1

MM203Mechanics of Machines: Part 1

2

Module

• Lectures

• Tutorials

• Labs

• Why study dynamics?

• Problem solving

3

Vectors

P

Q

R

PQQP

RQP

QRQRP P

−Q

R

P

QR

4

7

1

2

3

5

RQPRQP

3

5

3

5

4

Unit vectors - components

kjiv zyx vvv

1

03

0

14

3

4

?3

4

222zyx vvvv v

0

3

4

3

4

5

Direction cosines

• l, m, and n – direction cosines between v and x-, y-, and z-axes

v

vn

v

vm

v

vl zyx ,,

– Calculate 3 direction cosines for

4

3

8

6

1, 222 nmlnmlv kjiv

6

Dot (or scalar) product

• Component of Q in P direction

zzyyxx QPQPQP QP

cosPQQPQ

P

Q P

QPP

QQP

QP cos

7

Angle between vectors

QPQPQP

zzyyxx

nnmmll

PQ

QPQPQP

PQ

QPcos

?

?

ji

ii2PPP

8

Dot product

• Commutative and distributive

RPQPRQP

PQQP

9

Particle kinematics

• What is kinematics?

• What is a particle?

• Rectilinear motion - review

• Plane curvilinear motion - review

• Relative motion

• Space curvilinear motion

10

• Combining gives

• What do dv and ds represent?

Rectilinear motion

dt

dva

dt

dsv

dsadvv

11

Example

• The acceleration of a particle is a = 4t − 30 (where a is in m/s2 and t is in seconds). Determine the velocity and displacement in terms of time. (Problem 2/5, M&K)

12

Vector calculus

PP

P uudt

ud

kjiP zyx PPP

• Vectors can vary both in length and in direction

13

Plane curvilinear motion

• Choice of coordinate system (axes)– Depends on problem – how information is given

and/or what simplifies solution– Practice

14

Plane curvilinear motion

• Rectangular coordinates

• Position vector - r

x

y

i

j

jijirva

jijirv

jir

yx

yx

aayx

vvyx

yx

• e.g. projectile motion • ENSURE CONSISTENCY IN DIRECTIONS

15

Plane curvilinear motion• Normal and tangential coordinates• Instantaneous radius of curvature –

• What is direction of v?

x

y db

ds

b dds

dt

d

dt

dsv

bv

Note that d/dt can be ignored in this case – see M&K.

16

Plane curvilinear motion

e ten

ttv eev b

tt

t vvdt

vdee

eva

nt ee b

sva

vva

t

n

bb2

2

17

Example

• A test car starts from rest on a horizontal circular track of 80 m radius and increases its speed at a uniform rate to reach 100 km/h in 10 seconds. Determine the magnitude of the acceleration of the car 8 seconds after the start. (Answer: a = 6.77 m/s2). (Problem 2/97, M&K)

18

Example

• To simulate a condition of “weightlessness” in its cabin, an aircraft travelling at 800 km/h moves an a sustained curve as shown. At what rate in degrees per second should the pilot drop his longitudinal line of sight to effect the desired condition? Use g = 9.79 m/s2. (Answer: db/dt = 2.52 deg/s). (Problem 2/111, M&K)

b

19

Example

• A ball is thrown horizontally at 15 m/s from the top of a cliff as shown and lands at point C. The ball has a horizontal acceleration in the negative x-direction due to wind. Determine the radius of curvature of the path at B where its trajectory makes an angle of 45° with the horizontal. Neglect air resistance in the vertical direction. (Answer: = 41.8 m). (Problem 2/125, M&K)

x

A

B

C

50 m

40 m

20

Plane curvilinear motion

• Polar coordinates

rrer

rr eeee ,

x

y

r

eev

eerv

rr

rr

r

rr

21

Plane curvilinear motion

2

2

12 r

dt

d

rrra

rra

where

aa

r

rr

eea

22

Example

• An aircraft flies over an observer with a constant speed in a straight line as shown. Determine the signs (i.e. +ve, -ve, or 0) for

• for positions • A, B, and C.• (Problem 2/134, M&K)

x

y

ABC v

r

andrrr ,,,,,

23

Example

• At the bottom of a loop at point P as shown, an aircraft has a horizontal velocity of 600 km/h and no horizontal acceleration. The radius of curvature of the loop is 1200 m. For the radar tracking station shown, determine the recorded values of d2r/dt2 and d2/dt2 for this instant. (Answer: d2r/dt2 = 12.5 m/s2, d2/dt2 = 0.0365 rad/s2). (Problem 2/141, M&K)

P

r

600 km/h

1000 m

400 m

24

Relative motion

• Absolute (fixed axes)

• Relative (translating axes)

• Used when measurements are taken from a moving observation point, or where use of moving axes simplifies solution of problem.

• Motion of moving coordinate system may be specified w.r.t. fixed system.

25

Relative motion

• Set of translating axes (x-y) attached to particle B (arbitrarily). The position of A relative to the frame x-y (i.e. relative to B) is X

Y

O

B

A

rB

rA rA/B

y

x

jir yxBA /

26

Relative motion

• Absolute positions of points A and B (w.r.t. fixed axes X-Y) are related by

BAAB

ABAB

BABA

where

or

//

/

/

rr

rrr

rrr

27

Relative motion

• Differentiating w.r.t. time gives

• Coordinate systems may be rectangular, tangential and normal, polar, etc.

BABA

BABA

BABA

/

/

/

aaa

vvv

rrr

ABBA

ABBA

//

//

aa

vv

28

Inertial systems

• A translating reference system with no acceleration is known as an inertial system. If aB = 0 then

• Replacing a fixed reference system with an inertial system does not affect calculations (or measurements) of accelerations (or forces).

BAA /aa

29

Example

• A yacht moving in the direction shown is tacking windward against a north wind. The log registers a hull speed of 6.5 knots. A “telltale” (a string tied to the rigging) indicates that the direction of the apparent wind is 35° from the centerline of the boat. What is the true wind velocity? (Answer: vw = 14.40 knots). (Problem 2/191, M&K)

vw

50°

35°

30

Example

• To increase his speed, the water skier A cuts across the wake of the boat B which has a velocity of 60 km/h as shown. At the instant when = 30°, the actual path of the skier makes an angle b = 50° with the tow rope. For this position, determine the velocity vA of the skier and the value of d/dt. (Answer: vA = 80.8 km/h, d/dt = 0.887 rad/s). (Problem

2/193, M&K)

vB

A

b

B

31

Example• Car A is travelling at a constant

speed of 60 km/h as it rounds a circular curve of 300 m radius. At the instant shown it is at = 45°. Car B is passing the centre of the circle at the same instant. Car A is located relative to B using polar coordinates with the pole moving with B. For this instant, determine vA/B and the values fo d/dt and dr/dt as measured by an observer in car B. (Answer: vA/B = 36.0 m/s, d/dt = 0.1079 rad/s, dr/dt = −15.71 m/s). (Problem 2/201, M&K)

32

Space curvilinear motion

• Rectangular coordinates (x, y, z)

• Cylindrical coordinates (r, , z)

• Spherical coordinates (R, , )

• Coordinate transformations – not covered

• Tangential and normal system not used due to complexity involved.

33

Space curvilinear motion

• Rectangular coordinates (x, y, z) – similar to 2D

kjia

kjiv

kjiR

zyx

zyx

zyx

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Space curvilinear motion

• Cylindrical coordinates (r, , z)

keR zr r

keev zrr r

keea zrrrr r 22

z

y

x

z

r

R

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Space curvilinear motion

• Spherical coordinates (R, , )

cossin1

sin2cos

cos

cos

22

2

222

RRdt

d

Ra

RRdt

d

Ra

RRRa

where

aaa

RRR

R

RR

R

eeea

eeev z

y

x

R

R

36

Example

• A section of a roller-coaster is a horizontal cylindrical helix. The velocity of the cars as they pass point A is 15 m/s. The effective radius of the cylindrical helix is 5 m and the helix angle is 40°. The tangential acceleration at A is gcos. Compute the magnitude of the acceleration of the passengers as they pass A. (Answer: a = 27.5 m/s2). (Problem 2/171, M&K)

5 m

= 40°

A A

37

Example

• The robot shown rotates about a fixed vertical axis while its arm extends and elevates. At a given instant, = 30°, d/dt = 10 deg/s = constant, l = 0.5 m, dl/dt = 0.2 m/s, d2l/dt2 = −0.3 m/s2, and = 20 deg/s = constant. Determine the magnitudes of the velocity and acceleration of the gripped part P. (Answer: v = 0.480 m/s, a = 0.474 m/s2). (Problem 2/177, M&K)

P

Ox

y

38

Particle kinetics

• Newton’s laws

• Applied and reactive forces must be considered – free body diagrams

• Forces required to produce motion

• Motion due to forces

aF m

39

Particle kinetics

• Constrained and unconstrained motion

• Degrees of freedom

• Rectilinear motion – covered

• Curvilinear motion

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Rectilinear motion - example

• The 10 Mg truck hauls a 20 Mg trailer. If the unit starts from rest on a level road with a tractive force of 20 kN between the driving wheels and the road, compute the tension T in the horizontal drawbar and the acceleration a of the rig. (Answer: T = 13.33 kN, a = 0.667 m/s2). (Problem 3/5, M&K)

20 Mg 10 Mg

maF

41

Example

• The motorized drum turns at a constant speed causing the vertical cable to have a constant downwards velocity v. Determine the tension in the cable in terms of y. Neglect the diameter and mass of the small pulleys. (Problem 3/48, M&K)

3

2222

42 y

vbgyb

y

mT

• Answer:

m

vy

2b

42

Curvilinear motion

• Rectangular coordinates

• Normal and tangential coordinates

• Polar coordinates

yyxx maFmaF ,

ttnn maFmaF ,

maFmaF rr ,

43

Example

• A pilot flies an airplane at a constant speed of 600 km/h in a vertical circle of radius 1000 m. Calculate the force exerted by the seat on the 90 kg pilot at point A and at point A. (Answer: RA = 3380 N, RB = 1617 N). (Problem 3/63,

M&K)

600 km/h

44

Example

• The 30 Mg aircraft is climbing at an angle of 15° under a jet thrust T of 180 kN. At the instant shown, its speed is 300 km/h and is increasing at a rate of 1.96 m/s2. Also is decreasing as the aircraft begins to level off. If the radius of curvature at this instant is 20 km, compute the lift L and the drag D. (Lift and drag are the aerodynamic forces normal to and opposite to the flight direction, respectively). (Answer: D = 45.0 kN, L = 274 kN). (Problem 3/69, M&K)

T

45

Example

• A child's slide has a quarter circle shape as shown. Assuming that friction is negligible, determine the velocity of the child at the end of the slide ( = 90°) in terms of the radius of curvature r and the initial angle .

• Answer 0sin12 grv

46

Slide

• Does it matter what profile slide has?

• What if friction added?

47

Example• A flat circular discs rotates about a

vertical axis through the centre point at a slowly increasing angular velocity . With = 0, the position of the two 0.5 kg sliders is x = 25 mm. Each spring has a stiffness of 400 N/m. Determine the value of x for = 240 rev/min and the normal force exerted by the side of the slot on the block. Neglect any friction and the mass of the springs. (Answer: x = 118.8 mm, N = 25.3 N). (Problem 3/83, M&K)

xx

80 mm 80 mm

48

Work and energy

• Work/energy analysis – don’t need to calculate accelerations

• Work done by force F

• Integration of F = ma w.r.t. displacement gives equations for work and energy

rR+dr

O

A

A′

drF

a

r

rF

dds

where

dsFddU

acos

49

Work and energy

• Active forces and reactive forces (constraint forces that do no work)

• Total work done by force

• where Ft = tangential force component

dsFU

or

dzFdyFdxFdU

t

zyxrF

50

Work and energy

• If displacement is in same direction as force then work is +ve (otherwise –ve)

• Ignore reactive forces

• Kinetic energy

• Gravitational potential energy

221 mvT

mghVg

51

Example• A small vehicle enters the top of

a circular path with a horizontal velocity v0 and gathers speed as it moves down the path. Determine the angle b (in terms of v0) at which it leaves the path and becomes a projectile. Neglect friction and treat the vehicle as a particle. (Problem 3/87, M&K)

• Answer:

v0

b

gR

v

33

2cos

201b

52

Example• The small slider of mass m is

released from point A and slides without friction to point D. From point D onwards the coefficient of kinetic friction between the slider and the slide is k. Determine the distance s travelled by the slider up the incline beyond D. (Problem 3/125, M&K)

• Answer:

31

4

k

Rs

R

2R

30°

B

C

Am

D

53

Example• A rope of length r/2 and

mass per unit length is released with = 0 in a smooth vertical channel and falls through a hole in the supporting surface. Determine the velocity v of the chain as the last part of it leaves the slot. (Problem 3/173, M&K)

• Answer:

4

2grv

r

54

Linear impulse and momentum

• Integration of F = ma w.r.t. time gives equations of impulse and momentum.

• Useful where time over which force acts is very short (e.g. impact) or where force acts over specified length of time.

55

Linear impulse and momentum

• If mass m is constant then sum of forces = time rate of change of linear momentum

• Linear momentum of particle

• Units – kg·m/s or N·s

• Scalar form:

GvvF mdt

dm

vG m

zzyyxx GFGFGF ,,

56

Linear impulse and momentum

• Integrate over time

• Product of force and time is called linear impulse

• Scalar form

21

12

2

1

2

1

GFG

GGGF

t

t

t

t

dt

or

dt

.,12

2

1

etcmvmvdtF x

t

t xx

57

Linear impulse and momentum

• Note that all forces must be included (i.e. both active and reactive)

58

Linear impulse and momentum

• If there are no unbalanced forces acting on a system then the total linear momentum of the system will remain constant (principle of conservation of linear momentum)

m1v1

F

−F

m2v2

?

?

59

Impact

• How to determine velocities after impact?

m2

m1

(V1)n

(V2)n• Forces normal to contact surface. Fd is force during deformation period while Fr is force during recovery period.

• The ratio of the restoration impulse to the deformation impulse is called the coefficient of restitution

0

0

0

t

d

t

t r

dtF

dtFe

60

Impact

• For particle 1, (v0)n being the intermediate normal velocity component (of both particles) and (v1)′n being normal velocity component after collision

• Similarly for particle 2

nn

nnt

d

t

t r

vmvm

vmvm

dtF

dtFe

1101

0111

0

0

0

nn

nn

vmvm

vmvme

2202

0222

61

Impact

• Combining gives

• e = 0 for plastic impact, e = 1 for elastic impact

• Note that tangential velocities are not affected by impact

nn

nn

vv

vve

21

12

62

Example

• A 75 g projectile traveling at 600 m/ strikes and becomes embedded in the 50 kg block which is initially stationary. Compute the energy lost during the impact. Express your answer as an absolute value and as a percentage of the original energy of the system. (Problem 3/180, M&K)

600 m/s75 g

50 kg

63

Example

• The pool ball shown must be hit so as to travel into the side pocket as shown. Specify the location x of the cushion impact if e = 0.8. (Answer: x = 0.268d) (Problem 3/251, M&K)

xd

d/2

d/2

64

Example• The vertical motion of the 3 kg

load is controlled by the forces P applied to the end rollers of the framework shown. If the upward velocity of the cylinder is increased from 2 m/s to 4 m/s in 2 seconds, calculate the average force Rav under each of the two rollers during the 2 s interval. Neglect the small mass of the frame. (Answer: Rav = 16.22 N) (Problem 3/199, M&K)

v

PP

3 kg

65

Example• A 1000 kg spacecraft is

traveling in deep space with a speed vs = 2000 m/s when struck at its mass centre by a 10 kg meteor with velocity vm of magnitude 5000 m/s. The meteor becomes embedded in the satellite. Determine the final velocity of the spacecraft. (Answer: v = 36.9i + 1951j – 14.76k m/s) (Problem 3/201, M&K)

x

y

z

vm

5

4 2

vs

66

Cross (or vector) product

• Magnitude of cross-product

• Direction of cross-product governed by right-hand rule

sinPQQP Q

P

sinsin PQPQ PQ

67

Right-hand rule

• Middle finger in direction of R if thumb in direction of P and index finger in direction of Q.

• Use right-handed reference frame for x,y, and z.

RQP

68

Cross (or vector) product

• Distributive

P

QP×Q

Q×P=−P×Q

RPQPRQP

.,

.,

etc

etc

0ii

kji

69

Cross (or vector) product

• Derivative

kji

kji

QP

xyyxxzzxyzzy

zyx

zyx

QPQPQPQPQPQP

QQQ

PPP

QPQP

QP

dt

d

70

Angular impulse and momentum

• The angular momentum of a particle about any point is the moment of the linear momentum about that point.

• Units are kg·m/s·m or N·m·s

71

Angular impulse and momentum

• Planar motion

• There are 3 components of the angular momentum of P about arbitrary point O: i.e. about x-,y-, and z-axes.

P

Ox

y

r

mv

72

Angular impulse and momentum

• Since P is coplanar with x- and y-axes, it has no moment about these axes. It only has a moment about the z-axis.

P

Ox

y

r

mv

73

Angular impulse and momentum

• Is angular momentum of P about O positive or negative? – governed by right-hand rule

P

Ox

y

r

mv

74

Right-hand rule

• Curl fingers in. Rotation indicated by fingers is in direction of thumb. Is this positive or negative in this case?

75

Angular impulse and momentum

• Direction of component about z-axis is in z-direction

sin

2cos rmvrmvO

H

kH sinrmvO

P

Ox

y

r

mv

76

Angular impulse and momentum

P

Ox

y

r

mv

x

mvx

mvy

kH xyO mvymvx

77

Angular impulse and momentum

• Note

vrH mO

rvvr mm

., etcvzvymH yzx

78

Angular impulse and momentum

• The resultant moment of all forces about O is

• From Newton’s 2nd law

• Differentiate w.r.t. time

• Now• so

FrMO

vrM mO vrH mO

vrvvH mmO 0 vv m

OO HM

79

Angular impulse and momentum

• The moment of all forces on the particle about a fixed point O equals the time rate of change of the angular momentum about that point.

• If moment about O is zero then angular momentum is constant (principle of conservation of angular momentum).

• If moment about any axis is zero then component of angular momentum about that axis is constant.

., etcHM xOxO

80

Angular impulse and momentum

• Particle following circular path at constant angular velocity. Is angular momentum about O varying with time?

• Is angular momentum about O′ varying with time?

• Is component about z-axis varying with time?

m

OO′

x

y

z

81

Angular impulse and momentum

• i.e. change in angular momentum is equal to total angular impulse

12

2

1OO

t

t O

OO

dt HHM

HM

.,12

2

1

etcHHdtM xOxO

t

t xO

82

Angular impulse and momentum

• Example – ice skater

1 2

83

Example• Calculate HO, the angular

momentum of the particle shown about O (a) using the vector definition and (b) using a geometrical approach. The centre of the particle lies in the x-y plane. (Answer: HO = 128.7k N·m·s) (Problem 3/221, M&K) x

y

O

2 kg

6 m

8 m

7 m/s30°

84

Example• A particle of mass m moves with negligible

friction across a horizontal surface and is connected by a light spring fastened at point O. The velocity at A is as shown. Determine the velocity at B. (Problem 3/226, M&K)

A

B

O

vA = 4 m/s

vB

54°

65°

85

Example• Each of 4 spheres of mass m is treated

as a particle. Spheres A and B are mounted on a light rod and are rotating initially with an angular velocity 0 about a vertical axis through O. The other two spheres are similarly (but independently) mounted and have no initial velocity. When assembly AB reaches the position indicated it latches with CD and the two move with a common angular velocity . Neglect friction. Determine and n the percentage loss of kinetic energy. (Answer = 0/5, n = 80%). (Problem 3/227,

M&K)

2r

2r

r r

AB

D

C

O

0

86

Example• The particle of mass m is launched from point O with a

horizontal velocity u at time t = 0. Determine its angular momentum about O as a function of t. (Answer H0 = −½mgut2k). (Problem 3/233, M&K)

m

Ou

x

y

87

Relative motion

• Fixed reference frame X-Y

• Moving reference frame x-y

Y

XO

B

y

x

A

rA

rB

rA/B = rrel

rel

relBA

relBA

m

mm

aF

aaaF

aaa

88

Relative motion

• Special case – inertial system or “Newtonian frame of reference” with zero acceleration

• Note that work-energy and impulse momentum equations are equally valid in inertial system – but relative momentum/relative energy etc. will, in general, be different to those measured relative to fixed frame of reference.

89

Example• The ball A of mass 10 kg is attached to the light rod of length

l = 0.8 m. The rod is attached to a carriage of mass 250 kg which moves on rails with an acceleration aO as shown. The rod is free to rotate horizontally about O. If d/dt = 3 rad/s when = 90°, find the kinetic energy T of the system if the carriage has a velocity of 0.8 m/s. Treat the ball as a particle. (Answer: T = 112 J). (Problem 3/311, M&K)

aOO

Al

90

Example• The small slider A moves with negligible friction down the

tapered block, which moves to the right with constant speed v = v0. Use the principle of work-energy to determine the magnitude vA of the absolute velocity of the slider as it passes point C if it is released at point B with no velocity relative to the block. (Problem 3/316, M&K)

• Answer:

v

B

C

A

sin2cos2sin2 02

0 glvglvvA