Dr. Jagadeep Thota

MAE 364 Kinematics and Dynamics of Machines

Chapter 2: Motion

MAE 364: Motion JT

In this chapter we will be studying

motion of bodies AND motion of points

on the bodies.

Linear and angular displacement.

Linear and angular velocity.

Linear and angular acceleration.

Normal and tangential acceleration.

Relative motion

Line of transmission.

Overview

MAE 364: Motion JT

Points (even on a rotating body) will

ONLY have linear motion.

Bodies can have linear OR angular

motion.

***Note***

Very Important. You will need to remember this through out the semester.

MAE 364: Motion JT

Displacement of a point is the change of its position and is a

vector quantity.

Linear Displacement

• Consider the motion of point P along the path MN from position B to position C.

• This is linear displacement of point P.

• This can be expressed as a vector equation,

• Magnitude of the above vector (linear displacement),

• Direction of the vector w.r.t x-axis,

∆𝑠 = ∆𝑥 → ∆𝑦

∆𝑠 = (∆𝑥)2+ (∆𝑦)2

𝑡𝑎𝑛ψ = ∆𝑦

∆𝑥

MAE 364: Motion JT

Consider a body rotating about the fixed axis O and let P be

a point fixed in the body.

Angular Displacement

• As P moves to P’, the angular displacement of the line OP or the body itself is,

𝐴𝑛𝑔𝑢𝑙𝑎𝑟 𝐷𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 = ∆𝜃

MAE 364: Motion JT

Linear velocity is the time rate of change of linear

displacement.

Linear Velocity

• Let us assume point P moves from position B to position C in time ∆𝑡.

• The average velocity during this time interval is,

• The instantaneous linear velocity of point P, when it is at position B is,

• Linear velocity is directed tangent to the path of motion.

𝑉𝑎𝑣𝑔 = ∆𝑠

∆𝑡

𝑉 = 𝑑𝑠

𝑑𝑡

MAE 364: Motion JT

Consider a body rotating about the fixed axis O and let P be

a point fixed in the body.

Angular Velocity

• Angular displacement of the body (or line OP) is ∆𝜃.

• Let us assume the angular displacement of the body occurs in a time ∆𝑡.

• Then, the average angular velocity of the body during this time interval is,

𝜔𝑎𝑣𝑔 = ∆𝜃

∆𝑡

The instantaneous angular velocity of the body for the position OP is,

𝜔 = 𝑑𝜃

𝑑𝑡

MAE 364: Motion JT

Point P has a radius of rotation R equal to length OP.

V is the linear velocity of point P and is tangent to the path PP`.

Hence, perpendicular to the radius R.

Relation Between Linear and Angular Velocity

• Angular displacement, ∆𝜃, is very small and is expressed in radians.

• Therefore, the arc length PP` is,

• But, the arc length PP` is the linear displacement of point P.

𝑃𝑃′ = 𝑅. ∆𝜃

𝑑𝑠 = 𝑃𝑃′ = 𝑅. ∆𝜃

• Linear velocity is given as,

𝑉 = 𝑑𝑠

𝑑𝑡 ⇒ 𝑉 =

𝑅. ∆𝜃

𝑑𝑡 ⇒ 𝑉 = 𝑅

𝑑𝜃

𝑑𝑡

𝐵𝑢𝑡,𝑑𝜃

𝑑𝑡= 𝜔 = 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦

𝑽 = 𝑹𝝎

Therefore, the linear velocity is related to the angular velocity as shown,

MAE 364: Motion JT

Determine the velocity of point P if the square plate is

rotating at a rate of 80 rpm clockwise. Magnitude & Direction!

Solution: Convert 𝜔 from rpm to rad/s,

Find the radius of rotation, which is distance between the fixed point in the body about which the body is rotating to the point of interest which is P.

Example 1: Linear Velocity

𝜔 =2𝜋𝑛

60=

2𝜋(80 𝑟𝑝𝑚)

60 𝑠𝑒𝑐= 8.38 𝑟𝑎𝑑/𝑠

O

𝐷𝑖𝑎𝑔𝑜𝑛𝑎𝑙 𝑜𝑓 𝑡ℎ𝑒 𝑆𝑞𝑢𝑎𝑟𝑒 = 22 + 22 = 8 = 2 2 𝑖𝑛

R

𝑅𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛 = 𝑅 = 𝐷𝑖𝑎𝑔𝑜𝑛𝑎𝑙

2=

2 2

2= 2 𝑖𝑛

Velocity of point P,

𝑉 = 𝑅𝜔 = 2 8.38 = 11.85 𝑖𝑛/𝑠

Direction? Linear velocity is always

tangent to the path of motion OR perpendicular to radius of rotation.

Therefore, direction is ↑

V

MAE 364: Motion JT

Linear acceleration is the time rate of change of linear velocity.

Let the initial velocity of a point be Vo and the velocity after a time

interval ∆𝑡 be V.

Then, the average linear acceleration during the time interval is,

The instantaneous linear acceleration is,

Linear Acceleration

𝐴 = 𝑉 − 𝑉𝑜

∆𝑡=

∆𝑉

∆𝑡

𝑨 = 𝒅𝑽

𝒅𝒕

𝐴 = 𝑑

𝑑𝑡

𝑑𝑠

𝑑𝑡=

𝑑2𝑠

𝑑𝑡2 𝑨 =

𝒅𝟐𝒔

𝒅𝒕𝟐

MAE 364: Motion JT

Angular acceleration is the time rate change of angular velocity,

Angular displacement, velocity, and acceleration are either CW (clockwise) or CCW (counter-clockwise).

Throughout this course, CCW is considered +ve, and CW as –ve.

Angular Acceleration

𝜶 = 𝒅𝝎

𝒅𝒕

𝛼 = 𝑑

𝑑𝑡

𝑑𝜃

𝑑𝑡

𝜶 = 𝒅𝟐𝜽

𝒅𝒕𝟐

MAE 364: Motion JT

A point can have acceleration in a direction either

normal, tangential, or both to its path of motion.

If a point has curvilinear motion, it will have a

normal acceleration resulting from a change in

direction of its linear velocity.

If the magnitude of the linear velocity changes, the point will

have a tangential acceleration.

A point having rectilinear motion has no normal

acceleration since its linear velocity does not

change direction.

It will have a tangential acceleration if the linear velocity

changes in magnitude.

Normal and Tangential Acceleration

MAE 364: Motion JT

Velocity Diagram

Velocity is tangent to path

MAE 364: Motion JT

The tangential acceleration of a point is the time rate

change of the magnitude of its linear velocity.

Tangential Acceleration

𝐴𝑡 = 𝑑𝑉𝑡

𝑑𝑡

𝐴𝑡 = 𝑑

𝑑𝑡(𝑅𝜔)

𝐴𝑡 = 𝑅𝑑𝜔

𝑑𝑡

𝑨𝒕 = 𝑹𝜶

𝑆𝑖𝑛𝑐𝑒, 𝑉 = 𝑅𝜔

𝑆𝑖𝑛𝑐𝑒, 𝛼 = 𝑑𝜔

𝑑𝑡

MAE 364: Motion JT

Normal acceleration of a point is the time rate of change of

its velocity in a direction normal to the path.

Normal Acceleration

𝐴𝑛 = 𝑑𝑉𝑛

𝑑𝑡

∆𝑉𝑛 ≈ 𝑑𝑉𝑛 = 𝑉. 𝑑𝜃

From figure,

𝐴𝑛 = 𝑉.𝑑𝜃

𝑑𝑡= 𝑉

𝑑𝜃

𝑑𝑡

𝐴𝑛 = 𝑉𝜔

𝐴𝑛 = 𝑅𝜔2

𝐴𝑛 = 𝑉2

𝑅

𝑆𝑖𝑛𝑐𝑒, 𝜔 = 𝑑𝜃

𝑑𝑡

𝑆𝑖𝑛𝑐𝑒, 𝑉 = 𝑅𝜔

MAE 364: Motion JT

Magnitude of the resultant acceleration,

Direction of the resultant acceleration,

Resultant Acceleration

22 )()( tn AAA

n

t

A

A1tan

∅

𝐴𝑛

𝐴𝑡 𝐴

t

n

A

A1tan

∅

𝐴𝑡

𝐴𝑛 𝐴

MAE 364: Motion JT

The square body is rotating about its center as shown. The body is

rotating at a rate of 20 rad/s CW and it is speeding up at the rate of 400

rad/s/s. Determine the acceleration of point P. Magnitude and Direction!

Solution:

Example 2: Acceleration

Given,

𝜔 = 20 𝑟𝑎𝑑/𝑠 𝛼 = 400 𝑟𝑎𝑑/𝑠2

Since the path of point P is going to be curvilinear and the body is subjected to angular acceleration, the point P will have both normal and tangential acceleration.

Normal acceleration,

𝐴𝑝𝑛 = 𝑅. 𝜔2 = 2 20 2 = 566

𝑖𝑛

𝑠2 →

Tangential acceleration,

𝐴𝑝𝑡 = 𝑅. 𝛼 = 2 (400) = 566

𝑖𝑛

𝑠2 ↑

Acceleration magnitude,

𝐴𝑝 = 𝐴𝑝𝑛2

+ 𝐴𝑝𝑡 2

= 800𝑖𝑛

𝑠2

Acceleration direction,

𝜑 = 𝑡𝑎𝑛−1𝐴𝑝

𝑡

𝐴𝑝𝑛 = 𝑡𝑎𝑛−1 1 = 45° ∴ 𝑨𝒑= 𝟖𝟎𝟎 𝒊𝒏/𝒔𝟐

R

At

An

Ap

MAE 364: Motion JT

Absolute motion is the motion of a body/point in relation to

some other body/point which is at rest.

In the previous slides, we considered the motion of a point

in relation to some fixed point.

Thus we have been discussing absolute motion till now.

In mechanics and kinematics, we assume earth to be fixed.

Hence, the motion of a body in relation to the earth or

ground is absolute motion.

In vector form it is represented as,

Absolute Motion

𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑝𝑜𝑖𝑛𝑡 𝐵 = 𝑉𝐵

MAE 364: Motion JT

Motion of a body or point as observed from another point or

body.

The “other” body or point may be stationary.

The “other” body or point may be moving.

A body has motion relative to another body only if there is

difference in their absolute motions.

The velocity of a body M relative to a body N is the absolute velocity of M

minus the absolute velocity of N.

The absolute motion can also be written in relative motion format as shown

Relative Motion

𝑉𝑀/𝑁 = 𝑉𝑀 → 𝑉𝑁 𝑉𝑀 = 𝑉𝑁 → 𝑉𝑀/𝑁

𝑉𝐵 = 𝑉𝐵/𝑂 The above equation represents absolute velocity of point B relative to a fixed or stationary point O. Usually, when writing absolute motion we neglect the relative or second subscript term.

MAE 364: Motion JT

Consider two automobiles A and B travelling with velocities

of 60 km/h and 40 km/h.

Let 𝑉𝐴 and 𝑉𝐵 denote the absolute velocities of A and B

respectively.

The velocity of automobile A relative to automobile B is

given as,

In vector diagram form,

Relative Motion illustration

A 𝑉𝐴 = 60 𝑘𝑚/ℎ

B 𝑉𝐵 = 40 𝑘𝑚/ℎ

𝑉𝐴/𝐵 = 𝑉𝐴 → 𝑉𝐵 = 𝑉𝐴 → −𝑉𝐵 = 60 − 40 = 20 𝑘𝑚/ℎ

𝑉𝐴/𝐵 = 20 𝑘𝑚/ℎ

𝑉𝐴

−𝑉𝐵

O +

MAE 364: Motion JT

So, in the previous illustration, the velocity of A relative to B

is the velocity that A would appear to have to an observer in

B if the observer were to imagine that B were at rest.

To the observer in B, then A would appear to be moving to

the left at 20 km/h.

Now, let us look at the velocity of B relative to A,

In vector diagram format,

To an observer in A, the car B would appear to be moving to

the right at 20 km/h.

Relative Motion illustration

𝑉𝐵/𝐴 = 𝑉𝐵 → 𝑉𝐴 = 𝑉𝐵 → −𝑉𝐴 = 40 − 60 = −20 𝑘𝑚/ℎ

𝑉𝐵/𝐴 = −20 𝑘𝑚/ℎ

−𝑉𝐴

𝑉𝐵

O +

MAE 364: Motion JT

In a vector equation, terms can be transposed, provided

their signs are changed.

If the subscripts are reversed on a vector in a vector

equation, the sign of the vector must be changed.

Vector Concept

𝑉𝐴/𝐵 = 𝑉𝐴 → 𝑉𝐵

−𝑉𝐴 = −𝑉𝐴/𝐵 → 𝑉𝐵

𝑉𝐵 = 𝑉𝐴 → 𝑉𝐴/𝐵

−𝑉𝐵/𝐴= 𝑉𝐴 → 𝑉𝐵

−𝑉𝐴= 𝑉𝐵/𝐴 → 𝑉𝐵

𝑉𝐵 = 𝑉𝐴 → 𝑉𝐵/𝐴

MAE 364: Motion JT

Determine the velocity of P relative to point Q if the plate is

rotating at the rate of 100 rad/s CW. Magnitude & Direction!

Solution:

Given, 𝜔 = 100 𝑟𝑎𝑑/𝑠

Find the absolute velocity of point P,

Find the absolute velocity of point Q,

Velocity of P relative to Q,

Example 3: Relative Motion

𝑉𝑃 = 𝑉𝑃/𝑂 = 𝑅𝜔 = 𝑂𝑃. 𝜔 = 2 ∗ 100 = 141.4 𝑖𝑛/𝑠

𝑉𝑄 = 𝑉𝑄/𝑂 = 𝑅𝜔 = 𝑂𝑄. 𝜔 = 2 ∗ 100 = 141.4 𝑖𝑛/𝑠

𝑉𝑃/𝑄 = 𝑉𝑝 → 𝑉𝑄 = 𝑉𝑃 → −𝑉𝑄 = 𝑉𝑃2 + 𝑉𝑄

2 = 200 𝑖𝑛/𝑠

O

𝜃𝑃

𝜃𝑄

𝑉𝑃 −𝑉𝑄

𝑉𝑃/𝑄

45°

135°

∴ 𝑉𝑃/𝑄= 200𝑖𝑛

𝑠 O

+

MAE 364: Motion JT

Same question as Example 3.

Solution:

This is an alternate procedure for solving the previous example.

Since, we need to find the velocity of P relative to Q, the point Q can be

assumed to be a stationary point relative to P.

Then 𝑃𝑄 can be taken as the radius of rotation ‘R’.

Then using the relationship between linear and angular velocity, the

velocity of P relative to Q can be computed,

Direction: The linear velocity of a point is always

perpendicular to its ‘R’.

Example 4: Relative Motion

O R

VP/Q 𝑉𝑃/𝑄 = 𝑅. 𝜔 = 𝑃𝑄 ∗ 𝜔 = 2 100 = 200 𝑖𝑛/𝑠

∴ 𝑉𝑃/𝑄= 200𝑖𝑛

𝑠

MAE 364: Motion JT

In the below mechanisms,

Link 2 is called the driver as input motion is given to this link.

Link 4 is the driven member (output motion), and hence called the follower.

Link 3 is called the coupler, as it transmits motion from the driver to the

follower.

Mechanisms

Coupler link is Rigid

Coupler link is Flexible

No Coupler link; Direct Contact

MAE 364: Motion JT

Motion is transmitted from the driver to the follower along

the line of transmission.

Line of Transmission

Line of Centers

Line of Transmission

MAE 364: Motion JT

Angular Velocity Ratio

O2 O4

P

P is outside line segment O2O4 𝜔2 and 𝜔4 have SAME directions P is inside line segment O2O4 𝜔2 and 𝜔4 have OPPOSITE directions

𝝎𝟒

𝝎𝟐=

𝑶𝟐𝑷

𝑶𝟒𝑷

MAE 364: Motion JT

For the below mechanism, the angular velocity of the driver

is given to be 200 rpm CW. Find the angular velocity of the

follower.

Solution:

Example 5: Angular Velocity Ratio

O2 O4

P

216

32

4

2

2

4 PO

PO

CWrpm 4002 24

MAE 364: Motion JT

Line of Transmission

Line of Centers

Line of Transmission

Common Tangent Point of

Contact

MAE 364: Motion JT

Determine 𝜔4? Given, 𝜔2 = 200 𝑟𝑝𝑚 𝐶𝑊.

Example 6: Angular Velocity Ratio

1. 200 rpm cw

2. 200 rpm ccw

3. 800 rpm cw

4. 800 rpm ccw

5. 50 rpm cw

6. 50 rpm ccw

MAE 364: Motion JT

Homework 4 Text book problems 2-2, 7, 11, 15, 16, 19, 20

Due 10/02/2014

Exam 1 On 10/07/2014

Covers chapters 17, 19, 1, 2 & 3

Quiz on Chapter 3 next class (09/30/2014)

Conclusion