© samir n. shoukry, 2004, dynamics mae 242. quiz

© Samir N. Shoukry, 2004, Dynamics MAE 242

MAE 242: DYNAMICSMAE 242: DYNAMICS

Instructor Instructor Dr. Samir N. ShoukryDr. Samir N. ShoukryProfessor, MAE & CEEProfessor, MAE & CEE

assistants assistants K. Mc Bride Dhananjy Rao K. Mc Bride Dhananjy Rao

ony Oomen Praveenony Oomen Praveen

Class 2Class 2


QuizQuiz


QuizQuiz

2 2

1 1

3.5 10

2 4

2 2 2 2

2

(3.5 2 ) / 2 (10 4 )

10.2

s v

s v

dv dv ds dvSince a v

dt ds dt dsdv

a ks vds

ks ds v dv

ks ds vdv

k

k S

1 1

0

0 0

2 2 2 21 0 1 0

2 2 2 21 0 1 0

2 2 2 20 1 0 1

2 2 21 0 1

0

2 2 2

0

:

( ) / 2 ( ) / 2

( )

( )

( )

(4 0 ) 10.2 * 21.56

10.2

s v

s v

To find s note that ks ds v dv

k s s v v

ks ks v v

ks v v ks

v v kss

k

s ft

2 2

1 1

3.5 10

2 4

2 2 2 2

2

(3.5 2 ) / 2 (10 4 )

10.2

s v

s v

dv dv ds dvSince a v

dt ds dt dsdv

a ks vds

ks ds v dv

ks ds vdv

k

k S


2

62 6

0

2 0

2

, 6 11 /

(2 1) 2 [ ]

36 6 2 32 /

v

dva dv a dt when t s a m s

dt

dv t dt v t t

v ft s

2

2

6 3 22 6

0

1 0

3 2

(2 1) ; @ 0, 2 / 2

2

( 2) [ 2 ]3 2

6 62*6 67

3 2

s

dv t v t t c t v m s c

dsv t t

dt

t tds t t dt t

s m

12-11


A race starting from rest travels along a straight road and for 10 seconds has an acceleration as shown. Draw the v-t graph that describes the motion and find the distance traveled in 10 seconds.

12:46

12 6

18 6

2

6 , 12 / 18

6 10

6

6 24

(6 24)

3 24 54

10 , 36 / 114

v t

s t

whent s v m s and s m

For t dv a dt

dv dt

v t

ds t dt

s t t

whent s v m s and s m

2

0 0

3

3

0 0

4

0 6

6

1

18

18

1

72

v t

s t

For t

dv a dt

tdv dt

v t

tds dt

s t


0 2 4 6 8 10

10

20

30

40

v(m/s)

0

v= 6t-24

v=t3/18

t (s)


0 0

22

2

2

20 0

6 0.2 (6 0.2 ) (6 0.02 )

6 0.01 ......................(1)2

2(6 0.01 )

250 *ln 6 0.02 * 12 0.02 ......(2)

1012 0.02

(1) 20

v s

t s

dva v s v dv s ds v dv s ds

ds

vs s

ds dsv s s v dt

dt v

dsdt t s s s

s S

from when s

00, 322 /

(2) 2000, 31.9

v m s

from when s t s

12:22


Motion underMotion underConstant AccelerationConstant Acceleration

0

20 0

2 20 0

1

2

2 ( )

v v a t

s s v t a t

v v a s s

The following relations may be used ONLY if the acceleration is CONSTANTCONSTANT:

, , .o o os v a aretheinitial conditions of motion


2 2 20 0 0 0 0

0 0

Total distance =acceleration dist. + deceleration dist.

1For consta

..

nt acceleration: , , & 2 ( )2

During Acceleration & 0, then:

................

..(1)T A D

A

v v a t s s v t a t

S

v v

S S

v a

a s s

v s

2

2

2

0 0During Deceleration, the clock is reset and: & 0, :

1,

2

At the end of deceleation

1....(2) & ............(3)

2

10 ....(4), ....(5

2: ,

A A A A A

D A D D D

A

D A D D

A D D

A D

D D

v v s thus

v v a t and s v t a t

t

t s a t

v v a t and s v t a tt

2 2

2 2 22 2

1 1From 1, 3, & 5: , & 2 & 4

2 2

, 1000 , 1.5 / , 2 /

)

41.4 /2 2

41.4 41.448.3

1.5 2

T A A A D D D A D

A A AT T A D A

A D D

A AT A D

A D

S a t v t a t substitute for t t fom

v v vs s m a m s a m s v m s

a a a

v vt t t s

a a



Dr. Samir N. ShoukryDr. Samir N. ShoukryProfessor, MAE & CEEProfessor, MAE & CEE

Class 3Class 3

assistants assistants K. Mc Bride Dhananjy Rao K. Mc Bride Dhananjy Rao

ony Oomen Praveenony Oomen Praveen


Quiz 2 (5 minutes)Quiz 2 (5 minutes)

A car accelerates according to the relation a=0.02s m/s2. Determine its velocity when s=100 m if s=v=0 when t=0.


How Position May Be Described?How Position May Be Described?

New York

Morgantown

East

North

West

South

Point of reference that may be placed anywhere …, say Charleston, WV

East

North

West

South


W

M

N

rM/W

rN/W

East

North

West

South


East

North

West

South


W

M

N

rM/W

rN/W

Position is defined relative to a point on which we set a coordinate system Position is defined relative to a point on which we set a coordinate system


OROR

Position is defined by a vector “r” Position is defined by a vector “r” that originates from a reference pointthat originates from a reference point

Remember: A vector has a length and direction

New York

PathPath

N

EW

S

rM/N

Position VectorPosition Vector

Morganown

New York

Path

Position Vector

New York

Path

Position Vector

rN/M

MorganownN

EW

S

New York

Morgantown

East

North

West

South

New York

Morgantown

East

North

West

South

Morgantown

East

North

West

South

North

West

South

rM/W

rN/W


X

Y

Z

X

Y

Z

r r = = xx i i + y+ y j j r r = = xx i i + y+ y j j + + z z kk

X

Y

Z

xy

zγ r

β

X

Y

Z

xy

zγ r

β

r

x

y

Y

X

r

x

y

Y

X

Position is defined by the distance from Position is defined by the distance from three perpendicular lines from an origin three perpendicular lines from an origin OO..

Rectangular CoordinatesRectangular Coordinates

222 zyxr

L e n g t h o f r

D i r e c t i o n o f r

z

r

y

r

x

r ||cos,

||cos,

||cos

222 zyxr

L e n g t h o f r

D i r e c t i o n o f r

z

r

y

r

x

r ||cos,

||cos,

||cos

22 yxr

Len g t h o f r

D i r ec t i o n o f r)(tan 1

x

y

22 yxr

Len g t h o f r

D i r ec t i o n o f r)(tan 1

x

y


Rectangular Coordinates:Rectangular Coordinates:Velocity & AccelerationVelocity & Acceleration

r r = = xx i i + y+ y j j + + z z kk

v v = = vvxx i i + v+ vyy j j + + vvzz kk

a a = = aaxx i i + a+ ayy j j + + aazz kk

ddtv= dr

dt = (xi) (yj) (zk)+ ddt

ddt+ dx

dt i+ didtxd

dt(xi)= = = dxdt i vxi

222zyx vvvv

222zyx aaaa

X

Y

Z

X

Y

Z


Example: Given , Calculate the magnitude and direction of the velocity and acceleration vectors. Calculate the magnitude of the position vector.

2 32t t r i j k

2 2

22 2 2 1

2 6 , 2 & 6

6(2 ) (6 ) tan

2

x y

drt t v t v t

dt

tv t t v

t

v i j

2 2 1

2 12 , 2 & 12

12(2) (12 ) tan

2

x y

dt a a t

dtt

a t a

va i j

2t

26t

y

v

2 3

2 2 3 2 2

, 3 & 1

( ) (2 ) (1)

x y zr t r t r

r t t

2


Example: A particle accelerates according to the relation: Given the initial conditions at t=0: x, y, z =0 and vx=vy=1 & vz=0, calculate the particle velocity & position.

5yx v t a i j k

22

0 1

2 1

20 0

1

2 2 2

1 , , sinh sinh1

x

x x x xx x

vxx

x

t x

x x

dv dv dv dvdx dxa x v

dt dx dt dt dx dx

vxx dx v dx

dx dxv x v dt t x x t

dt x

0 1 0 0

ln , 1y

y yy y

y

v yt ty t t t

y yy

dv dva v dt

dt v

dvdt t v v e dy e dt y e

v

2 2 3

0 0 0 0

5 5

55 2.5 , 2.5

6

z

zz z

v t z t

z z

dva t dv t dt

dt

dv t dt v t dz t z t



Instructor Instructor Dr. Samir N. ShoukryDr. Samir N. ShoukryProfessor, MAE & CEEProfessor, MAE & CEE

assistants assistants Dr. G. William, CEE Dr. G. William, CEE Mr. K. Mc Bride, MAEMr. K. Mc Bride, MAE

Class 4Class 4


Quiz 3 (5 minutes)Quiz 3 (5 minutes)A particle moves according to the relation r=5t3 i +2t2 j Calculate the magnitude and direction of its acceleration when t =0.5 second.


v0xv

0 yv

0v

yv

0x xv v

0y

fy

maxy

R

0x

0x xv v

upxdownx

x

ya gj j

0xa i

Motion of A ProjectileMotion of A Projectile

0

20 0

2 20 0

1

2

2 ( )

v v a t

s s v t a t

v v a s s

0

20 0

2 20 0

:

1

2

2 ( )

y y

y

y y

In y Direction

v v g t

y y v t g t

v v g y y

0

20 0

2 20 0

0

0 0

:

1

2

2 ( )

0

x x x

x x

x x

x

x x

x

In x Direction

v v a t

x x v t a t

v v a x x

If a

v v

x x v t


A skier leaves the ramp A at angle with the horizontal. He strikes the ground at point B, Determine his initial speed vA and the time of flight from A to B.

25A

3100( ) 60

5

-64

4100( ) 80

5

cosA Av

sinA Av

20 0

2

Inetial and final positions are given. Thus let us use the

1position relation S=S

2: 80 0 ( cos 25) 0 ..........(1)

1: 64 0 ( sin 25) ...........(2)

2. 1& 2

A

A

A

v t at

In X direction v t

InY direction v t gt

Solve Eq v

19.41 / 4.547m s AND t s


x

y

15cos 60 7.5 /ft s

15sin 60 12.99 /ft s

20 0

Initial position and velocity are given. The relation between

x and y components of he final position is given.

1Apply in x and y directions.

2: 0 7.5 ............(1)

s s v t at

In x direction x t

In y directi

2

2

1: 0 12.99 ...........(2)

2

: 0.05 ............(3)

& 5.15 1.33

on y t gt

Also we have y x

Solve for x y x ft and y ft


y5 m

5 5

10 2814

5 5 5

14/

5ft s

20 0

2

, .

1:

228

: 10 0 0.79865

14 1: 0 1.8718 5 1.8718 3.1282

25

: De

Initial velocity and position are given also given x coordinate of the final position

Apply s s v t at

In x direction t t s

In y direction y t gt y m h m

Another Methode

2 2

termine the distance the squirrel drops in time t.

1.5*9.81*.7986 3.1282

2h gt m


The boy at A attempts to throw a ball over the roof of a barn such that it is launched at an angle of 40 degrees. Determine the speed vA at which he must throw the ball so that it reaches its maximum height at C. Also, find the distance d where the boy must stand so that he can make the throw.

Given: Y0 = 1m, At max. height, vY = 0, Y = 8m and v0x = v0cos400; v0y = v0sin400

(1) X = v0xt; (2) vx = v0x (constant velocity in x-direction);

(3) Y = Y0 1/2gt2 + v0Yt (4) vY = v0Y gt (constant acceleration -g in y-direction)

At the top of the trajectory vy = 0 and from (4) we get: v0y = gt

Substituting the above into (3) we get 7 = ½ g t2 and solving the above to get the time to reach maximum elevation. t = 1.195 seconds Substituting for v0Y at the top of the trajectory, we get v0sin40o = 9.81(1.195)

which is solved for the initial velocity.v0 = 18.24 m/sec

Substituting into (1) we get the total distance X.X = 18.24(cos40)(1.195) = 16.7 mWhich gives d = X 4 = 12.70m


The ball is thrown from the tower with a velocity of 20 ft/s as shown. Determine the x and y coordinates to where the ball strikes the slope. Also, determine the speed at which the ball hits the ground.Initial Conditions:v0 = 20 ft/s v0x = 20(3/5) = 12 ft/s v0y = 20(4/5) = 16 ft/s

theta = 26.60 , X0 = 0 Y0 = 80 ft

Motion in the x-direction at a constant velocity. (1) vx = v0x = 12 ft/s (2) X = vxt = 12t

Motion in the y-direction at a constant acceleration a = -g. Integrating for velocity(3) vy = v0y - gt = 16 - 32.2(t)

and the y-displacement is obtained by integrating the latter, which givesY = -1/2gt2 + v0yt + Y0

or (4) Y = -16.1t2 +16t + 80The x and y coordinates of the ball when it hits the ground can be written as.(5) x = 20 + Rcos26.6o ; (6) y = Rsin26.6o

Equations (2), (4) (5) and (6) are four equations in four unknowns. They can be solved for t - the time when the ball hits the ground. Thus t = 2.7 sec.Substituting into (2) and (4) provides the x and y coordinates of the ball when it hits the ground. x = 32.4 ft; y = 6.56 ftTo solve for the speed, we substitute t = 2.7s into equation (3) which gives.vx = 12 ft/s and vy = 70.9 ft/s

The total speed is found from v = (vx2 + vy

2)1/2 ; v = 71.9ft/s

© samir n. shoukry, 2004, dynamics mae 242. quiz

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