NEWTON’S LAWSCONCEPTS OF MOTION PHY1012F Kinematics Gregor Leigh [email protected]

NEWTON’S LAWS CONCEPTS OF MOTION

PHY1012

F

Kinemati

cs

Gregor [email protected]

NEWTON’S LAWS CONCEPTS OF MOTIONPHY1012F

2

KINEMATICS

Learning outcomes:At the end of this chapter you should be able to…

Interpret, draw and convert between position, velocity and acceleration graphs for 1-d motion.

Use an explicit problem-solving strategy for kinematics problems.

Apply appropriate mathematical representations (equations) in order to solve numerical kinematics problems involving motion in one dimension.


3

The positive end of the x-axis points to the right; The positive end of the y-axis points upwards.

MOTION IN ONE DIMENSION

We shall standardise on the following sign conventions:

y

x


4

x < 0; y > 0

Positions left of the y-axis have negative x values; Positions right of the y-axis have positive x values.Positions below the x-axis have negative y values;Positions above the x-axis have positive y values.


We shall standardise on the following sign conventions:

y

x

x > 0; y > 0

0x = 0; y > 0

x > 0; y < 0x < 0; y < 0

x < 0; y = 0


5

Vectors pointing to the right (or up) are positive; Vectors pointing to the left (or down) are negative.


We shall standardise on the following sign conventions for representing directions:

y

x

0v

0a

NB: The signs represent the directions. The magnitudes of vectors can never be negative!


6


In 1-d the relationship between acceleration and velocity simplifies to…

When is zero, velocity remains constant.

If and have the same sign, the object is speeding up.

If and have opposite signs, the object is slowing down.

a

a

v

a

v


7

The slope at a point on a position-vs-time graph of an object is

A the object’s speed at that pointB the object’s acceleration at that point C the object’s average velocity at that point D the object’s instantaneous velocity at that pointE the distance travelled by the object to that point


8

POSITION GRAPHS

Plotting a body’s position on a vertical axis against time on the horizontal axis produces a position-vs-time graph, or position graph…

v 1 frame per minute

x (m)0 200100 300 400

x (m)

t (min)

200

400

2 4 60

0


9

POSITION GRAPHS

Plotting a body’s position on a vertical axis against time on the horizontal axis produces a position-vs-time graph, or position graph…

v

x (m)

t (min)

200

400

2 4 60

0

1 frame per minute

x (m)0 200100 300 400


10

INTERPRETING POSITION GRAPHS

It is essential to remember that motion graphs are abstract representations of motion – they are NOT pictures!The following graph represents the motion of a car along a straight road… Describe the motion of the car.

x (km)

t (min)

–10

20 40 600

–20

10

80

After 30 min the car stops for 10 min at a position 20 km to the left of the origin.The car reaches the origin once more at 80 min.At 40 min the car starts moving back to the right.During the first 30 min the value of x changes from +10 km to 20 km, indicating that the car is moving to the left.

At t = 0 the car is 10 km to the right of the origin.


11

UNIFORM MOTION

Straight-line motion in which equal displacements occur during any successive equal-time intervals is called uniform motion.

Motion diagram:

Position graph:

s (m)

t (s)60 2 4

2

4

6

0

s = 4 m

t = 6 s

avg slope of position graphv

avgsvt


12

UNIFORM MOTION

Straight-line motion in which equal displacements occur during any successive equal-time intervals is called uniform motion.

Motion diagram:

Position graph:

An object’s straight line motion is uniform if and only if its velocity vx (or vy) is constant and unchanging.

avg

avg slope of position graph

svt

v

s (m)

t (s)60 2 4

2

4

6

0

t = 4 s

s = 6 m


13

POSITION GRAPHS OF UNIFORM MOTION

Body B is… travelling to the left/right and is going slower/faster than A.

Body A is travelling to the right at constant speed…

x (m)

t (s)

A

x (m)

t (s)

B

(Assume same axes scales.)


14


Summary:

Zero slope zero velocity. (Object is stationary.)

Steeper slopes faster speeds.

Negative slope negative velocity (ie vx is left/vy is

down).

The sign (negative or positive) refers only to the direction of the velocity, and has nothing to do with its magnitude.


15


Summary:

Zero slope zero velocity. (Object is stationary.)

Steeper slopes faster speeds.

Negative slope negative velocity (ie vx is left/vy is

down).

The sign (negative or positive) refers only to the direction of the velocity, and has nothing to do with its magnitude.

The slope is a ratio of intervals, x/t, not coordinates, x/t.

Remember: Position graphs are abstract representations! We are concerned with the physically meaningful slope [in m/s], not the actual slope of the graph on paper.


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0 1 2 3 4

Describe carefully (and quantitatively) the motion of the basketball player depicted by this position graph.

x (m)

t (s)

2

4

6

0


17

THE MATHEMATICS OF UNIFORM MOTION

The velocity of a uniformly moving object tells us the amount by which its position changes during each second.

f i ss s v t

ssvt

s (m)

t (s)

sf

si

tfti

t

s

(For uniform motion)

f i

f i

s st t


BobBob

PHY1012F

18

Bob leaves home in Chicago at 09:00 and travels east at a steady 100 km/h. Susan, 680 km to the east in Pittsburgh, leaves at the same time and travels west at a steady 70 km/h. Where will they meet?

Pictorial representation:

(x0)B, (vx)B, t0

(ax)B = 0

(x0)S, (vx)S, t0(x1)B, (vx)B, t1

(x1)S, (vx)S, t1

(ax)S = 0

MULTI-REPRESENTATIONAL PROBLEM-SOLVING

(x0)B = 0 km(x0)S = +680 km

(vx)B = +100 km/h(vx)S = –70 km/h

t0 = 0 h t1 is when (x1)B = (x1)S

(x1)B = ?

x0


19



Physical representation:

Bv

Chicago

Sv

PittsburghmeetB 0a

S 0a


20


Graphical representation:


x (km)

200

0tmeet

t (h)0

400

600Susan

Bob


21

680 km

100 km/h 70 km/h


Mathematical representation:

(x1)B = (x0)B + (vx)B(t1 – t0)

(x1)S = (x0)S + (vx)S(t1 – t0)

They meet when (x1)B = (x1)S

(x1)B = (vx)Bt1 = 100 km/h 4.0 h = 400 km east of Chicago


sf = si + vst

= (vx)Bt1

= (x0)S + (vx)St1

i.e. (vx)Bt1 = (x0)S + (vx)St1

0 S1

B Sx x

xt

v v

4.0 hours


22

INSTANTANEOUS VELOCITY

Adjusting the time interval between “movie frames” of the horizontally orbiting tennis ball alters the average velocity vectors and the information they convey…

1v

0v

2v3v

4v

5v

6v

7v


23


Adjusting the time interval between “movie frames” of the horizontally orbiting tennis ball alters the average velocity vectors and the information they convey…

0v

1v

2v

3vThe longer the time

interval, the less “real” the representation:(Especially if the “background” information is removed.)


24


Conversely, the shorter the time interval, the more the vectors tend to show us what the ball is really doing – rather than merely depicting “average” behaviour for that time interval.

As t gets smaller and smaller, approaches a limit – a constant value representing the instantaneous velocity at that point in time.

avgs

tv

0limst

s dsvt dt

Mathematically:


25

ctn1. Multiply the expression by the existing

index.

2. Subtract 1 from the index.

n –1dudt

DERIVATIVES

0limst

s dsvt dt

In general (using an arbitrary function as a template), if u = ctn, to find the derivative of u (with respect to t)…

d du dwu wdt dt dt

The derivative of a sum is the sum of the derivatives:

dsdtThe limit, , is called

the derivative of s with respect to t.

E.g.

212s ut at

i.e.

v u at

1 1 2 1121 2ds ut at

dt


26


The same process of shrinking the time interval to determine the instantaneous velocity at one particular time can also be applied to linear motion.

In this case, however, it is more helpful to make use of a position graph rather than a motion diagram…


27

12 14 164 6 8


The motion diagram represents an object whose speed is NOT constant, but increases uniformly each second:v 1 frame per second

s (m)0 2 10

(Plotting uniformly accelerated motion against time results in a parabolically-shaped position graph.)

2 4 60t (s)

8

4

8

0

s (m)

12

16

s

s


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s (m)

t (s)

The ratio gives the average velocity, , for that particular time interval, represented graphically by the slope of the dotted line.

st

avgv

The larger t, the less detailed the information…

s

s s

ttt


29


s (m)

t (s)

Conversely, if we let the time interval either side of time t shrink towards the limit (t0), we get the instantaneous velocity at time t.

On a position graph this corresponds to the slope of the tangent to the curve at time t.

t

s

t0

limst

s dsvt dt

Mathematically:


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INSTANTANEOUS vs AVERAGE VELOCITY

s (m)

t (s)

Note that (for uniform acceleration) the average velocity for a whole time interval is the same as

t

the instantaneous velocity at time t in the middle of the interval…

…as illustrated by the parallel slopes of the dotted lines on the position graph.


31

xt

vx (m/s)

–10

2 4 60

5

8t (s)

For the first 3 s the

is

POSITION GRAPHS VELOCITY GRAPHS

Velocity is equivalent to the slope of a position graph.

x (m)

t (s)

–102 4 6

0

–20

10

8

20 10 m 10 m/s3 0 s

xt

E.g. A car travels along a straight road…

velocityslope


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Between 3 s and 4 s the

is



x (m)

–102 4 6

0

–20

10

8

vx (m/s)

–10

2 4 60

5

8

20 20 m0 m/s

1 sxt


t (s)

t (s)

velocityslope


33

x

t


is



x (m)

–102 4 6

0

–20

10

8

vx (m/s)

–10

2 4 60

5

8

0 20 m5 m/s

8 4 sxt


t (s)

t (s)

velocityslope


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increases steadily from zero to 7 m/s.



x (m)

t (s)

–102 4 6

0

–20

10

8


velocityslope

vx (m/s)

2 4 60

8t (s)

4

8


35

From 3 s to 6 s the

remains a steady 7 m/s.



x (m)

t (s)

–102 4 6

0

–20

10

8


velocityslope

vx (m/s)

2 4 60

8t (s)

4

8


36


quickly decreases to 0…



x (m)

t (s)

–102 4 6

0

–20

10

8


velocityslope

vx (m/s)

2 4 60

8t (s)

4

8

…and remains there.


37

THREE-MINUTE PAPER

On a smallish piece of paper (which you’re going to fold in half), answer the following questions:

What have you learnt so far?

What still confuses you the most?

Are you having fun? (If not, why not?!)


38

FINDING POSITION FROM VELOCITY

In the previous chapter we showed that a body’s position can be determined from its velocity using .

f is s v t

Graphically, the change in position (s = vt) is given by

the area “under” a velocity graph:

tv

vs (m/s)

2 4 60

8t (s)

4

8

During the time interval 2 s to 8 s the body travels a distance

8 m/s 8 2 s

48 m

v t

s


39


In the previous chapter we showed that a body’s position can be determined from its velocity using .

f is s v t

Graphically, the change in position (s = vt) is given by

the area “under” a velocity graph:vs (m/s)

2 4 60

8t (s)

4

8

Even if the velocity varied (uniformly) during the time interval, s could still be determined by summing the “bits”:1 2 3s s s s

s1 s3s2


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If the velocity varies non -uniformly during the interval…

vs (m/s)

ti tf

t (s)

The total area under the graph is approximately

…we can approximate the motion with a series of constant velocity intervals.

s1 s2

t t

(vs)1

(vs)2 1 2s ss v t v t

2

1s k

ks v t

1 2s s s


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s3

s4

s1 s2


Once again we apply calculus, shrinking the t’s to obtain more and more accurate approximations…

vs (m/s)

until, as t 0,

1 2 1 2N s s s Ns s s s v t v t v t

f

0 1 i

limtN

s skt k tv t v dt

t (s)ti tf

f

f ii

t

st

s s v dt So

1

N

s kk

s v t

i.e.


42

INTEGRALS

is called the integral of vs dt from ti to tf.

Since it has two definite boundaries (ti and tf), it

is known as a definite integral.

f

i

t

st

v dt

In general (using an arbitrary function as a template),

The integral of a sum is the sum of the integrals:

f f f

i i i

t t t

t t tu w dt udt w dt

1 1f i

1 1

n nct ctn n

ff 1

i i1

tt nn

t t

ctct dtn


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f

i

t

t

udt +1

To integrate u = ctn …

ctn

DIFFERENTIATION AND INTEGRATION FOR DUMMIES

To differentiate u = ctn …

1. Multiply the expression by the existing index.

2. Subtract 1 from the index.

n –1dudt

ctn1. Add 1 to the index.

2. Divide the expression by the new index.

3. Evaluate the integral at the upper limit,

and...

4. subtract the lower limit value of the

integral.

n +1

f

i

t

t

1f

1

nctn

1i

1

nctn


44


A body which starts at position xi = 30 m at time ti,

moves according to vx = (– 5t + 10) m/s.

Where does the body turn around?At what time does the body reach the origin?

2 4 6t (s)

vx (m/s)

0

10

–20

–10


45

0

10t

dt0

5t

t dt

0

30 5 10t

t dt


A body which starts at position xi = 30 m at time ti,

moves according to vx = (– 5t + 10) m/s.

Where does the body turn around?At what time does the body reach the origin?

f i0

t

xx x v dt

and

2f

530 10 m2

x t t Substitute t = 2 and solve for x. Substitute x = 0 and solve for t.

0 0

30 5 10t t

t dt dt

2

0

52

tt 2 25 5 0

2 2t 25

2t

010

tt 10 0t 10t


46


is

VELOCITY GRAPHS ACCELERATION GRAPHS

Acceleration is equivalent to the slope of a velocity graph.

26 0 m/s 1 m/s6 0 s

vt


accelerationslope3 6 9

012

t (s)

vx (m/s)

–6

6

1

ax (m/s2)

–1

–2

3 6 90

12t (s)


accelerationacceleration

PHY1012F

47

For the last 6 s the

is

VELOCITY GRAPHS ACCELERATION GRAPHS

Acceleration is equivalent to the slope of a velocity graph.

26 6 m/s 2 m/s12 6 s

vt


slope3 6 9

012

t (s)

vx (m/s)

–6

6

1

ax (m/s2)

–1

–2

3 6 90

12t (s)


48

SUMMARY OF GRAPHS OF MOTION

t

s

Constant +ve velocity

t

as

0

t

vs

vis

t

s

Increasing +ve velocity

t

as

0

t

s

t

as

0

t

vs

vfs

vis

vis

t

vs

vfs

Decreasing +ve velocity


49

SUMMARY OF GRAPHS OF MOTION

t

s

Constant –ve velocity

t

as

0

t

s

Increasing –ve velocity

t

as

0

t

s

t

as

0

tvs

vis

vfs

Decreasing –ve velocity

t

vs

0t

vs

vfs

vis


50

KINEMATIC EQUATIONS OF CONSTANT a

By definition, vs (m/s)

t (s)

vfs

vis

tfti t

vs =

ast

f is s ss

v v va

t t

Hence: vfs = vis + ast

sf = si + area under v-graph between ti and tf .

½as(t)2

vist

Hence: sf = si + vist + ½as(t)2

vis

And, substituting t = (vfs – vis )/as: vfs2 = vis

2 + 2ass


51

PROBLEM-SOLVING STRATEGY FOR CONSTANT ACCELERATION KINEMATICS PROBLEMS

Use the particle model. Make simplifying assumptions.

Draw a physical representation (motion diagram).

Draw a pictorial representation.

Draw a graphical representation if appropriate.

Use a mathematical representation (using the equations of motion with appropriately modified subscripts) to solve.

Assess your solution: Is it complete? Is it reasonable?


52

A student is running at a constant speed of 5 m/s in an attempt to catch a Jammie Shuttle. When she is 11 m from the bus, it pulls away with a constant acceleration of 1 m/s2. From this point, how long does it take her to catch up to the bus if she keeps running with the same speed? Physical representation:

Wv

Jv

W 0a

Ja

x


53

A student is running at a constant speed of 5 m/s in an attempt to catch a Jammie Shuttle. When she is 11 m from the bus, it pulls away with a constant acceleration of 1 m/s2. From this point, how long does it take her to catch up to the bus if she keeps running with the same speed? Physical representation:

Wv

Jv

W 0a

Ja

x


54

A student is running at a constant speed of 5 m/s in an attempt to catch a Jammie Shuttle. When she is 11 m from the bus, it pulls away with a constant acceleration of 1 m/s2. From this point, how long does it take her to catch up to the bus if she keeps running with the same speed? Pictorial representation:

x (m)

0

(x0)W, (v0x)W, t0

(ax)W (x0)J, (v0x)J, t0 (x1)J, (v1x)J, t1

(x1)W, (v1x)W, t1

(ax)J

(x0)W = 0 m

(x0)J = +11 mt0 = 0 s

(v0x)W = (v1x)W = +5 m/s

(v0x)J = 0 m/s(ax)W = 0 m/s2

(ax)J = +1 m/s2

(x1)W = (x1)J = ?(v1x)J = ?

t1 = ? is when (x1)W = (x1)J


55

A student is running at a constant speed of 5 m/s in an attempt to catch a Jammie Shuttle. When she is 11 m from the bus, it pulls away with a constant acceleration of 1 m/s2. From this point, how long does it take her to catch up to the bus if she keeps running with the same speed? Graphical representation:

x (m)

t (s)

J

W0

10

20

30

40


56

A student is running at a constant speed of 5 m/s in an attempt to catch a Jammie Shuttle. When she is 11 m from the bus, it pulls away with a constant acceleration of 1 m/s2. From this point, how long does it take her to catch up to the bus if she keeps running with the same speed? Mathematical representation:

(x1)W = (x0)W + (v0x)W t + ½ (ax)W t2

(x1)J = (x0)J + (v0x)J t + ½ (ax)J t2

She catches the shuttle when (x1)W = (x1)J

i.e. 5t = 11 + ½t2

½t2 – 5t + 11 = 0

t = 3.3 s or t = 6.7 s

sf = si + vist + ½as(t)2

= 0 + 5t + (½ 0 t2) = 5t

= 11 + 0t + (½ 1 t2) = 11 + ½t2


FREE FALL

The motion of a body moving under the influence of gravity only, and no other forces, is called free fall.

(We often ignore air resistance for slow-moving, massive objects.)

Consequently…

Two objects dropped from the same height in the absence of air resistance will hit the ground simultaneously, at the same speed.

free fallaAny two objects in free fall experience

the same acceleration, .

57


58

FREE FALL

Notes: g = 9.80 m/s2 is magnitude of the acceleration due to gravity. It is therefore never negative!

In our convention, afree fall = –g.

g = 9.80 m/s2 is the average value for the surface of the Earth.

“Free fall” refers also to objects which have been projected upwards – not only to those which are literally falling downwards.


59

FREE FALL

A kingfisher hovers 30 m directly above a boy with a catapult. If the boy launches a stone straight up at 25 m/s, how long does the stone take to hit the bird?

y

0 y0, v0y, t0

y1, v1y, t1

ay

y0 = 0 m t0 = 0 s

y1 = +30 m

v0y = +25 m/s

ay = –g = –10 m/s2

t1 = ? (= t)


60

FREE FALL


vy (m/s)

15

25

20

–5

10

0

5

1 2 3t (s)


61

FREE FALL


y

0 y0, v0y, t0

y1, v1y, t1

ay

y0 = 0 m t0 = 0 s

y1 = 30 m

v0y = 25 m/s

ay = –g = –10 m/s2

t1 = ? (= t)

y1 = y0 + v0yt + ½ay(t)2

30 = 0 + 25t + ½ (–10)t2

5t2 – 25t + 30 = 0

t2 – 5t + 6 = 0

(t – 2)(t – 3) = 0

t = 2 s or t = 3 s ?!


62

FREE FALL


1v

2v

0v

3v

a

stop/start

5v

4v


63

FREE FALL


vy (m/s)

15

25

20

–5

10

0

5

1 2 3t (s)


64

Sign chosen by inspection.(In cases where the acceleration points left, as = –g sin )

MOTION ON AN INCLINED PLANE

The acceleration down (i.e. parallel to) this frictionless plane which is inclined at an angle to the horizontal is…

sv

sa

a

a

free fall

a

g

as = g sin

a

v


65

INSTANTANEOUS ACCELERATION

For non-uniformly accelerated motion we can define instantaneous acceleration similarly to the way we defined instantaneous velocity…

0lim s s

st

v dva

t dt

Mathematically, as

Graphically, as the slope of the tangent to the velocity curve at a specific instant of time t.

vs (m/s)

t (s)t


66

KINEMATICS

Learning outcomes:At the end of this chapter you should be able to…

Interpret, draw and convert between position, velocity and acceleration graphs.

Use an explicit problem-solving strategy for kinematics problems.

Apply appropriate mathematical representations (equations) in order to solve numerical kinematics problems.

Documents

NEWTON’S LAWSCONCEPTS OF MOTION PHY1012F Kinematics Gregor Leigh [email protected]