2:1

MM203Mechanics of Machines: Part 2

2:2

Kinetics of systems of particles

• Extension of basic principles to general systems of particles– Particles with light links– Rigid bodies– Rigid bodies with flexible links– Non-rigid bodies– Masses of fluid

2:3

Newton’s second law

• G – centre of mass

• Fi – external force, fi – internal force

• i – position of mi relative to G

O

Gri

_r

imi

mi

F1

F2

F3

f1

f2

f3

2:4

Newton’s second law

• By definition

• For particle i

• Adding equations for all particles

ii

i

ii

mm

m

m

rr

rr

iim rfffFFF 321321

iim rfF

2:5

Newton’s second law

• Differentiating w.r.t. time

• gives

• Also

• so

• (principle of motion of the mass centre)

iimm rr

iimm rr

0 f

arF mm

2:6

Newton’s second law

• Note that is the acceleration of the instantaneous mass centre – which may vary over time if body not rigid.

• Note that the sum of forces is in the same direction as the acceleration of the mass centre but does not necessarily pass through the mass centre

., etcamF xx

2:7

Example

• Three people (A, 60 kg, B, 90 kg, and C, 80 kg) are in a boat which glides through the water with negligible resistance with a speed of 1 knot. If the people change position as shown in the second figure, find the position of the boat relative to where it would be if they had not moved. Does the sequence or timing of the change in positions affect the final result? (Answer: x = 0.0947 m). (Problem 4/15, M&K)

B C A

CBA

1 knot

0.6 m 1.8 m 2.4 m

x 1.8 m1.2 m 1.2 m

2:8

Example• The 1650 kg car has its mass centre at G. Calculate the

normal forces at A and B between the road and the front and rear pairs of wheels under the conditions of maximum acceleration. The mass of the wheels is small compared with the total mass of the car. The coefficient of static friction between the road and the rear driving wheels is 0.8. (Answer: NA = 6.85 kN, NB = 9.34 kN). (Problem 6/5, M&K)

2:9

Work-energy

• Work-energy relationship for mass i is

• where (U1-2)i is the work done on mi during a period of motion by the external and internal forces acting on it.

• Kinetic energy of mass i is

ii TU 21

2

212

21

iiiii mvmT r

2:10

Work-energy

• For entire system

2211

21

21

TUT

or

TU

or

TU ii

2:11

Work-energy

• Note that no net work is done by internal forces.

• If changes in potential energy possible (gravitational and elastic) then

• as for single particle

22221111 egeg VVTUVVT

2:12

Work-energy

• For system

• Now

• and note that

• so

2

21

iivmT

ii ρvv

iiiv vv 2

iiiii

iii

mmvm

mT

ρvρ

ρvρv

2

212

21

21

2:13

Work-energy

• Since i is measured from G,

• Now

0 iim ρ

0

ii

iiii

mdt

d

mm

ρv

ρvρv

2:14

Work-energy

• Therefore

• i.e. energy is that of translation of mass-centre and that of translation of particles relative to mass-centre

2

212

21

iii mvmT ρ

2:15

Example

• The two small spheres, each of mass m, are rigidly connected by a rod of negligible mass and are released in the position shown and slide down the smooth circular guide in the vertical plane. Determine their common velocity v as they reach the horizontal dashed position. Also find the force R between sphere 1 and the guide the instant before the sphere reaches position A. (Answer: v = 1.137(gr)½, R = 2.29mg). (Problem 4/9, M&K)

m

m

y

x45°

A

1

2

12

2:16

Rigid body

• Motion of particles relative to mass-centre can only be due to rotation of body

• Velocity of particles due to rotation depends on angular velocity and the distance to centre of rotation. Where is centre of rotation?

• Need to examine kinematics of rotation

2

212

21

iii mvmT ρ

2

212

21

iimvmT ρ

2:17

Plane kinematics of rigid bodies

• Rigid body– distances between points remains unchanged– position vectors, as measured relative to coordinate

system fixed to body, remain constant

• Plane motion– motion of all points is on parallel planes– Plane of motion taken as plane containing mass centre– Body treated as thin slab in plane of motion – all points on

body projected onto plane

2:18

Kinematics of rigid bodies

2:19

Translation

• All points move in parallel lines or along congruent curves.

• Motion is completely specified by motion of any point – therefore can be treated as particle

• Analysis as developed for particle motion

2:20

Kinematics of rigid bodies

2:21

Rotation about fixed axis

• All particles move in circular paths about axis of rotation

• All lines on body (in plane of motion) rotate through the same angle in the same time

• Similar to circular motion of a particle

• where riO is distance to O, the centre of rotation, and IO is mass moment of inertia about O

OiOiiiii IrmrmmT 22122

2122

212

21 ρ

2:22

Mass moment of inertia

• Mass moment of inertia in rotation is equivalent to mass in translation

• Rotation and translation are analogous

OIT 221

2:23

General plane motion

• Combination of translation and rotation

• Principles of relative motion used

2:24

Rotation

• Angular positions of two lines on body are measured from any fixed reference direction

12

12

12

12

2:25

Rotation

• All lines on a rigid body in its plane of motion have the same angular displacement, the same angular velocity, and the same angular acceleration

• Angular motion does not require the presence of a fixed axis about which the body rotates

2:26

Angular motion relations

• Angular position, angular velocity, and angular acceleration

• Similar to relationships between s, v, and a.

• Also, combining relationships and cancelling out dt

,

dd

2:27

Angular motion relations• If constant angular acceleration

• Direction of +ve sense must be consistent

• Analogous to rectilinear motion with constant a

• Same procedures used in analysis

2

21

00

02

02

0

2

tt

t

2:28

Example• The angular velocity of a gear is controlled according to =

12 – 3t2 where , in radians per second, is positive in the clockwise sense and where t is the time in seconds. Find the net angular displacement from the time t = 0 to t = 3 s. Also find the total number of revolutions N through which the gear turns during the 3 seconds. (Answer: = 9 rad, N = 3.66 rev). (Problem 5/5, M&K)

2:29

Kinetic energy of rigid body

Ivmrmvm

mvmT

iGi

ii

2212

2122

212

21

2

212

21

ρ

IvmT 2212

21

• If rotation about O

IOGmT 22122

21

O

G

OG

2:30

Parallel axis theorem

• Now (P.A.T.)

• and so

2OGmIIO

OIT 221

2:31

Radius of gyration

• Radius of gyration

• Mass moment of inertia of point mass m at radius of gyration is the same as that for body

• P.A.T.

mkIorm

Ik 2

222 OGkk GO

2:32

Work done on rigid body

GG

F

dr

rF dU

2:33

Work done by couple

• Couple is system of forces that causes rotation but no translation

• Moment about G

• Moment about O

221121 rFrF FrFrM

F

FG

r1

r2

O

d

r1

2121 rrFrdFdrFM

2:34

Work done by couple

• Moment vector is a free vector

• Forces have turning effect or torque

• Torque is force by perpendicular distance between forces

• Work done

• positive or negative

MdU

2:35

Forces and couples

• Torque is

• Also unbalanced force

F1

G

r1

r2F2

2211 rFrF

21 FFF

2:36

Work-energy principle

• When applied to system of connected bodies only consider forces/moments of system – ignore internal forces/moments.

• If there is significant friction between components then system must be dismembered

2:37

Example• A steady 22 N force is applied

normal to the handle of the hand-operated grinder. The gear inside the housing with its shaft and attached handle have a combined mass of 1.8 kg and a radius of gyration about their axis of 72 mm. The grinding wheel with its attached shaft and pinion (inside housing) have a combined mass of 0.55 kg and a radius of gyration of 54 mm. If the gear ratio between gear and pinion is 4:1, calculate the speed of the grinding wheel after 6 complete revolutions of the handle starting from rest. (Answer: N = 3320 rev/min). (Problem 6/119, M&K)

2:38

Rotation about fixed axis

• Motion of point on rigid body

rar

vvra

rv

t

n

2

2

2:39

Vector notation• Angular velocity

vector, , for body has sense governed by right-hand rule

• free vector

2:40

Vector notation

• Velocity vector of point A

• What are magnitude and direction of this vector?

• Note that

rωrv

vωr

2:41

Vector notation

• Acceleration of point

rαvω

rωrωω

rωrωva

2:42

Vector notation

• Vector equivalents

• Can be applied in 3D except then angular velocity can change direction and magnitude

rαa

rωωa

rωv

t

n

2:43

Example• The T-shaped body rotates about

a horizontal axis through O. At the instant represented, its angular velocity is = 3 rad/s and its angular acceleration is = 14 rad/s2. Determine the velocity and acceleration of (a) point A and (b) point B. Express your results in terms of components along the n- and t- axes shown. (Answer: vA = 1.2et m/s, aA = −5.6et + 3.6en m/s2, vB = 1.2et + 0.3en m/s, aB = −6.5et + 2.2en m/s2). (Problem 5/2, M&K)

2:44

Example• The two V-belt pulleys form an

integral unit and rotate about the fixed axis at O. At a certain instant, point A on the belt of the smaller pulley has a velocity vA = 1.5 m/s, and the point B on the belt of the larger pulley has an acceleration aB = 45 m/s2 as shown. For this instant determine the magnitude of the acceleration aC of the point C and sketch the vector in your solution. (Answer: aC = 149.6 m/s2). (Problem 5/16, M&K)

2:45

Linear impulse and momentum

• Returning to general system: Linear momentum of mass i is

• For system (assuming m does not change with time)

iii m vG

0iiii

iiii

mmdt

dm

mm

vρv

ρvvG

vG m

2:46

Linear impulse and momentum

• Differentiating w.r.t. time

• Same as for single particle – only applies if mass constant

• Same for rigid body

FvG m

2:47

Example

• The 300 kg and 400 kg mine cars are rolling in opposite directions along a horizontal track with the speeds shown. Upon impact the cars become coupled together. Just prior to impact, a 100 kg boulder leaves the delivery chute and lands in the 300 kg car. Calculate the velocity v of the system after the boulder has come to a rest relative to the car. Would the final velocity be the same if the cars were coupled before the boulder dropped? (Answer: v = 0.205 m/s). (Problem 4/11, M&K)

30°

1.2 m/s

0.6 m/s 0.3 m/s

400 kg300 kg

100 kg

2:48

Angular impulse and momentum

• Considered about a fixed point O and about the mass centre.

O (fixed)

G

ri_r

imi

2:49

Angular impulse and momentum• About O

• First term is zero since vi × vi =0 so

• - sum of all external moments (net moment of internal forces is zero)

iiiiiiO

iiiO

mm

m

vrvrH

vrH

iiiiiO m FrarH

OO MH OO MH

2:50

Angular impulse and momentum

• About O: same as for single particle. As before, does not apply if mass is changing.

• About G

GG MH

GG MH

2:51

Example

• The two balls are attached to a light rod which is suspended by a cord from the support above it. If the balls and rod, initially at rest, are struck by a force F = 60 N, calculate the corresponding acceleration aˉ of the mass centre and the rate d2/dt2 at which the angular velocity of the bar is changing. (Answer: aˉ = 20 m/s2, d2/dt2 = 336 rad/s2). (Problem 4/17, M&K)

F

2 kg

1 kg

175 mm

75 mm

150 mm

250 mm

2:52

Rigid body

• Angular momentum

ωvrH OiiiO Im

ωH IG

vrωH mI OGO /

OO IH

IHG

OGvmIHO

• Planar motion

2:53

Rigid body

• Angular acceleration

αMH OOO I

αMH IGG

vrαH mdt

dI OGO /

OO IH

IHG

OGamIHO

2:54

Kinetic diagrams

2:55

Kinetic diagrams - translation

• Alternative moment equation for rectilinear translation

damMM PA ,0

2:56

Kinetic diagrams - translation

• Alternative moment equation for curvilinear translation

AtBA damMM ,0

2:57

Example• The cart B moves to the right with

acceleration a = 2g. If the steady-state angular deflection of the uniform slender rod of mass 3 m is observed to be 20°, determine the value of the torsional spring constant K. The spring, which exerts a moment M = K on the rod, is undeformed when the rod is vertical. The values of m and l are 0.5 kg and 0.6 m, respectively. Treat the small end sphere of mass m as a particle. (Answer: K = 46.8 N·m/rad). (Problem 6/16, M&K)

2:58

Example• The mass of gear A is 20 kg

and its centroidal radius of gyration is 150 mm. The mass of gear B is 10 kg and its centroidal radius of gyration is 100 mm. Calculate the angular acceleration of gear B when a torque of 12 N·m is applied to the shaft of gear A. Neglect friction. (Answer: B = 25.5 rad/s2 (CCW)). (Problem 6/46, M&K)

2:59

Example• The 28 g bullet has a

horizontal velocity of 500 m/s when it strikes the 25 kg compound pendulum, which has a radius of gyration of kO = 925 mm. If the distance h = 1075 mm, calculate the angular velocity of the pendulum with its embedded bullet immediately after the impact. (Answer: = 0.684 rad/s) (Problem 6/174, M&K)

2:60

Plane kinematics: absolute motion

• Absolute motion analysis– Get geometric relationships– Get time derivatives to determine velocity and

acceleration– Straightforward if geometry is straightforward– Must be consistent with signs

2:61

Example• Point A is given a constant acceleration a to the right starting from rest

with x essentially 0. Determine the angular velocity of link AB in terms of x and a. (Problem 5/24, M&K)

• Answer:

224

2

xb

ax

2:62

Example• The wheel of radius r rolls without slipping, and its centre O has a

constant velocity vO to the right. Determine expressions for the velocity v and acceleration of point A on the rim by differentiating its x- and y-coordinates. Represent your result graphically as vectors on your sketch and show that v is the vector sum of two vO vectors. (Problem 5/25, M&K)

• Answer:

Otowardsr

va

vv

O

O

2

,sin12

2:63

Example• One of the most common mechanisms is the slider-crank. Express the

angular velocity AB and the angular acceleration AB of the connecting rod AB in terms of the crank angle for a given constant crank speed 0. Take AB and AB to be positive counter-clockwise. (Problem 5/54, M&K)

• Answer:

23

22

2

2

2

20

22

2

0

sin1

1sin

sin1

cos

lr

lr

l

r

lrl

r

AB

AB

2:64

Plane kinematics: relative velocity

• Two points on same rigid body.• Motion of one relative to the other must be circular

since distance between them is constant.

2:65

Plane kinematics: relative velocity

BABA /vvv

BABA // rωv

ABv BA /

• Relative linear velocity is always in direction perpendicular to line joining points

2:66

Plane kinematics: relative velocity

2:67

Plane kinematics: relative velocity

• Relative velocity principles may also be used in cases where there is constrained sliding contact between two links – A and B may be on different links

2:68

Example

• Determine the angular velocity of the telescoping link AB for the position shown where the driving links have the angular velocities indicated. (Answer: AB = 0.96 rad/s). (Problem 5/61, M&K)

2:69

Example

• For an interval of its motion the piston rod of the hydraulic cylinder has a velocity VA = 1.2 m/s as shown. At a certain instant = = 60°. For this instant determine the angular velocity BC of the link BC. (Answer: BC = 15.56 rad/s). (Problem 5/66, M&K)

2:70

Example

• The mechanism is designed to convert from one rotation to another. Rotation of link BC is controlled by the rotation of the curved slotted arm OA which engages pin P. For the instant represented = 30° and , the angle between the tangent to the curve at P and the horizontal, is 40°. If the angular velocity of OA is as shown, determine the velocity of the point C. (Answer: vC = 4.33 m/s). (Problem 5/85, M&K)

2:71

Relative acceleration

tBAnBABBABA /// aaaaaa

ABva

ABAB

va

BAtBA

BAnBA

//

22

//

• May need to know velocities first

2:72

Example

• If OA has a CCW angular velocity 0 = 10 rad/s (giving BC = 5.83 rad/s and AB = 2.5 rad/s), calculate the angular acceleration of link AB for the position where the coordinates of A are x = −60 mm and y = 80 mm. Link BC is vertical for this position. (Answer: AB = 2.5 rad/s2). (Problem 5/137, M&K)