kinetic diagrams of A:

y

WA

x

T N

mAaA

=

F = mk N

Apply the equations of motion to A: +

 Fy  m a y  0 N  WA  mA g

F  mk N  mk WA  mk mA g

+   Fx  m ax

T  F  mA a A T  m k m A g  mA a A

(2)

EXAMPLE (continued)

Now consider the kinematics. Constraint equation: sA

Datums

A sB

B

- sA + 2 sB = constant or - vA + 2 vB = 0 Therefore - aA + 2 aB = 0 aA = 2 aB (3)

(Notice aA is considered positive to the right and aB is positive downward.)

EXAMPLE (continued) Now combine equations (1), (2), and (3).

aB

mB  2m k mA  g mB  4 mA

a A  2 aB

aB  2.616 m / s 2 a A  5.232 m / s 2

VA2  V02A  2 a A s A  s A0  V  2  2  5.232  4  2 A

2

VA  6.772 m / s 

Lecture

2

The inclined plane is a plane surface set at an angle, other than a right angle, against a horizontal surface. The inclined plane permits one to overcome a large resistance by applying a relatively small force through a longer distance than the load is to be raised.

Components of the Weight on the Ramp

Wx  W sin (  )  Wy  W cos (  )

Now instead of talking about the horizontal & vertical components we have components in the direction of the ramp (x) and in the perpendicular direction (y)!

Notice, it is convenient to resolve the weight of the object on an incline into components. The x-axis is in the direction of the incline (motion/acceleration), y is perpendicular to the inclined surface.

Example:

Inclined Plane (Ramp)

Your brother (30 kg) slides downhill (=300) on a sled

(10kg)

(a) Find the acceleration of the brother and the sled (b) Find the contact force of the surface on the sled. (c) Explain.

Solution To solve the problem, we had to decide what objects we are looking at. In this problem: a brother-sled is the system.

 Wx  W sin(30 )  0  Wy  W cos(30 ) 0

We had to resolve the weight of the system into components first!

  F  m a  ext    N W  m a   Fx  max    Fy  ma y  mg sin( )  m a x    mg cos( )  N  0  g sin( )  a x  m g cos( )  N 2 0  ( 10 m / s ) sin( 30 )  ax   2 0  ( 40 kg )( 10 m / s ) cos( 30 )N 

 5m / s  a x   346 Newtons  N 2

Solution Reality Check: Units and Special Cases Units make sense! When the angle of an inclined plane is zero the acceleration along the incline is zero: ax = gsin()= gsin(00)=0

When an inclined plane is vertical, the acceleration along it must equal g (free fall): ax = gsin(900) = g The answer makes sense!

Forces in Everyday Life:

Friction and

Tension

I: Frictional Forces: Observations • Frictional forces are contact forces • No surface is smooth • Extremely smooth surfaces stick as well • Whatever you have you will always have some friction • Frictional forces depend on many things, we will explore it in detail…

Static and Kinetic Friction

Static (v = 0) and kinetic friction (non zero v) are two related by different types of friction! We denote them and fs and fk.

Kinetic versus Static Friction

• Kinetic friction is the frictional force existing during relative motion of surfaces • Static friction exist when the surfaces are not moving relatively to each other.

Exploring Kinetic Friction What can kinetic friction depend on?      

Quality of the surfaces Area of Contact Weight of the object Normal force Relative speed of motion Temperature of objects…

II: Model for Kinetic and Static Friction • Kinetic Friction

f k  mk mg

• Static Friction

f s  ms mg

or

or

f k  mk N

f s  ms N

Example : Incline with Friction You are pushing a 10-kg shopping cart up a 300 incline with a constant force of F = 80N parallel to the incline. If mk=0.2, find cart’s acceleration.

Problem Solving Strategy y

1. Choose coordinate axes

N

2. Resolve the forces into

x a=?

Fperson f

components 3. Apply Newton’s 2nd law to each one of the dimensions: x and y 4. Solve the equations

W

5. Check IF the answers make sense

Applying Newton’s 2nd Law     Fext  Fnet ma     N surface /cart  f surface/cart  FEarth/cart  Fperson/cart  macart      F  W  N  f  ma   Fx  m a x    Fy  m a y  m g sin( )  F  m k N  m a x   m g cos( )  N  0   m g sin( )  F  m k m g cos( )   m a x    N  m g cos( )

Applying Newton’s 2nd Law axcart   g sin( )  mk g cos( )  F / mcart



  



  

a x   10 m / s 2 sin 300  0.2 10 m / s 2 cos 300  80 N  / 10 kg  a x  5 m / s 2  1.72 m / s 2  8 m / s 2 a x  1.38 m / s 2

So What does it Mean? 1- No Friction force ( f ) and no pushing force ( F ) :

ax   g sin  2- No pushing force ( F ) [ only friction and weight ]: a x   g sin   m k g cos   a x   g m k cos   sin  

If g sin   m k g cos   m k  tan   a  0 Notice, FRICTION changes its sign: its direction is up now – against gravitational force

3- No friction :

ax   g sin( )  F / mcart

–

String (cable) Tension • We will deal with massless and nonstretching strings. In this case the tension across the string is uniform. • We will denote it as T

Example: Pulley and Inclined Planes Frictionless inclined plane and frictionless pulley:

x

Find a - ?

x



Problem Solving Strategy     Fext  Fnet  m  M  a

mg  Mg sin   m  M  a mg  Mg sin  a mM When friction is involved the problem gets trickier: you have to figure out first where the object is going to move and then to include friction in the direction opposite to relative motion. If you need to find string tension, look at each mass separately!

1. Choose coordinate axes

2. Resolve the forces into components 3. Apply Newton’s 2nd law to each one of the dimensions: x and y 4. Solve the equations 5. Check IF the answers make sense

Problem The two blocks A and B shown are originally at rest. Neglecting the masses of the pulleys and assuming that the coefficients of friction between block A and the incline are ms = 0.25 and mk = 0.20, determine the acceleration of block B. 100 kg A

30

O

B

160 kg

Problem Block A has a mass of 25 kg and block B a mass of 15 kg. The coefficients of friction between all surfaces are ms = 0.20 and mk = 0.15. Knowing that  = 25o and that the magnitude of the force P applied to block A is 250 N, determine: (a) the acceleration of block A , (b) the tension in the cord.

P

A B 

Lecture

3

Circular Motion

Looking for Acceleration… Where to look for acceleration:

  a  DV Dt

Changing magnitude of velocity

Changing direction of velocity

Changing magnitude and direction of v

Speeding up or slowing down along a straight line

Moving with constant speed along a curved line

Speeding up or slowing down along a curved line

EQUATIONS OF MOTION:

CYLINDRICAL COORDINATES This approach to solving problems has some similarity to the normal & tangential method. However, the path may have attributes that make it desirable to use cylindrical coordinates (e.g., a function of ).

EQUATIONS OF MOTION: CYLINDRICAL COORDINATES

The equations of motion in cylindrical coordinates (using r,  , and z coordinates) may be expressed in scalar form as:

   m  r  2r  

2   Fr  mar  m r  r 

 F  ma  F  ma  m z z

z

EQUATIONS OF MOTION (continued)

If the particle is constrained to move only in the r –  plane (i.e., the z coordinate is constant), then only the first two equations are used (as shown below). The coordinate system in such a case becomes a polar coordinate system.



2   Fr  mar  m r  r 





  2r  F  ma  m r    



NORMAL & TANGENTIAL COORDINATES (Review) When a particle moves along a curved path, it may be more convenient to write the equation of motion in terms of normal and tangential coordinates.

The normal direction (n) always points toward the path’s center of curvature. In a circle, the center of curvature is the center of the circle. The tangential direction (t) is tangent to the path, usually set as positive in the direction of motion of the particle.

EQUATIONS OF MOTION FOR NORMAL & TANGENTIAL COORDINATES The equation of motion,

F = ma,

may be written in terms of the n & t coordinates as

Ft ut + Fn un = m at ut + m an un Here Ft & Fn are the sums of the force components acting in the t & n directions, respectively. This vector equation will be satisfied provided the individual components on each side of the equation are equal, resulting in the two scalar equations:

Ft = mat

and

Fn = man

NORMAL AND TANGENTIAL ACCELERATIONS (Review)

The tangential acceleration, at = dv/dt, represents the time rate of change in the magnitude of the velocity. The normal acceleration, an = v2/r, represents the time rate of change in the direction of the velocity vector. Remember, an always acts toward the path’s center of curvature.

Recall, if the path of motion is defined as y = f(x), the radius of curvature at any point can be obtained from

  dy  1      dx  r 2 d y dx 2

2

  

3

2

SOLVING PROBLEMS WITH n-t COORDINATES • Use n-t coordinates when a particle is moving along a known, curved path.

• Establish the n-t coordinate system on the particle. • Draw free-body and kinetic diagrams of the particle. The normal acceleration (an) always acts “inward” (the positive ndirection). The tangential acceleration (at) may act in either the positive or negative t direction. • Apply the equations of motion and solve. • It may be necessary to employ the kinematic relations:

at = dv/dt = v dv/ds

and/or

an = v2/r

Circular Motion: Observations • Whenever an object is moving along a curved path with constant speed, the direction of velocity is changing therefore acceleration must be present! • The presence of acceleration means that the net force acting on an object is NOT zero.

• What are the magnitude and direction of this net force?

    Fext  Fnet  ma

y

dV at  dt V2 aN  R

When an object is moving along a circular path with a constant speed, the net force acting on it has to be directed toward the center of a curvature (circle). In this case, only radial acceleration is involved. The magnitude of the net force has to equal the mass of the object multiplied by the square of its speed and divided by the radius of the curvature.

If the object is moving with changing speed along the circular path the object will have both radial and tangential accelerations. Radial acceleration (for turning) and tangential acceleration for changing the speed!

Looking for Acceleration… Where to look for acceleration: a = Dv/Dt Changing magnitude of velocity

Changing direction of velocity

Changing magnitude and direction of v

Speeding up or slowing down along a straight line

Moving with constant speed along a curved line

Speeding up or slowing down along a curved line

Tangential a: at

Radial a: aR

aR and at

Example: Bridges & Speed Bumps

Evaluate the maximum safe speed of the car which allow you to stay on the bump ( N > 0 )

Example 2: Bridges & Speed Bumps 2

v  Fext  m R v2 mg  N  m R NR v  gR  m N  0  vCr 

gR

The curvature of the bridge/bump and your speed determines if you are going to stay on the bump (N>0) or if you are going to bump: fly off it!

Example 3: Car Rounding a Curve – Unbanked

A curve in a Round/About has a radius of 45 m. Evaluate the maximum safe speed of the car which prevents skidding on the curve.

Car Rounding a Curve – Unbanked



  Fext  ma

  v2 v2 x : f  m x : mk N  m R  R    y : N  mg  0  y : N  mg   v2 x : m k mg  m  v  m k gR ; R  45m; m k  0.5; v  53.5km / h R On a wet pavement : m k  0.1  v  24 km / h The max speed is independent of the mass of the car! Friction is what makes your car turn: slow down when wet!

Example 4: Car Rounding a Curve – Banked

A curve in a speed track has a radius R and a banking angle . Evaluate the maximum safe speed of the car which prevents skidding on the curve.

Car Rounding a Banked Curve – without friction



  Fext  ma

x : n  normal direction

2  mg v sin( )  m  v2 x : R  x : N sin( )  m  cos( )   R   y : N cos( )  mg  0  y : N  mg   cos( )  v2 x : g tan( )   v  Rg tan( ) ; R  50m;  200 ; v  48 km / h R

The max speed is independent of the mass of the car! This solution does not include friction!

Example 5: Car Rounding a Banked Curve y – with friction – W = mg

n

= m an 

f N





Car Rounding a Banked Curve – with friction



  Fext  ma

2  v n : N sin( )  m N cos   m  v2  n : N sin( )  f cos   m  R  R  mg  y : N cos( )  f sin( )  mg  0 y : N    cos( )  m sin( ) 

mg v2 sin   m cos    m n: cos( )  m sin( ) R v

sin   m cos  Rg cos( )  m sin( )

sin   m cos  v2 g  cos( )  m sin( ) R

; R  50m;   20 0 ; m  0.5; v  74 km/h

f

This solution include friction!

Problem 1: The bob of a 2 m pendulum describes an arc of circle in a vertical plane. If the tension in the cord is 2.5 times the weight of the bob for the position shown, find the velocity and acceleration of the bob in that position.

2m 30 o

The weight of the bob is W=mg; the tension in the cord is thus 2.5 mg.

Recalling that an is directed toward O and assuming at as shown, we apply Newton’s second law and obtain :

n

T = 2.5 mg

m an

= m at t W = mg

30 o

+

 Ft  mat

mg sin 30  mat o

at  g sin 30o   4.90 m / s 2

at  4.90 m / s 2 +

 Fn  man

2.5 mg  mg cos 30o  m an an  1.634 g   16.03 m / s 2 an  16.03 m / s 2

Since an = v 2/R we have v2 = R an = (2m)(16.03 m/s2) V =  5.66 m/s

V = 5.66 m/s

(up or down)

Problem 2: A single wire ACB passes through a ring at C attached to a sphere which revolves at a constant speed v in the horizontal circle. Knowing that the tension is the same in both portions of the wire, determine the speed v.

A

30 o B 45 o C

5 kg 1.6 m

Recalling that an is directed toward the centre of rotation, we apply Newton’s second law and obtain :

y T T

=

n:x

45 o

W = mg 30 o

m an

   Fx  ma

mv2 T sin 30  sin 45  R

   Fy  0

T cos 30  cos 45  mg  0 T cos 30  cos 45  mg

Divide (1) by (2) :

sin 30  sin 45 v2  cos 30  cos 45 Rg v  3.47 m / s

(1)

(2)

Problem 3: A

A single wire ACB passes through a ring at C attached to a sphere which revolves at a constant speed v in the horizontal circle. Knowing that the tension is the same in both portions of the wire, determine the speed v.

30 o

1.5 m

45 o B

C 3 kg

Recalling that an is directed toward the centre of rotation, we apply Newton’s second law and obtain :

y T

45 o

=

n:x T

W 30 o

m an

   Fx  ma

mv2 T sin 30  sin 45  R

   Fy  0

T cos 30  cos 45  mg  0 T cos 30  cos 45  mg

Divide (1) by (2) :

sin 30  sin 45 v2  cos 30  cos 45 Rg v  10.57 m / s

(1)

(2)