Topic 1 - Handout_1

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Mechanical Engineering Science 1 - Dynamics

Rectilinear Motion, Projectiles and Relative Motion

Contents

1.1 Introduction

1.2 Displacement, Velocity and Acceleration1.3 Derivation of Constant Acceleration Formulae1.4 Motion in One Dimension

1.4.1 Horizontal Motion1.4.2 Vertical Motion

1.5 Projectiles1.5.1 Constant Acceleration Formulae for a Projectile1.5.2 Worked Examples

1.6 Relative Motion1.7 Summary

Prerequisite knowledge

• Knowledge of the difference between vector and scalar quantities.

• Calculus - definite integrals.

• Some familiarity with constant acceleration formulae would be useful.

Learning objectives

By the end of this Topic, you should be able to:

• Derive the three constant acceleration formulae from the calculus definitions of

acceleration and velocity;

• Apply these equations to describe the motion of a body with uniform acceleration

moving in one dimension;

• Apply the calculus definitions of acceleration and velocity to the motion of a bodywith variable acceleration;

• Apply constant acceleration formulae to describe the motion of a particle with

uniform acceleration moving in two dimensions;

• Solve problems involving relative displacements, velocities and accelerations.




1.1 Introduction

Rectilinear motion deals with the motion of an object (or body) in a straight line (for

example, the motion of the land-speed record car or a snooker ball). You should already be familiar with motion where the acceleration is constant and with the vector nature of

the quantities displacement, velocity and acceleration. For an object moving in astraight line with uniform acceleration, the average velocity over a period of time is given by

( )

2

vu +

where u and v are the initial and final velocities, respectively. The displacement s after atime t is then given by

( )t

vu s

2

+=

This relationship is, equivalent to the scalar equation

d = v x t

Where d is the distance and v is the average speed.

Distance is a scalar quantity and it is always positive, its vector equivalent, displacement

can be one, two or three dimensional depending on the situation and it is possible to have

negative displacements. For example the displacement of a rock dropped off of a cliff could be described as a negative number if the direction of positive displacement isvertically up.

2




Figure 1.1.1 – Rock falling off of a vertical clif.

This is often the case for problems involving gravity as gravitational acceleration is often

taken to be -9.81 m/s2. The negative sign indicating that the acceleration is towards the

ground.

Outline of material covered in this topicWe will begin by considering a body accelerating non-uniformly. The motion of a bodyundergoing constant acceleration will then be considered. This will lead to three constant

acceleration formulae relating, acceleration, initial velocity, final velocity, displacement

and time. These relationships will then be applied to motion in one and two dimensions.Finally the concept of relative motion will be introduced and applied to a number of

scenarios.

1.2 Displacement, Velocity and Acceleration

The velocity v of a particle can be defined as the rate of change of displacement s(Measured in metres, m) with respect to time t (measured in seconds, s). Velocity has the

units of metres per second (m/s). Mathematically we can write

dt

dsv =

20 m s

Direction of

positive displacement

Origin

3




The acceleration a of a particle can be defined as the rate of change of velocity with

respect to time. Acceleration has the units of metres per second per second, m/s2. This can

be expressed by the equation,

2

2

dt

sd

dt

dva ==

By rearranging the above equation and using the chain rule of differentiation a can begiven in terms of the velocity and displacement as,

ds

dvv

ds

dv

dt

dsa ==

This is sometimes a useful thing to do if the acceleration is expressed as a function of

displacement and/or velocity. See the paper exercise in the following section.

Example

Consider a body with the following equation of motion.

s = t 2 + 5t

where s denotes displacement and t denotes time.

The velocity and acceleration relations with respect to time can be calculated bydifferentiating the displacement-time equation.

52 +== t dt dsv

Differentiating again gives the acceleration,

2==dt

dva

The displacement ( s), velocity (v) and acceleration (a) can be calculated for t =0.0, 0.5,

1.0, and 2.0 s and displacement-time, velocity-time and acceleration-time graphs can bedrawn.

t(sec) 0 0.5 1 2

s(m) 0 2.75 6 14

v(m/s) 5 6 7 9

a(m/s2) 2 2 2 2

4




Displacement-time

0

5

10

15

0 0.5 1 1.5 2 2.5

time (sec)

D i s t a n c e

( m )

Velocity-time

0

2

4

6

8

10

0 0.5 1 1.5 2 2.5

time (sec)

v e l o c i t y ( m / s e c )

Acceleration-time

0

0.5

1

1.5

2

2.5

0 0.5 1 1.5 2 2.5

time (sec)

A c c

e l e r a t i o n ( m / s e c

2 )

5

s = t 2+5t

v = 2t+5

a = 2




The results of this example represent uniformly accelerated motion. Gravity is an

example of such motion where a is constant with respect to t . When the accelerationvaries with time then the resulting motion is not uniform.

Example Problem:

Suppose the acceleration of a body is directly proportional to time. Derive equations

for displacement, and velocity as a function of time.

Solution

First a constant of proportionality must be assumed which will be given the symbolk . Mathematically the equation of motion for acceleration is

kt

dt

dva ==

Separating the variables gives

∫ ∫ = dt adv

∫ ∫ = dt t k dv

Akt

v +=⇒2

2

With the motion starting at time t = 0 and v = 0 the constant of integration, A, can

be evaluated to give the velocity of the body.

0 = 0 + A

Similarly the displacement can be determined from

dt

dsv =

Separating variables gives

∫ ∫ ∫ == dt t k dt vds 2

2

1

Bkt

s +=⇒6

3

6




With the initial conditions set at t = 0, s = 0 and v = 0 the constant of integration B, can

be evaluated.

0 = 0 + B

Therefore the equations of motion are

kt a =

2

2kt

v =

6

3kt

s =

Paper Exercise

A particle moves so that its acceleration a is inversely proportional to its velocity,

(v),

va

1∝

or equivalently

v

k a =

Where k is a constant.

Derive equations, which will enable both the velocity and displacement ( s) to be

calculated, given that at t = 0 the displacement is zero and the particle has initial velocityu.

Time: 10 minutes

Hint – You are going to have to use the ‘chain rule’

dsdvv

dsdv

dt dsa ==

1.3 Derivation of Constant Acceleration Formulae

7




Learning objective : To derive the three constant acceleration formulae

from the calculus definitions of acceleration and

velocity.

The equations describing motion with a constant acceleration can be derived from thedefinitions of acceleration and velocity.

We will start with the definition of instantaneous acceleration, acceleration is the rate of change of velocity with respect to time

dt

dva =

We are looking at motion where a is a constant. To find the velocity after time t , we willintegrate this expression over the time interval from t= 0 to

t = t (please excuse the clumsy notation).

∫ ∫ ∫ ==t v

u

t

dt adt adv00

Remember a is constant!

Carrying out these integrations:

[ ] [ ] t v

u t av 0=

at uv =−

at uv += (1.1)

Equation 1.1 gives us the velocity v after time t , in terms of the acceleration a and theinitial velocity u. Velocity is defined as the rate of change of displacement. We will now

use this definition to derive the second constant acceleration formula:

dt

dsv =

We can substitute for v using Equation 1.1

t avdt

ds+=

Integrating over the time interval from 0 to t,

8




( )∫ ∫ += s t

dt t auds0 0

[ ]t

st at u s

0

2

0

2

1

+=

2

2

1t at u s += (1.2)

Equation 1.2 gives us the displacement s after time t , in terms of the acceleration a and

the initial velocity u. To obtain the third constant acceleration formula, we first rearrangeequation 1.1.

a

uvt

−=

Substituting this expression for t into Equation 1.2 gives us

2

2

1

−+

−=

a

uva

a

uvu s

2222222 uuvvuuv sa +−+−=

sauv 222 += (1.3)

We have obtained three equations relating displacement s, time elapsed t , acceleration a

and the initial and final velocities u and v. Note that with the exception of t , these are all

vector quantities.

at uv += 2

2

1t at u s += sauv 2

22 +=

Although we are dealing with vector quantities, we are only considering the special case

of motion in a straight line. It is vital to ensure we assign the correct positive or negativesign to each of the quantities s, u, v and a.

1.4 Motion in One Dimension

Learning objective: To apply the constant acceleration formulae to

describe the motion of a particle moving in one

dimension.

1.4.1 Horizontal Motion

9




The constant acceleration formulae can be used to solve problems of motion in One

Dimension with constant acceleration. In all cases we will be ignoring the effects of air resistance.

Example Problem:

A car accelerates from rest at a rate of 4.0 m /sec2

1. What is its velocity after 10 s?2. How long does it take to travel 72 m?

3. How far has it travelled after 8.0 s?

Solution:

We can list the data given to us in the question,

u = 0 m/ sec (the car starts from rest)

a = 4.0 m/ sec2

1. The time elapsed is ten seconds t = 10 s and we wish to find v. So with u, a and t

known and v unknown, the correct constant acceleration formula to use is

at uv +=

Velocity - Time Curve

0

10

20

30

40

50

0 2 4 6 8 10 12

Time/ sec-1

V e l o c i t y / m s

e c - 1

2. In part 2 u, a and s are known, and t is the unknown, the constant acceleration formula

to use is,

2

2

1t at u s +=

10

v = u + at

v = 0 + (4 x 10)

v= 40 m sec-1




u = 0 m/ sec (the car is still at rest)a = 4.0 m/ sec2 s = 72 m

t = ?? sec

Displacement - Time Curve

0

10

20

30

40

50

60

70

80

0 2 4 6 8

Time/ sec-1

D i s p l a c e m e n t / m

3. Finally in part 3 s is unknown and, u, a and t are given, so again the appropriate

constant acceleration formula to use is,

2

2

1t at u s +=

This time using different data,

u = 0 m/ sec

a = 4.0 m/ sec2 t = 8.0 s

s = ?? m _____ __ ___ _ _

11

2

2

1t at u s +=

72 = 0 + (0.5x4xt2)

72 = 2t2

t2 = 36

t = 6 sec

2

2

1t at u s +=

s = 0 + (0.5x4x82)

s = 128 m




The same strategy should be used in solving all problems involving bodies undergoing

constant acceleration. List the data given to you in the question. This will ensure that

you use the correct values when you perform any calculations, and should also make itclear to you which of the constant acceleration formulae to use. It is often useful to make

a sketch diagram with arrows, to ensure that any vector quantities are being measured in

the correct direction.

Learning Points

Always use the same procedure to solve a constant acceleration problem in onedimension

• Sketch a diagram.

• List the data.

• Select the appropriate constant acceleration formulae.

• Rearrange constant acceleration formula if necessary.

• Substitute data.

1.4.2 Vertical Motion

When dealing with freely-falling bodies, the acceleration of the body is the acceleration

due to gravity, g = 9.81 m /sec2. If any force other than gravity is acting in the vertical

plane, the body is no longer in free-fall, and the acceleration will take a different value.Problems should be solved using exactly the same method used to solve horizontal

motion problems.

Example Problem:

A student drops a stone from a second floor window, 15 m above the ground.

1. How long does it take for the stone to reach the ground?

2. With what velocity does it hit the ground?

Solution:

When dealing with motion under gravity, we must take care with the direction we chooseas the positive direction. Here, if we take a as a positive acceleration, then v and s will

also be positive in the downward direction.

We are told that

u = 0 m/seca = g = 9.8 m/ sec2

s = 15 m

1. To find t , we use

12






vector is then positive when the object is travelling upwards and negative when it is

returning to the ground. In this situation the acceleration is always negative, as it is

always directed towards the ground.

1.5 Projectiles

In the previous section motion of a body in 1 dimension was considered, such as a stone

falling off a wall under the influence of gravity, g

Figure 1.5.1

A number of constant acceleration formulae were derived, relating initial and finalvelocity, displacement, acceleration and time.

In this section the motion of a body in two dimensions is considered, such a body iscalled a projectile. The two dimensions are defined by the horizontal co-ordinate

direction and vertical co-ordinate direction.

Constant acceleration formulae for a projectile will be described. In addition expressions

for quantities such as the time of flight, maximum height of a projectile, maximum

horizontal distance of a projectile or range will be derived.

We shall be considering the motion of a projectile (golf ball, bullet, football, etc.) which

does not experience any resistive force, i.e. air drag or skin friction., is acted on by

gravity alone, and moves in two dimensions (there is no motion in a third axis); in other

words, an ideal projectile.

1.5.1 Constant Acceleration Formulae for a Projectile

Consider the projectile shown below, fired with an initial velocity u at an angle θ to the

horizontal plane. The co-ordinate frame is taken as the Cartesian x-y axes.

u initial velocity

v initial velocity

s

14




Projectile Trajectory

x

y

Figure 1.5.2

The projectile will have two components of displacement, velocity and acceleration. To

derive the constant acceleration formulae in two dimensions for a projectile we consider

the projectiles motion in the x co-ordinate direction first. It is assumed there is no air resistance to the projectile so the acceleration in the x direction is zero.

a x = 0

As the acceleration in the x direction is zero the x component of the velocity, v x – is

constant, equal to its initial value,

v x = u cos θ

The displacement in the x co-ordinate direction is just the x component of velocity x

time.

s x = v x t

s x = (u cos θ ) t

15

U

θ

g

a x v x s x

s y v ya y




The projectiles motion in the vertical or y co-ordinate direction can be derived similarly

except we have the added complication that the acceleration is not zero, it is -g ,

acceleration due to gravity.

a y = -g

The minus sign is introduced as gravity acts in the negative y direction. The initial

velocity component in the y direction is,

v y = u sin θ

Using the constant acceleration formulae introduced in section 1.4 for 1 dimensional

motion,

v y = u sin θ - gt

Similarly the 1 dimensional constant acceleration formulae derived in section 1.4 can beused to derive an expression for the projectile height,

( ) 2

2

1sin t g t u s y −= θ

ö õø÷_ù ûú _ü0ýÿþ ó________ _______

The constant acceleration formulae in two dimensions are summarised below,

x-components:

a x = 0

v x = u cos θ

s x = (u cos θ ) t

y-components:

a y = -g

v y = u sin θ - gt

( ) 2

2


16




Example Problem:

A boy throws a stone. If the stone has an initial trajectory of 45o and an initial speed of 20m/s how far and how high will the stone be after 1 second?

Solution:

The initial horizontal component of velocity is

v x = u cos θ = 20 cos 45o = 14.1 m/sec

The horizontal distance is

s x = (u cos θ ) t = 14.1 x 1= 14.1 m

Using the vertical displacement formula,

( ) 2

2

1sin t g t u s y −= θ = (20 x sin 45o ) x 1 – 0.5 x 9.81 x 12

= 9.2 m

Paper Exercise

Calculate the vertical and horizontal displacements for a range of times (no more than 6

points) and plot the results. From these estimate the maximum height of the stone. (hint:

The stone is in the air for nearly 3 seconds)

Time: 15 minutes

1.5.2 Worked Examples

Worked Example 1 – Bullet fired from a gun

A bullet is fired at an angle of 30o to the horizontal at a speed of 100m/s. What is the time

of impact with the ground, assuming the bullet misses its target and the terrain is flat also

calculate the range of the bullet (how far has it gone)?

Solution:

A good starting point is to list the data given in the question,

θ = 30o

u = 100 m/sec

When the bullet hits the ground,

17




s y = 0.

Substituting the above three values into the equation for the vertical displacement

( ) 2

2


0 = (100 sin 30o ) t – 0.5 (9.81) t 2

This is a quadratic equation in t which can be factored,

0 = t ( 50 – 4.095 t )

t = 0 or t = 50/4.905 = 10.2 sec

So

t flight = 10.2 sec

An equivalent question to “To calculate the range of the bullet” is “What is the horizontaldisplacement at the time of flight ?”, Which when you have just calculated the time of

flight is an easy question to answer as we know all of the variables on the right hand side

of the horizontal displacement equation,

s x = (u cos θ ) t

s x = (100 cos 30ο

) 10.2

s x = 883.3 m

Worked Example 2 – Projectile on sloped ground

Calculate the initial velocity and the time to impact of the projectile shown in the figure

below.

18




Projectile Trajectory

-12

-10

-8

-6

-4

-2

0

2

4

0 10 20 30

x /m

y / m

Figure 1.5.3

Solution:

From inspection of the diagram, the projectile impacts on the slope at the point (25, -10).The horizontal displacement satisfies the equation,

s x = (u cos θ ) t

and the vertical displacement satisfies the equation,

( ) 2

2


When s x = 25,

o

u

t

30cos

25=

When s y = -10, substituting for time in the horizontal displacement equation. (Note the

negative sign is very important!)

−=−

oo

o

u g

u

u

30cos

25

2

1

30cos

30sin2510

22

2

19




Tidying the equation up and rearranging it to be explicit in initial speed,

4.24

5.40872 =u

u = 12.94 m/s

Substitute for u in the equation for s x gives

25 = (12.94 cos30 ) t

t = 2.23 sec

We could just as easily have made the substitution

ot u

30cos25=

into the horizontal displacement equation and found the time of flight first, in fact this is

a good exercise you might want to try?

1.6 Relative Motion

Thus far we have considered the motion of particles or bodies in one or two dimensions

in a fixed frame of reference. For example when considering projectiles in the previoussection displacements were measured from the launch point of the projectile. In the real

world it is often useful to consider the motion of a particle or body relative to another particle or body, and indeed all motion is relative. Whether we are crossing the street,taking a flight on our holidays or simply travelling on the bus, the motion of one object

can be determined relative to another. In relative motion, relative displacements, relative

velocities and relative accelerations can be calculated. Therefore we shall be primarilydealing with vector quantities.

1.6.1 Relative Motion in One Dimension

Consider the motion of two planes labelled A and B,

20




Figure 1.6.1

Plane A has a velocity of 200 m/s and plane B has a velocity of 50 m/s in the samedirection as plane A. Then the velocity of A relative to B is

v A – v B = 200 – 50 = 150 m/sec

or the relative velocity can be denoted

v AB = 150 m/sec

In a vector diagram this would be represented as

Figure 1.6.2

In addition at time zero seconds suppose the two planes have a displacement of zero

metres. The equations of linear motion discussed in section 1.3 can be applied.

s A = v A t

s B = v B t

s AB = ( v A – v B ) t = v AB t = s A - s B

AB

v A

= 200 m/s v B

= 50 m/s

v A

v AB v

B

21




Therefore after 10 seconds plane A has travelled 2,000 m, plane B has travelled 500 m

and the distance between the two planes is 1,500 metres.

There are two particularly interesting conditions,

s AB = 0 “Plane A and plane B are in the same place !”

v AB = 0 “Plane A and plane B have the same velocity”

1.6.2 Relative Motion in Two Dimensions

The extension to two dimensions is an application of trigonometry. This can best be

appreciated by considering an example.

Example

A ship A is steaming due north at 16 km/h and another ship B is steaming due west

at 12km/h.

Figure 1.6.3

A16 km/h

B

12 km/h

22




We are interested in calculating the relative velocity of ship A with respect to ship B.

Denoting the two velocities of the ships A and B as v A and v B respectively. The relativevelocity of ship A with respect to ship B, v AB satisfies the equation,

v AB = v A – v B

and can be represented in a vector diagram.

Figure 1.6.4

The magnitude of v AB is,

hkmvvv B A AB

/2012162222

=+=+=

and the bearing angleθ

is given by

o

A

B

v

v9.36tan 1 =

−= −

θ

_ _____!#"#$ _&%'_ ÿ _ __ þ _ __()_+*,)-

In some situations a relative vector is known and other vectors remain to be determined.

Example

A man travelling north at 14 km/h finds that the wind appears to blow from the west. For

a second man travelling at a speed of 28 km/h (he is a fast walker) the wind appears to

come from the Northwest. What is the velocity of the wind?

Solution

If we label the first man as M1, and the second as M2 the information given can be

summarised,

vM1 = 14 in the northerly direction

v A

-v B

v ABθ

23




vW,M1 = ? in the west direction

vM2 = 28 in the northerly direction

vW,M2 = ? in the north west direction

AsvW,M1 = vW - vM1

vW = vM1 + vW,M1

This can be represented graphically as

Figure 1.6.5

We know the direction of the relative velocity but not the speed. The wind velocity wedo not know anything apart from it is equivalent to adding the velocities of the man and

the velocity of the wind relative to the man together, hence the dotted arrows.

Similarly for the second man the relative velocity equations and velocity diagram are,

vW,M2 = vW - vM2

vW = vM2 + vW,M2

vM1

vW,M1

vW

24




Figure 1.6.6

If we superimpose the two velocity triangles on top of each other the problem turns into

one of trigonometry, see the figure below,

Figure 1.6.7

From trigonometry

vW,M1 = 14 tan 45o = 14

vM2

vW,M2

vW

45o

vW,M2

14

vW,M1

vW

28

45o

25




and from Pythagoras’s theorem

hkmvW /8.19141422 =+=

With a north easterly direction.

1.6.3 Using Cartesian Co-ordinate Systems

As well as applying vector diagrams to solve problems involving relative quantities,

problems can be formulated in terms of Cartesian co-ordinates.

Figure 1.6.8

The velocity and acceleration vector can be represented in the same way.

Example

Two ships are observed from a coastguard station at 10.00 hours, at which time they have

the following displacement and velocity vectors.

Ship 1: sA = (1,3) vA = (1,2)

Ship 2: sB = (1,2) vB = (5,6)

When is the distance between the two ships at a minimum?

x

y

s1

= ( x1

, y1)

x1

y1

26




Solution

Note this is potentially confusing as a velocity vector looks exactly the same as adisplacement vector. As the velocities are constant the accelerations are zero and the

displacement and velocity vectors can be used to write down the displacement as a

function of time as,

s =s0 + v t

Where s0 represents the displacement at time zero. Graphically the initial set up can be

represented as shown in the figure below.

0

1

2

3

4

5

6

7

8

9

0 1 2 3 4 5 6 7

Figure 1.6.9

So

s A(t) = (1,3)+(1,2) t

s B(t) = (1,2)+(5,6) t

When t = 1

27

A

B

v A

v B




s A(1) = (1,3)+(1,2) = (2,5)

s B(t) = (1,2)+(5,6) = (6,8)

When t = 2

s A(1) = (1,3)+(1,2) x2 = (3,7)

s B(t) = (1,2)+(5,6) x2 = (11,14)

At time t

s A(t) = (1+t , 3+2t )

s B(t) = (1+5t , 2+6t )

The relative displacement is (Note it does not matter if you calculate A relative to B or

vice versa)

s AB(t) = s A(t) – s B(t)= (1+t , 3+2t ) - (1+5t , 2+6t ) = ( -4 t , 1-4t )

The distance between the two ships is equivalent to the magnitude of the relative

displacement.

( ) 1832)41(4 222 +−=−+−= t t t t s AB

To find the minimum distance we could plot the relative displacement vs. time?

28




Distance vs Time

0

0.2

0.4

0.6

0.8

1

1.2

1.4

0 0.1 0.2 0.3

t

s A B

Figure 1.6.9

From this we see the minimum relative displacement occurs when t is approximately 0.15

and the distance is approximately 0.7.

To do better than this we have to tackle the equation analytically

18322

+−= t t s AB

The minimum distance occurs at the minimum of the quadratic equation

183222

+−= t t s AB (It is easier to work with the distance squared

rather than the distance)

The minimum value occurs when

0

2

=dt

sd AB

864

2

−= t dt

sd AB

So

29




8

1

min=t

Therefore the distance between the ships is at a minimum at (10 + 1/8) hours = 10h 7 1/2

minutes. Substitute for t value into the distance formula to give

18

18

8

132

22

+

−

= AB s

2

12 = AB s

707.0= AB

s

1.7 Summary

The material covered in this Topic should have added to your understanding of motion of

bodies. By the end of this Topic you should be able to:

• Use calculus to solve variable acceleration problems

• Derive the three constant acceleration formulae from the calculus definitions of acceleration and velocity.

at uv += 2

2

1t at u s += sauv 2

22 +=

• Apply these equations to describe the motion of a particle with uniformacceleration moving in one dimension.

• find the horizontal and vertical components of the velocity of a body moving in

two dimensions.

• apply the constant acceleration formulae to describe the motion of a projectile

moving in two dimensions.

a x = 0

v x = u cos θ

s x = (u cos θ ) t

a y = -g

v y = u sin θ - gt

( ) 2

2


30



• Solve problems involving the relative motion of objects using both vector

diagrams and Cartesian co-ordinates.

Remember when considering an object’s (A) velocity with respect to another

object (B) the appropriate relationship is

v AB = v A – v B

and the equivalent vector diagram is

Figure 1.7.1

v A

-v B

v ABθ

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