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Final Exam with Basic Formula - Dynamics | MEM 238, Exams of Dynamics

Material Type: Exam; Class: Dynamics; Subject: Mechanical Engr & Mechanics; University: Drexel University; Term: Summer 2013;

Typology: Exams

2013/2014

Uploaded on 09/03/2014

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NAME______________________________
MEM238- Dynamics
Summer 2013
FINAL EXAM SOLUTIONS
(Closed Book, Closed Notes, Show All Work, 5 Problems)
Basic Formulas:
2 2 2
(Law of sines)
sin sin sin
c 2 cos (Law of cosines)

a b c
a b ab
-1 -1 -1
2 2 2
For vector : cos , cos , cos ,
cos cos cos 1 (Relation between direction cosines)
cos (Dot product)

A
AB
y
xz
A
AA
A A A
AB
n)Restitutio ofnt (Coefficie
)( )(
)( )(
Momentum) and ImpulseAngular of (Principle
Momentum) and ImpulseLinear of (Principle)()(
(Power)
Moment) CoupleConstant a of(Work
Spring) ElasticLinear a of(Work
2
1
2
1
2
2
2
1
Energy) and Work of (Principle
Force) a of(Work cos
on)Accelerati Transverse and (Radial 2 ,
2
Velocity) Transverse and (Radial ,
on)Accelerati Normal and l(Tangentia ,
2
,
2
,
,
Motion)Angular -on Accelerati(Constant )(2 ,
2
1
,
Motion)Linear -on Accelerati(Constant )(2 ,
2
1
,
Motion)Angular - ipsRelationsh Kinematic ial(Different , ,
Motion)Linear - ipsRelationsh Kinematic ial(Different , ,
11
22
21
21
2
21
1
21
2
222
222
2
1
G
2
1
G
2
1
2
1
B
B
oo
t
to
t
t
M
s
r
r
n
t
ococooco
ococooco
v
A
vA
vv
e
dt
mdtm
dt
dU
P
MU
ksks
s
U
TUT
dsFdU
rrarra
rvrv
v
ava
r
n
ar
t
a
rv
ttt
ssavvtatvsstavv
dd
dt
d
dt
d
vdvads
dt
ds
v
dt
dv
a
s
HMH
vFv
vF
rF
rrαa
rωv
r
r
pf3
pf4
pf5

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MEM238- Dynamics

Summer 2013

FINAL EXAM SOLUTIONS

(Closed Book, Closed Notes, Show All Work, 5 Problems)

Basic Formulas:

2 2 2

(Law of sines)

sin sin sin

c 2 cos (Law of cosines)

a b c

a b ab

-1 -1 -

2 2 2

For vector : cos , cos , cos ,

cos cos cos 1 (Relation between direction cosines)

cos (Dot product)

  

  

A

A B

y x z

A A A

A A A

AB

(CoefficientofRestitutio n)

( ) ( )

(PrincipleofAngularImpulseandMomentum)

( ) ( ) (PrincipleofLinearImpulseandMomentum)

(Power)

(WorkofaConstantCoupleMoment)

(WorkofaLinearElasticSpring)

(Principle ofWorkandEnergy)

cos (WorkofaForce)

, 2 (RadialandTransverseAcceleration)

, (RadialandTransverseVelocity)

, (TangentialandNormalAcceleration)

, 2 ( )(ConstantAcceleration-AngularMotion) 2

, 2 ( )(ConstantAcceleration-LinearMotion) 2

, , (DifferentialKinematicRelationships-Angular Motion)

, , (DifferentialKinematicRelationships-LinearMotion)

1 1

2 2

1 2

1 2

1 1 2 2

1 2

2

2 2 2

2 2 2

2

1

G

2

1

G

2

1

2

1

B

B

o o

t

o t

t

t

M

s

r

r

t n

o c o o c o c o

o c o o c o c o

v A

v

A

v v

e

dt

m dt m

dt

dU P

U M

ks ks s

U

T U T

U d F ds

a r r a r r

v rv r

v a va

r n

ra t

a

v r

t t t

v v ats s vt at v v a s s

d d dt

d

dt

d

ads vdv dt

ds v dt

dv a

s

H M H

v F v

Fv

F r

a αr r

v ωr

r

r

Problem 1

For the disk shown,  0 = 1rad/s and r = 0.5m. Determine (a)(9pts) the angular velocity

AB of link AB and (b)(9pts) the velocity v B of collar B for the configuration shown in the

figure. If the angular velocity  0 of the disk is constant, determine (c)(6pts) the angular

acceleration  AB

of link AB.

Problem 3

An 8lb uniform rectangular door opens upward and is counterbalanced with a spring as

shown in the figure. The modulus and undeformed length of the spring are k = 24lb/ft and

l o = 23in., respectively. If the door has an angular velocity of 3rad/s counterclockwise

when it is vertical (  = 0

o

), determine the angular velocity of the door when it is

horizontal (  = 90

o

). Model the door as a uniform, slender rod. (20pts)

Problem 4

The horizontal velocity of the landing airplane is 50 m/s, its vertical velocity (rate of

descent) is 2 m/s, and its angular velocity is zero. The mass of the airplane is 12 Mg, and

the moment of inertia about its center of mass is 1 X 10

5

kg m

2

. Neglect the horizontal

force exerted on the wheels by the runway. (a)(3pts) Draw the impulse-momentum

diagram for the airplane. Determine (b)(10pts) the angular velocity of the airplane and

(c)(10pts) the linear velocity of its center of mass just after it touches down if its wheels

don’t stay in contact with the runway and the coefficient of restitution of the impact

is e = 0.4.

Problem 5

The large rotor has a mass of 60kg and a radius of gyration about its vertical axis of

200mm. The small rotor is a solid circular disk with a mass of 8kg and is initially rotating

with an angular velocity  1 = 80rad/s with the large rotor at rest. A spring-loaded pin P

which rotates with the large rotor is released and bears against the periphery of the small

disk, bringing it to a stop relative to the large rotor. Neglect any bearing friction and

calculate the final angular velocity of the assembly. (8pts)