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48640

Machine Dynamics

Dr Dongbin Wei

School of Electrical, Mechanical and Mechatronic Systems

Faculty of Engineering and Information Technology

University of Technology, Sydney

Office: Building 2, Level 6, Room 619

Email: dongbin.wei@uts.edu.au

WEEK 1

Where are we?

Statics: The study of bodies in equilibrium.

Dynamics: 1. Kinematics – concerned with the

geometric aspects of motion 2. Kinetics – concerned with the

forces causing the motion

Mechanics: The study of how bodies react to forces acting on them.

Particle Rigid Body

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Recommended Text Book:

R.C. Hibbeler, Mechanics for

Engineers: Dynamics, SI Edition,

13/E, Prentice Hall

or

R.C. Hibbeler, Engineering

Mechanics: Dynamics in SI Units

Pack, 12 Edition, Prentice Hall

3

Reference:

J.L. Meriam and L.G. Kraige, Engineering Mechanics –

Dynamics in SI Version (7th Edition), Wiley

Planar Kinematics of Rigid Body (Chapter 16) Displacement, Velocity, Acceleration

Planar Kinetics of Rigid Body Kinetic Equation of Motion (Chapter 17) Acceleration, Force Work and Energy (Chapter 18) Force, Velocity, Displacement Impulse and Momentum (Chapter 19) Force, Velocity, Time

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Step 1: Define the problem - Specify required outputs

Step 2: Accumulate data – Problem inputs

• Free body diagram (very important and useful)

- establishing appropriate coordinate system

• Kinematic or kinetics diagram) if necessary

• Initial conditions

• List and express all known parameters

• List and express all unknown parameters

Step 3: Select appropriate theory / principle

• Choose useful equations

• Recognize the conditions and assumptions

Step 4: Estimate the answer bounds

Step 5: Solve the problem

• use SI units (N, mm, MPa and kN, m, kPa)

Step 6: Verify and check results

Procedure for solving problems

Objectives: Students will be able to: • Analyze the kinematics of a rigid body undergoing planar

translation or rotation about a fixed axis.

Contents:

• Types of Rigid-Body Motion

• Planar Translation

• Rotation about a Fixed Axis

PLANAR RIGID BODY MOTION: TRANSLATION & ROTATION

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16.1 Planar Rigid-Body Motion

• When all the particles of a rigid body move along paths which are equidistant from a fixed plane, the body is undergoing planar motion

• Three types of rigid body planar motion:

1) Translation

2) Rotation about a fixed axis

3) General Plane Motion

1) Translation

When every line segment on the body remains parallel to its original direction during the motion

• Rectilinear translation - when two particles paths of motion are along equidistant straight lines

• Curvilinear translation - when paths of motion are along curves lines are equidistant

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2) Rotation about a fixed axis

• When a rigid body rotates about a fixed axis, all the particles of the body, except on the axis of rotation, move along circular paths

3) General Plane Motion

• When a body is subjected to general plane motion, there is a combination of translation and rotation

An example of bodies undergoing the three types of motion

rectilinear translation.

rotation about a fixed axis.

curvilinear translation, since it will remain horizontal as it moves along a circular path.

2

The wheel and crank undergo

The piston undergoes

The connecting rod 1 undergoes The connecting rod 2 undergoes general plane motion, as it will both translate and rotate.

v v

v v

1

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Passengers on this amusement ride are subjected to

Does each passenger feel the same acceleration?

If the angular motion of the rotating arms is known, how can we determine the velocity and acceleration experienced by the passengers? Why would we want to know these values?

Applications

curvilinear translation since the vehicle moves in a circular path but they always remains upright.

Gears, pulleys and cams, which rotate about fixed axes, are often used in machinery to generate motion and transmit forces. The angular motion of these components must be understood to properly design the system.

To do this design, we need to relate the angular motions of contacting bodies that rotate about different fixed axes. How is this different from the analyses we did in dynamics of a particle?

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16.2 Planar Translation Position

By vector addition

Velocity

Magnitude of 𝑟 𝐵/𝐴 is constant due

to rigid body

Acceleration

Time derivative of velocity equation yields

ABAB /rrr

AB vv

AB aa

x

z

y

O

𝒋

𝒊 𝒌

𝑖 , 𝑗 and 𝑘 represent the unit vectors along x, y and z respectively

16.3 Rotation about a Fixed Axis

Angular Motion

Only lines or bodies undergo angular motion. (A point is without dimension and cannot have angular motion.)

Angular Position

Angular position of r is defined by 𝜃

Angular Displacement (vector)

Change in the angular position measured

as a differential 𝑑𝜽

• Direction: right hand rule

• Units: degrees, radians, revolutions

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Angular Velocity (vector)

Time rate of change in the angular position

• 𝑑𝜽 occurs during an instant of time dt so its magnitude ω

Angular Acceleration (vector)

Time rate of change of the angular velocity

• Magnitude α

dt

d ( +)

( +) 𝜶 =𝒅𝝎

𝒅𝒕=𝒅𝟐𝜽

𝒅𝒕𝟐

Angular Acceleration

• Differential relation between the angular acceleration, angular velocity and angular displacement is

• Only if Angular Acceleration is a constant 𝜶𝒄

dd ( +)

)(2

2

1

020

2

200

0

c

c

c

tt

t

( +)

( +)

( +)

𝜃0, ω0 represent the initial position and velocity of the body at t = 0.

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KINEMATIC RELATIONS: RECTILINEAR MOTION of A particle

• Differentiate position to get velocity and acceleration.

v = ds/dt ; a = dv/dt or a = v dv/ds

• Integrate acceleration for velocity and position.

so and vo represent the initial position and velocity of the particle at t = 0.

Velocity:

t

o

v

v o

dt a dv s

s

v

v o o

ds a dv v or t

o

s

s o

dt v ds

Position:

Recall

• In the case of CONSTANT ACCELERATION

More generally, using Vector Calculation to determine magnitude and direction of velocity of P

Motion of Point P on Rigid Body Rotating about A Fixed Axis

Point P travels along a circular path of radius r and centre at point O

• Position

P is defined by the position vector 𝒓

• Velocity

– Magnitude: 𝑣 = 𝜔𝑟

– Direction: tangent to circular path (it may be determined by observation in simple cases)

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𝒗 = 𝝎 × 𝒓𝑷

Right hand rule: from direction of 𝝎 𝐭𝐨 d𝐢𝐫𝐞𝐜𝐭𝐢𝐨𝐧 𝐨𝐟 𝒓𝑷 - Thumb tangents to the path in the direction of motion

Vector Calculation - the cross product of vector to determine magnitude and direction of velocity of P

• Acceleration

– Acceleration of P

𝒂 =𝒅𝒗

𝒅𝒕=𝒅(𝝎 × 𝒓𝒑)

𝒅𝒕

=𝒅𝝎

𝒅𝒕× 𝒓𝒑 +𝝎×

𝒅𝒓𝒑

𝒅𝒕

= 𝜶 × 𝒓𝒑 +𝝎× 𝝎× 𝒓𝒑

= 𝜶 × 𝒓𝒑 +𝝎× 𝝎× 𝒓𝒑

= 𝜶 × 𝒓 − 𝝎𝟐𝒓

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• Acceleration

– Acceleration of point P can be expressed conveniently by Normal and Tangential Components

– Vector form

– Magnitude of acceleration is

Change direction of velocity

Change magnitude of velocity

rat

ran

2

22nt aaa

Magnitude of at and an

𝒂 = 𝒂𝒕 + 𝒂𝒏 = 𝜶 × 𝒓 − 𝝎𝟐𝒓

12.7 Curvilinear Motion: Normal and Tangential Components

Acceleration

• Acceleration of the particle a can be written as

• Magnitude of acceleration of the particle a is

2

and or wherev

avdvdsava

uauaa

ntt

nntt

22nt aaa

Recall

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Procedure for Analysis

Angular Motion

• Establish the positive sense of direction along the axis of rotation

• When relationship is known between any two of the four variables α, ω, 𝜃 and t, we can use

dt

d

dt

d

dd

Procedure for Analysis

Angular Motion

• When body’s angular acceleration is constant 𝛼𝑐

• Sense of α, ω and 𝜃 is determined from algebraic signs of their numerical quantities

)(2

2

1

020

2

200

0

c

c

c

tt

t

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Procedure for Analysis

Motion of Point P

• Velocity of P and its two components of acceleration can be determined from the scalar equations

• If geometry of the problem is difficult to visualize, we use vector equations

rat rv ran

2

rra

rra

rrv

2)(

Pn

Pt

P Using right hand rule to determine the direction of

𝑣 , 𝑎𝑡 and 𝑎𝑛

READING QUIZ

1. If a rigid body is in translation only, the velocity at points A and B on the rigid body _______ .

A) are usually different B) are always the same C) depend on their position D) depend on their relative position

2. If a rigid body is rotating with a constant angular velocity about a fixed axis, the velocity vector at point P is _______.

A) rp B) rp

C) drp/dt D) All of the above.

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CONCEPT QUIZ

2. A Frisbee is thrown and curves to the right. It is experiencing

A) rectilinear translation. B) curvilinear translation.

C) pure rotation. D) general plane motion.

1. A disk is rotating at 4 rad/s. If it is subjected to a constant angular acceleration of 2 rad/s2, determine the acceleration at B.

A) (4 i + 32 j) m/s2 B) (4 i - 32 j) m/s2

C) (- 4 i + 32 j) m/s2 D) (- 4 i -32 j) m/s2

O A

B

x

y

2 m

2 rad/s2

A cord is wrapped around a wheel which is initially at rest. If a force is applied to the cord and gives it an acceleration a = (4t) m/s2, where t is in seconds, determine as a function of time (a) the angular velocity of the wheel, and (b) the angular position of the line OP in radians.

Example I (Example 16.1)

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Solution Part (a) The wheel is subjected to rotation about a fixed axis passing through point O.

Acceleration of Point P has both tangential and normal components.

Angular acceleration of the wheel is

2rad/s20

)2.0()4(

)(

t

t

ra tP

+

Solution Part (a) Integrating with the initial condition that

ω = 0, t = 0

+

rad/s10

20

)20(

2

00

t

dttd

tdt

d

t

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Solution Part (b) Angular position 𝜃 of OP can be found from ω = 𝑑𝜃/dt

Integrating with the initial condition

𝜃 = 0 at t = 0,

rad33.3

10

)10(

3

0

2

0

2

t

dttd

tdt

d

t

+

Given: The motor turns gear A with a constant angular acceleration, αA = 4.5 rad/s2, starting from rest. The cord is wrapped around pulley D which is rigidly attached to gear B.

Find: The velocity of cylinder C and the distance it travels in 3 seconds.

Example II

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1) The angular acceleration of

gear B (and pulley D) can be related to αA .

2) The acceleration of cylinder C can be determined by using the equations of motion for a point on a rotating body since (at)D at point P is the same as ac.

3) The velocity and distance of C can be found by using the constant acceleration equations.

Plan:

Solution:

Since gear B and pulley D turn together, αD = αB = 1.5 rad/s2

1) Gear A and B will have the same speed and tangential component of acceleration at the point where they mesh. Thus,

at = αArA = αBrB (4.5)(75) = αB (225) αB = 1.5 rad/s2

2) Assuming the cord attached to pulley D is inextensible and does not slip, the velocity and acceleration of cylinder C will be the same as the velocity and tangential component of acceleration along the pulley D.

aC = (at)D = aD rD = (1.5)(0.125) = 0.1875 m/s2

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3) Since aC will be constant. The constant acceleration equation for rectilinear motion can be used to determine the velocity and displacement of cylinder C when t = 3 s (s0= v0 = 0):

vc = v0 + aC t = 0 + 0.1875 m/s2(3 s) = 0.563 m/s

sc = s0 + v0 t + (0.5) aC t2 = 0 + 0 + (0.5) 0.1875 m/s2 (3 s)2 = 0.844 m

GROUP PROBLEM SOLVING Given: Gear A is given an angular acceleration αA= 4t3 rad/s2, where t is in seconds, and (A)0= 20 rad/s.

Find: The angular velocity and angular displacement of gear B when t = 2 s.

1) Apply the kinematic equation of variable angular acceleration to find the angular velocity of gear A.

2) Find the relationship of angular motion between gear A and gear B in terms of time and then use 2 s.

Plan:

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Solution:

1) Motion of Gear A : Applying the kinematic equation

2) Since gear B meshes with gear A,

B = 36 (0.05 / 0.15) = 12 rad/s

B = 46.4 (0.05 / 0.15) = 15.5 rad

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ATTENTION QUIZ

1. The fan blades suddenly experience an angular acceleration of 2 rad/s2. If the blades are rotating with an initial angular velocity of 4 rad/s, determine the speed of point P when the blades have turned 2 revolutions (when ω = 8.14 rad/s).

A) 4.27 m/s B) 5.31 m/s

C) 6.93 m/s D) 8.0 m/s

2. Determine the magnitude of the acceleration at P when the blades have turned the 2 revolutions.

A) 0 m/s2 B) 1.05 m/s2

C) 34.79 m/s2 D) 34.80 m/s2

Three various calculation methods can be adopted to determine the velocity and/or acceleration of a rigid body undergoing general plane motion.

• Absolute Motion Analysis for determining both velocity and acceleration

• Relative Motion Analysis Velocity Acceleration

• Instantaneous Centre (IC) of zero velocity for determining velocity

Two or more methods may be applicable for a certain problem, however, they have various difficulties.

PLANAR RIGID BODY MOTION: GENERAL PLANE MOTION

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ABSOLUTE MOTION ANALYSIS

Objective: Students will be able to: 1. Determine the velocity and acceleration of a rigid body

undergoing general plane motion using an absolute motion analysis.

Contents:

• General Plane Motion

APPLICATIONS

As a part of the design process for the truck, an engineer had to relate the velocity at which the hydraulic cylinder extends and the resulting angular velocity of the bin.

The dumping bin on the truck rotates about a fixed axis passing through the pin at A. It is operated by the extension of the hydraulic cylinder BC. The angular position of the bin can be specified using the angular position coordinate and the position of point C on the bin is specified using the coordinate s.

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The large window is opened using a hydraulic cylinder AB. The position B of the hydraulic cylinder rod is related to the angular position 𝜃 , of the window.

A designer has to relate the translational velocity at B of the hydraulic cylinder and the angular velocity and acceleration of the window? How would you go about the task?

16.4 Absolute Motion Analysis

• A body subjected to general plane motion undergoes a simultaneous translation and rotation.

• One way to define these motions is to use a rectilinear position coordinate s to locate the point along its path and an angular position coordinate 𝜃 to specify the orientation of the line.

𝑠 = 𝑓(𝜃)

• By direct application of the time-differential equations

v = ds/dt ω = 𝑑𝜃/𝑑𝑡

a = dv/dt 𝛼 = dω/dt

the motion of the point and the angular motion of the line can be related.

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The position of the piston, s, can be defined as a function of the angular position of the crank, . By differentiating s with respect to time, the velocity of the piston can be related to the angular velocity, , of the crank. This is necessary when designing an engine.

The stroke of the piston is defined as the total distance moved by the piston as the crank angle varies from 0 to 180°. How does the length of crank AB affect the stroke?

s

Procedure for Analysis

Position Coordinate Equation

• Locate point P using a position coordinate s, measured from a fixed origin and along the straight-line path of motion of point P.

• Measure from a fixed reference line the angular position 𝜃 of a line lying in the body.

s = f(𝜽)

Time Derivatives

• First time derivative of s = f(𝜃) to get a relationship between v and ω; second time derivative to get a relationship between a and α.

• Chain rule of calculus must be used when taking the derivatives of the position coordinate equation.

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READING QUIZ 1. A body subjected to general plane motion undergoes

a/an

A) translation.

B) rotation.

C) simultaneous translation and rotation.

D) out-of-plane movement.

2. In general plane motion, if the rigid body is represented by a slab, the slab rotates

A) about an axis perpendicular to the plane.

B) about an axis parallel to the plane.

C) about an axis lying in the plane.

D) None of the above.

CONCEPT QUIZ

1. The position, s, is given as a function of angular position, , as s = 10 sin 2 . The velocity, v, is

A) 20 cos 2 B) 20 sin 2

C) 20 cos 2 D) 20 sin 2

2. If s = 10 sin 2, the acceleration, a, is A) 20 sin 2 B) 20 cos 2 − 40 2 sin 2

C) 20 cos 2 D) -40 sin2

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The end of rod R maintains contact with the cam by means of a spring. If the cam rotates about an axis through point O with an angular acceleration α and angular velocity ω, determine the velocity and acceleration of the rod when the cam is in the arbitrary position 𝜃.

Example I (Example 16.3)

Solution Position Coordinate Equation

𝜃 and x are chosen to relate the rotational motion of OA to the rectilinear motion of the rod.

Since

Time Derivatives Using chain rule of calculus

cos2rx

)cossin(2)(cos2sin2

sin2)(sin2

2

radt

dr

dt

dr

dt

dv

rvdt

dr

dt

dx

cosrCBOC Fixed origin

i.e. s

R: Focused point

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The large window is opened using a hydraulic cylinder AB. If the cylinder extends at a constant rate of 0.5 m/s, determine the angular velocity and angular acceleration of the window

at the instant 𝜃 = 30°

Example II (Example 16.5)

Solution Position Coordinate Equation The angular motion of the window can be obtained using the coordinate 𝜃 and motion along the hydraulic cylinder is defined using a coordinate s.

When θ = 30°,

cos45

cos)1)(2(2)1()2(

2

222

s

s(1)

m239.1s

Fixed origin

s

B: Focused point

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Solution Time Derivatives Taking the time derivatives of Eq. 1,

Since vs = 0.5 m/s, then at θ = 30°

)(sin2)(

)sin(402

svs

dt

d

st

dss

(2)

rad/s620.0

Solution Time Derivatives Taking the time derivatives of Eq. 2 yields,

Since as = dvs/dt = 0, then

)(sin2)(cos2

)(sin2)(cos2

22

ss

ss

sav

dt

d

dt

d

dt

dvsv

dt

ds

2

22

rad/s415.0

30sin2)620.0(30cos20)5.0(

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Given: A circular cam is rotating clockwise about O with a constant 𝜔.

Find: The velocity and acceleration of the follower rod A as a function of 𝜃.

Example III

Set the coordinate x to be the distance between O and the rod A. Relate x to the angular position . Then take time derivatives of the position equation to find the velocity and acceleration relationships.

Plan:

Solution:

Relate x, the distance between O and the rod, to 𝜃.

𝑥 = 𝑒𝑐𝑜𝑠𝜃 + 𝑟

Take time derivatives of the position to find the velocity and acceleration.

x

Notice 𝜃 =-ω So 𝑣𝐴 = 𝑥 = 𝑒𝜔𝑠𝑖𝑛θ 𝑎𝐴 = 𝑥 = −𝑒𝜔2𝑐𝑜𝑠θ

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Given: Crank AB rotates at a constant velocity of 𝜔 = 150 rad/s.

Find: The velocity of point P when 𝜃 = 30°.

Define x as a function of 𝜃 and differentiate with respect to time.

Example IV

Plan:

vP = -0.2(sin)𝜃 + (0.5)[(0.75)2

– (0.2sin)2]-0.5(-2)(0.2sin)(0.2cos)𝜃

vP = -0.2 sin – [0.5(0.2)2 sin2 ] / (0.75)2 – (0.2 sin )2

At = 30°, = 150 rad/s and vP = -18.5 m/s ←

Solution:

Relate x the distance between A and P to 𝜃

𝑥𝑝 = 0.2𝑐𝑜𝑠𝜃 + 0.752 − (0.2𝑠𝑖𝑛𝜃)2

Take time derivatives of the position to find the velocity

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GROUP PROBLEM SOLVING

Given: The hydraulic cylinder AB shortens at a constant rate of 0.15 m/s so that girder G of a bascule bridge is raised.

Find: The angular velocity of the bridge girder when 𝜃 = 60.

Set the coordinate s to be distance AB. Then relate s to the angular position 𝜃 . Take time derivative of this position relationship to find the angular velocity.

Plan:

Solution:

Relate s, the distance AB, to q applying the law of cosines to the geometry.

AB2 = BC2+AC22(BC)(AC) cos(180-)

s2 = 32 + 52 2 (3) (5) cos (180-)

= 34 – 30 cos (180-)

s2 = 34 +30 cos

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ATTENTION QUIZ

2. If vA=10 m/s and aA=10 m/s2, determine

the angular acceleration, , when = 30.

A) 0 rad/s2 B) -50.2 rad/s2

C) -112 rad/s2 D) -183 rad/s2

1. The sliders shown below are confined to move in the

horizontal and vertical slots. If vA=10 m/s, determine the

connecting bar’s angular velocity when = 30.

A) 10 rad/s B) 10 rad/s

C) 8.7 rad/s D) 8.7 rad/s