Fundamental Notes

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Fundamentals Exam review material

ME274

ME 274: Basic Mechanics II

Dynamics Fundamentals: Mathematics, Vectors and Point Kinematics

1. Scalar kinematics a) Differentiation b) Integration c) Sample problems 2. Vector operations a) Addition and subtraction b) Dot (scalar) products c) Cross (vector) products d) Moments about points e) Projections of vectors f) Writing a vector in terms of a new set of unit vectors g) Sample problems 3. Vector kinematics a) Cartesian components b) Path components c) Polar components d) Sample problems 4. Solutions of sample problems (to be provided later)

Fundamentals Exam review material

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Scalar kinematics: differentiation Suppose that a point P travels on a straight-line path (rectilinear motion) whose position on the path is given by the distance variable s. The speed v of P is given by: v = ds/dt.

ME274

v

s

Case 1: On this path, we are given the speed v of point P in terms of time, t: v = v(t). From this we want to determine the acceleration of the point. The acceleration can be found by directly differentiating v with respect to t; that is, dv a= dt

Case 2: On this path, we are given the speed v of point P in terms of its position s: v = v(s). Here we need to employ the chain rule of differentiation (see below). Chain rule of differentiation Suppose y = y(x) where x = x(t). The derivative of y with respect to t can be found from the chain rule of differentiation to be:

dy dy dx = dt dx dt

Using the chain rule of differentiation, we see that: dv dv ds dv a= = =v dt ds dt ds where v = ds/dt.

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Fundamentals Exam review material

ME274

Scalar kinematics: integration Again, suppose that a point P travels on a straight-line path (rectilinear motion) whose position on the path is given by the distance variable s. The speed v of P is given by: v = ds/dt.

v

s

Case 1: On this path, we are given the acceleration of point P in terms of time, t: a = a(t). From this we want to determine the speed of the point at some instant in time t. The acceleration can be found by directly integrating a with respect to t; that is, a=

dv dt

t

" a (t ) dt =

!

0

v

"

t

dv

v( 0 )

!

v ( t ) = v ( 0 ) + " a ( t ) dt 0

Case 2: On this path, we are given the acceleration a of point P in terms of its position s: v = v(s). Here, prior to using separation of variables, we need to use the chain rule of differentiation to produce: dv dv ds dv a= = =v dt ds dt ds where v = ds/dt. Now, we can write: v

v dv = a ds

!

"

v( 0 )

s

v dv = " a ds 0

s

!

1 2 1 2 v # v ( 0 ) = " a ds 2 2 0

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Fundamentals Exam review material

ME274

Scalar kinematics: sample problems Example 1 Suppose that the speed of P is given by: v ( t ) = 3t 2 . Find the acceleration of P. Example 2 Suppose that the speed of P is given by: v s = 5sin 3s . Find the acceleration of P.

()

( )

Example 3 Suppose that the acceleration of P traveling on a straight-line path is given by: a ( t ) = 6 sin ( 2t ) , with v 0 = 4 . Find the speed of P as a function of time.

()

Example 4 Suppose that the acceleration of P traveling on a straight-line path is given by: a ( s ) = 8 s 3 with v s = 0 = 5 . Find the speed of P as a function of position s.

(

)

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Fundamentals Exam review material

ME274

Vector operations

a+b

Addition The addition of two vectors a and b is found through the “parallelogram” rule: move the tail of b to the head of a with a + b being the vector extending from the tail of a to the new position of the head of b (as shown in the figure to the right).

b a

If a and b are given in terms of sets of xyz components as:

a = a x i + a y j + az k b = bx i + by j + bz k then the components of the vector a + b are found from the scalar sum of the components of a and b:

(

)

a + b = ( ax + bx ) i + ay + by j + ( az + bz ) k

Subtraction The subtraction of two vectors a and b is found through the “parallelogram” rule on a and -b: move the tail of -b to the head of a with a - b being the vector extending from the tail of a to the new position of the head of -b (as shown in the figure to the right). If a and b are given in terms of sets of xyz components as:

b a -b

a-b

a = a x i + a y j + az k b = bx i + by j + bz k then the components of the vector a - b are found from the scalar difference of the components of a and b:

(

)

a ! b = ( ax ! bx ) i + ay ! by j + ( az ! bz ) k

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Fundamentals Exam review material

ME274

Dot (scalar) product The dot product of two vectors a and b is the following SCALAR:

b θ

a i b = a b cos!

a

where θ is the smallest angle between vectors a and b. If a and b are given in terms of sets of xyz components as:

a = a x i + a y j + az k b = bx i + by j + bz k then the scalar quantity a i b is found by1:

(

)(

a i b = ax i + ay j + az k i bx i + by j + bz k

(

)

) ( )( = ax bx ( i i i ) + ax by ( i i j ) + ax bz ( i i k ) + aybx ( j i i ) + ayby ( j i j ) + aybz ( j i k ) + az bx ( k i i ) + az by ( k i j ) + az bz ( k i k )

)

(

= ( ax i ) i bx i + by j + bz k + ay j i bx i + by j + bz k + ( az k ) i bx i + by j + bz k

)

= ax bx (1) + ax by ( 0 ) + ax bz ( 0 ) + aybx ( 0 ) + ayby (1) + aybz ( 0 ) + az bx ( 0 ) + az by ( 0 ) + az bz (1) = ax bx + ayby + az bz

HINT: In finding the dot product, you do not need to go through all of above steps; simply recall that the dot product of a and b is found by the sum of the products of the like components of a and b.

1

From the above dot product definition, note that: i i i = j i j = k i k = 1 and

ii j = j i k = k ii = j ii = k i j =ii k = 0.

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Fundamentals Exam review material

ME274

Cross (vector) product The cross product of two vectors a and b is given by a VECTOR whose magnitude is given by:

axb curl of fingers

a ! b = a b sin "

and whose direction is found through the “right hand rule”: sweep the fingers of your right hand FROM a TO b (through the smallest angle between vectors a and b) and your right thumb points in the direction of a ! b .

b θ

a

If a and b are given in terms of sets of xyz components as:

a = a x i + a y j + az k b = bx i + by j + bz k then the vector quantity a ! b is found by2:

(

) (

a ! b = ax i + ay j + az k ! bx i + by j + bz k

(

)

) ( ) ( = ax bx ( i ! i ) + ax by ( i ! j ) + ax bz ( i ! k ) + aybx ( j ! i ) + ayby ( j ! j ) + aybz ( j ! k ) + az bx ( k ! i ) + az by ( k ! j ) + az bz ( k ! k ) = ax bx ( 0 ) + ax by ( k ) + ax bz ( " j ) +

)

(

= ( ax i ) ! bx i + by j + bz k + ay j ! bx i + by j + bz k + ( az k ) ! bx i + by j + bz k

aybx ( "k ) + ayby ( 0 ) + aybz ( i ) +

()

az bx j + az by ( "i ) + az bz ( 0 )

(

)

(

)

= aybz " az by i " ( ax bz " az bx ) j + ax by " aybx k i = ax bx

2

j ay by

k az bz

From the above cross product definition, note that: i ! i = j ! j = k ! k = 0 ,

(

)

(

)

i ! j = " j ! i = k , j ! k = " k ! j = i and k ! i = " ( i ! k ) = j .

)

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ME274

Moments about points Recall that the moment about a point O due to a force F acting at point P is found through the cross product:

M O = r P /O ! F

F

where r P /O is the position vector from O to P.

rP/O P

O

Vector projections The projection of vector b onto the line of action of a second vector a is found by placing vectors a and b tail-to-tail and drawing a line from the head of b that is perpendicular to a (as shown in the figure to the right). From this figure, we see that this projection is given by:

" projection of b onto a" = b cos! Say we determine a unit vector ea = a / a (where ea = 1 ) that points in the same direction as a. Using the above, we see that:

" projection of b onto a" = b ea cos! = b i ea

b θ

a

projection of b onto the line of action of a

Fundamentals Exam review material

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Writing a vector in terms of a new set of unit vectors Suppose that you are given a vector a in terms of its XY components (unit vectors I and J): a = a X I + aY J

( aX and aY

ME274

J j

are KNOWN )

θ

and you desire to know the same vector a in terms of its xy components (unit vectors i and j):

a = ax i + ay j

( ax and ay are UNKNOWN )

where the coordinate axes are shown in the figure above. One way to do this is to first find i and j in terms of their I and J components (see projections in the above figure):

i = cos! I + sin ! J j = ! sin " I + cos" J Now we can use dot products to determine ax and ay through:

ax = a i i

= ( a X I + aY J ) i i

= ( a X I + aY J ) i ( cos! I + sin ! J ) = a X cos! + aY sin ! and ay = a i j = ( a X I + aY J ) i j

= ( a X I + aY J ) i ( ! sin " I + cos" J ) = !a X sin " + aY cos"

i θ

I

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Fundamentals Exam review material

ME274

Vector operations – sample problems Given two vectors a and b in terms of xyz components:

( ) b = ( !2 i ! 4 j + 6 k ) ft a = i ! 3 j + 5 k ft

Example 5: Find a + b Example 6: Find a - b Example 7: Find a • b Example 8: Find a x b Example 9: Find b x a Example 10: Find the projection of a onto b Example 11: Find the projection of b onto a Example 12: Consider a vector c given in XY components as: c = ( !2 I + 4J ) ft / sec . Find the xy components of c where the xy coordinate axes are oriented with respect to the XY axes as shown below. Example 13: Consider a vector d given in xy components as: d = 3i ! 5 j ft / sec 2 . Find the XY components of d where the XY coordinate axes are

(

)

oriented with respect to the xy axes as shown below.

J j 36.87°

i 36.87°

I

Fundamentals Exam review material

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ME274

Vector kinematics The following are the three sets of vector kinematical equations for describing the motion (velocity and acceleration) of a point P moving in a plane using Cartesian, path and polar coordinates: path of P

Cartesian coordinates ! + y! j v = xi

a = !! xi + !! yj

j P y i

x Path coordinates v = vet ! t+ a = ve

path of P

v2 e ! n

C

et

en

ρ

Note: Since the rate of change of speed v! is the projection of the acceleration vector a onto the unit tangent vector, we can find v! from:

P s

! v$ v! = a i et = a i # & " v%

path of P

Polar coordinates ! r + r!! e! v = re

(

)

er

a = !! r " r!! 2 er + ( r!!! + 2 r!!! ) e! P θ

r eθ

O

Fundamentals Exam review material

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Vector kinematics – sample problems

ME274

()

Example 14: A point P travels on a path given by y x = 0.3 + 0.5x 2 (both x and y in meters). The x-component of velocity of P is known to be x! = 4 m / sec = constant . Find the velocity and acceleration vectors for point P in Cartesian components when x = 2 meters. Example 15: At one instant in time, the speed of a point P traveling on a plane is 10 2 m/sec with this speed decreasing at a rate of 3 m/sec . The radius of curvature for the path of P at this instant is 50 meters. Find the acceleration of P in terms of its path variables. Example 16: Point P travels on a path given in polar coordinates as: r = 3sin ! , where r is in feet, θ is in radians and !! = 2 rad / sec = constant. Find the velocity and acceleration of P in terms of its polar coordinates when θ = π/2. Example 17: Make a sketch of the velocity and acceleration vectors for Example 14. Include the unit tangent and unit normal vectors in your sketch. Is the speed increasing or decreasing? Example 18: Make a sketch of the velocity and acceleration vectors for Example 16. Include the unit tangent and unit normal vectors in your sketch. Is the speed increasing, decreasing or constant?

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Fundamentals Exam review material

ME274

SOLUTIONS for sample problems Example 1

()

Suppose that the speed of P is given by: v t = 3t 2 . Find the acceleration of P. The acceleration of P on its straight-line path is found from: a =

dv = 6t . dt

Example 2 Suppose that the speed of P is given by: v s = 5sin 3s . Find the acceleration of P.

()

a=

( )

dv dv ds dv = = v = !" 5 sin ( 3s ) #$ !"15 cos ( 3s ) #$ = 75 sin ( 3s ) cos ( 3s ) dt ds dt ds

Example 3 Suppose that the acceleration of P traveling on a straight-line path is given by: a t = 6 sin 2t , with v 0 = 4 . Find the speed of P as a function of time.

()

( )

() v

dv = 6 sin ( 2t ) dt

"

!

v( 0 )

t

dv = " 6 sin ( 2t ) dt

!

0

t

v ( t ) = v ( 0 ) + " #$ 6 sin ( 2t ) %& dt 0

= v (0) '

6 t cos ( 2t ) 0 = 4 ' 3 #$ cos ( 2t ) ' 1%& = 7 ' 3cos ( 2t ) 2 t

v ( t ) = v ( 0 ) + % !" 6 sin ( 2t ) #$ dt 0

= v (0) &

6 t cos ( 2t ) 0 = 4 & 3 !" cos ( 2t ) & 1#$ = 7 & 3cos ( 2t ) 2

Example 4 Suppose that the acceleration of P traveling on a straight-line path is given by: a ( s ) = 8 s 3 with v s = 0 = 5 . Find the speed of P as a function of position s.

(

)

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Fundamentals Exam review material

a= v

"

v( 0 )

dv dv ds dv = =v dt ds dt ds

ME274

!

s

v dv = " a ds

!

0

s

1 2 1 2 v # v ( 0 ) = " a ds 2 2 0 Therefore, s

v = v ( 0 ) + 2 ! a ds 2

0

s

= v 2 ( 0 ) + 2 ! 8s 3 ds 0

( 5 )2 + 2

=

8 4 s 4

s s=0

= 25 + 4s 4

Given two vectors a and b in terms of xyz components:

( ) b = ( !2 i ! 4 j + 6 k ) ft a = i ! 3 j + 5 k ft

Example 5: Find a + b a + b = i ! 3 j + 5 k + !2 i ! 4 j + 6 k = ! i ! 7 j + 11 k ft

(

) (

) (

)

Example 6: Find a - b a ! b = i ! 3 j + 5 k ! !2 i ! 4 j + 6 k = 3i + j ! k ft

(

) (

Example 7: Find a • b

(

)(

) (

)

) ( )( ) ( )( ) ( )( )

a i b = i ! 3 j + 5 k i !2 i ! 4 j + 6 k = 1 !2 + !3 !4 + 5 6 = 40 ft 2 Example 8: Find a x b

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Fundamentals Exam review material

(

) (

a ! b = i " 3 j + 5 k ! "2 i " 4 j + 6 k i

j

k

ME274

)

(

)

= 1 "3 5 = 2 i " 16 j " 10 k ft 2 "2 "4 6

Example 9: Find b x a b ! a = "2 i " 4 j + 6 k ! i " 3 j + 5 k

(

i

j

) (

k

(

)

)

= "2 "4 6 = "2 i + 16 j + 10 k ft 2 1 "3 5

Example 10: Find the projection of a onto b " !2 i ! 4 j + 6 k % 40 2 a i eb = i ! 3 j + 5 k i $ ft '= 56 # 22 + 42 + 62 &

(

)

Example 11: Find the projection of b onto a " i ! 3j +5 k % 40 2 b i e a = !2 i ! 4 j + 6 k i $ ft '= 2 2 2 46 # 1 +3 +6 &

(

)

Example 12: Consider a vector c given in XY components as: c = !2 I + 4J ft / sec . Find the xy components of c where the xy coordinate axes are

(

)

oriented with respect to the XY axes as shown to the right. From the figure, we see that:

i = cos36.87° I + sin 36.87° J = 0.8 I + 0.6 J

J j 36.87°

i

Therefore:

( ) ( ) = ( !2 ) ( 0.8) + ( 4 ) ( 0.6 ) = 0.8 ft / sec

cx = ci i = !2 I + 4J i 0.8 I + 0.6 J

Also from the figure, we see that: j = ! sin 36.87° I + cos36.87° J = !0.6 I + 0.8 J

36.87°

I

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Fundamentals Exam review material

ME274

Therefore:

(

) ( ) = ( !2 ) ( !0.6 ) + ( 4 ) ( 0.8) = 4.4 ft / sec

c y = ci j = !2 I + 4J i !0.6 I + 0.8 J

In summary:

(

) (

)

c = cx i + c y j = 0.8i + 4.4 j ft / sec

Example 13: Consider a vector d given in xy components as:

(

)

d = 3i ! 5 j ft / sec 2 . Find the XY components of d where the XY coordinate axes are oriented with respect to the xy axes as shown to the right.

J j

From the figure, we see that: I = cos36.87° i ! sin 36.87° j = 0.8i ! 0.6 j

36.87°

i

Therefore:

(

)(

d X = d i I = 3i ! 5 j i 0.8i ! 0.6 j

( )( ) ( )(

)

)

= 3 0.8 + !5 !0.6 = 5.4 ft / sec 2 Also from the figure, we see that: J = sin 36.87° i + cos36.87° j = 0.6 i + 0.8 j

Therefore:

(

)(

dY = d i J = 3i ! 5 j i 0.6 i + 0.8 j

( )( ) ( )( )

)

= 3 0.6 + !5 0.8 = 2.2 ft / sec 2 = !2.2 ft / sec 2

In summary:

(

)

(

)

d = d X I + dY J = 5.4 I + ! 2.2J ft / sec 2

36.87°

I

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Fundamentals Exam review material

ME274

()

Example 14: A point P travels on a path given by y x = 0.3 + 0.5x 2 (both x and y in meters). The x-component of velocity of P is known to be x! = 4 m / sec = constant . Find the velocity and acceleration vectors for point P in Cartesian components when x = 2 meters.

dy dy dx = = x x! = 2 4 = 8 m / sec dt dx dt 2 d!y d !! y= = xx! = x!x! + x!! x = 4 + 2 0 = 16 m / sec 2 dt dt

( ) ( )( )

y! =

( )

( ) ( )( )

Therefore,

( ) !! + !! a = xi y j = (16 j ) m / sec 2

! + y! j = 4i + 8 j m / sec v = xi

Example 15: At one instant in time, the speed of a point P traveling on a plane is 10 2 m/sec with this speed decreasing at a rate of 3 m/sec . The radius of curvature for the path of P at this instant is 50 meters. Find the acceleration of P in terms of its path variables. ! t+ a = ve

v2 102 e n = "3 et + e = "3et + 2e n m / sec 2 ! 50 n

( )

(

)

Example 16: Point P travels on a path given in polar coordinates as: r = 3sin ! , where r is in feet, θ is in radians and !! = 2 rad / sec = constant. Find the velocity and acceleration of P in terms of its polar coordinates when θ = π/2.

dr dr d! = = 3cos! !! = 3 0 2 = 0 dt d! dt d ! r!! = 3! cos! = 3!!!cos! " 3!! 2 sin ! = 0 " 3 2 dt

(

r! =

(

)

) ( )( )( )

( )( )2 (1) = "12 ft / sec

Therefore, ! r + r!! e! = 0 er + 3 2 e! = 6 e! ft / sec v = re

( ) ( )( ) ( ) 2& # a = ( r!! " r!! 2 ) er + ( r!!! + 2r!!! ) e! = % "12 " ( 3) ( 2 ) ( er + ( 0 + 0 ) e! $ ' = ( "24er ) ft / sec 2

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Fundamentals Exam review material

ME274

Example 17: Make a sketch of the velocity and acceleration vectors for Example 14. Include the unit tangent and unit normal vectors in your sketch. Is the speed increasing or decreasing?

( ) v2 ! t + en a = (16 j ) m / sec 2 = ve !

et

v = 4i + 8 j m / sec = vet

a v

The unit tangent vector et must point in the same direction as the velocity vector. The unit normal vector en is perpendicular to et and is directed in a way that the

θ

en

v!

v2 / !

component of a in the en is positive. Note that the projection of a onto et is positive. Therefore, the rate of change of speed is positive, and the speed is increasing.

Example 18: Make a sketch of the velocity and acceleration vectors for Example 16. Include the unit tangent and unit normal vectors in your sketch. Is the speed increasing, decreasing or constant?

( )

er

v = 6 e! ft / sec = vet

(

)

! t+ a = "24er ft / sec 2 = ve

v2 e # n

v et, eθ

The unit tangent vector et must point in the en θ same direction as the velocity vector. The unit a normal vector en is perpendicular to et and is directed in a way that the component of a in the en is positive. Note that the projection of a onto et is zero. Therefore, the rate of change of speed is zero, and the speed is instantaneously constant.