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10 February 2005
Dear Recipient,
You are privileged to obtain a partial collection of answers for MEG 207 Dynamics. This
document was compiled from PDF files collected during Fall of 2002 and Spring 2003. I have
complied them into one document with Bookmarks to aid you in quickly finding the solutions you
need. Unfortunately this isn’t a complete compilation of all of the answers, but comprehensive
enough to help complement your studies.
Engineering Mechanics, Dynamics, Second Edition.
ISBN: 0-471-05339-2
Happy Studies!
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Chapter 14
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Chapter 15
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Chapter 16
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Chapter 17
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Old Exams
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Name:___________________
Last
First
UNLV, DEPARTMENT OF MECHANICAL ENGINEERING
MEG 207, SPRING 2002, FIRST TEST
Closed Book, one page of handwritten notes allowed. Enter the answer for each question
into the space provided. Enter SI units in all answer spaces with brackets ( ).
2
1. (15 points) A vehicle traveling at 108 km/h suddenly decelerates at a rate of 3 m/s .
a) Determine the time needed for the vehicle to come to rest.
b) Determine the distance traveled between the beginning of braking, and the full stop.
1(a) v0 := 108⋅
1000
in m/s
a := −3
3600
v0 + a⋅ t General equation. Solving for t when v=0
gives: t := v0 and
t = 10
−a
1
2
2
1 (b) v*dv = a*dx or a ⋅ distance
⋅ ( 0 − v0)
−v0
distance
:=
distance = 150
2
2⋅a
v ( t)
Answers: a)
Tstop =
10
b)
dStop =
150
(
s
units)
( m
units)
2. (20 points) A ball of mass m is thrown horizontally from a bridge 30 m above ground.
It touches ground at distance d = 15 m. Determine the ball's initial velocity. No friction.
v0
g
A (x0,y0)
y
h = 30 m
x
horiz. distance
d = 15 m
B
Problem 2
d := 15 h := 30 Angle is zero.
g := 9.81
Horizontal: d = v0*t. Vertical: y(t) = y0 -1/2*g*t^2. At point B, y = 0.
Given
d v0⋅ t
0
h − 0.5⋅ g ⋅ t
2
res := Find( t , v0)
Answer
v0 =
6.065
res =
 2.473 
 6.065 
(
m/s
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3. (20 points) Pin P moves at constant speed of 3 m/s in
a counterclockwise sense around the circular slot with
radius r = 2 m. Determine (a) the angular velocity of P.
(b) the total acceleration vector ( Use polar coordinates:
er and eθ directions) of pin P when θ = 30 degrees.
(c) the magnitude and angle of the resultant acceleration
vector in Cartesian x-y coordinates.
P
y
2m
R=
θ
x
(b) no radial or angular accel. In neg. er direction we
2
2
have r*ω = - 2*2.25 m/s
(c) Resultant Acceleration is purely inward, thus:
o
o
ax = -4.5*cos 30 and ay = -4.5*sin 30
4.5
Answer
ω =
v/r = 3/2 = 1.5
a =
rad/s
-4.5
e
0
2
4.5 m/s at -150 de rees
a =
eθ
2
m/s )
(ma nitude and an le in x- coordinates)
20
4. (25 points) Wind-driven rain is falling
with a speed of 30 m/s at an angle of 200
to the vertical as shown at left.
Determine the angle ψ at which the rain
is seen by passengers inside the bus. The
bus is traveling at 72 km/h.
0
a
R
in
ψ
y
v = 72 km/h
Vrain = VBus + VRain/bus
Vbus = 20m/s
x-and y-components of VRain/bus:
v R/B,x = -Vbu ---VR*sin(20deg) = -30.2 m/s
VRain/bus
v R/B,y =---VR*cos(20deg) = -28.19 m/s
-1
= tan (29.19/30.2) = 43 deg.
An le
seen b movin
assen ers =
ψ
43 de .
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5. (20 points) In the pulley system shown at left, the cable is attached at C. Mass B
moves to the left at vB = 3 m/s, and accelerates
y
2
B
also to the left at aB = 0.5 m/s . Using the x-y
x
C frame with origin at C, determine:
(a) the velocity of A
(b) the acceleration of A
A
L = 2xB(t) +(Y0 --- yA(t))
Differentiation gives:
2 vB = vA
and
2 aB = aA
Answer
vA =
aA =
6 m/s
2
1 m/s
(
)
(
)
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Na me:___________________
La st
F ir st
U NLV, DE P AR TMENT OF MEC H ANIC AL ENG INEE R ING
MEG 207, Spring 2002, Third Test
Closed B ook, one pa ge of ha ndw ritten n otes a llowed. E nter t he a nswer for each question
into the space provided. Ent er SI units in all answer spaces with brackets ( ).
1. (25 points) The thin, uniform rod (Length L, mass m,
2
I n er t i a : I G = m L / 12 . )
rotates
about fixed point A as it is released from rest in the horizontal position shown. At the
instant immediately following the release, determine
(a ) th e a ngula r a ccelerat ion of th e rod.
(b) the a ccelera tion, a B , of th e point B .
Sum of moments a bout A:
J
m
mg *L/2 = I A*α
B
A
I A = 2I G + m*(L/2)2 = (1/12+ 1/4)*m L 2 =
i
Unit vectors g
= m L /3
L
I n ser t in g in t o t h e fir st eq ua t ion
gives:
2
mg *L/2 = mL /3 *α
after simplification and regrouping we get:
αRod = 3g/(2L)
Part (b):
A is center of rotation. Thus a B = L*α = 3g/2
Answe r s: a ) αRod = 3g/(2L)
b) a B
= - 3g/2 in ---j-di re cti on
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2. (30 points) A uniform plat e of steel of length L = 1 m a nd m a ss m = 200 kg is placed
o nt o a t r u c k b e d a s sho wn. The p la t e c a nno t slide a t p o int A, b u t c a n r o t a t e a b o u t A.
Determine the maximum allowable acceleration of the truck so that the plate does not
rotate.
B
Bx
At ma x. a ccel, cont a ct force B x = 0.
J
From free-body diagram:
L
Ax = m*a x
Ay = m*g
i
mg
60 0
Ax*L /2*sin (α) --- Ay *L /2*cos (α) = I g*α = 0
atruck
A
Ax
Truck
Inserting the top two equations into the third :
Ay
m*a x = m*g* cos(α)/sin(α)
a ma x = g* cot(α) = 5.664 m /s
Answer aTruck,max =
5.664
2
(
m/s
2
Unit s)
3. (20 points) The s prin g (k = 1200 N/m) is in it ia lly
k = 1.2 kN/m
g
compressed by 2 m. The 3-kg block shown is not
µk = 0.2
attached
to the
spring.
After surfa
release
rest, the
block tr a vels
along
th e rough
ce (µfrom
k = 0.2).
Determine t he position x final a t w hich th e block comes
to rest.
Mass = 3 kg
2 meters
Spring is compressed
at time of release
Uncompressed
spring position
x
Forces doing work:
Spring: F srping
Friction F frict
ds
ds
2
T1 + U 1-2 = T2 w i t h T1 = T2 = 0. Thus U 1-2 = ½ *k*(2meters) ---µk*m*g*x = 0
2
x = 2*1200/(0.2*3kg&9.81m /s ) = 407.7 m
An sw er x final =
407.7
(
m
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4. (25 Points) A parcel of ma ss m is relea sed at a n initia l velocity v 0 = a s shown.
(a ) De t er mine t he minimu m velocit y a t B (immedia t e ly b efor e r ea c hing t he u pp er
conveyor) so tha t the pa rcel does not drop at B .
(b) Determine the minimum velocity v 0 at point A so that the parcel reaches point B at
th e minimum velocity computed in part (a ).
G iven: Radius r , ma ss m, g. No friction. Neglect the pa ckage t hickness!
(Use Energy method)
Answer (a)
In order not to fall
v 2/r
off a t B , g r a vit y
B
must be offset by the
C
m
centripetal
g
acceleration:
2
g
v /r = g
r v
A
or
0
2
v , min = g*r
P a r t (b )
2
T1 + U 1-2 = T2 w i t h T1 = ½ *m *v 0 a n d T2 = ½ *m*g*r
O n ly g r a v it y d oes w o rk : m g
ds
Thus
2
½ *m*v 0 - 2r*m*g = ½ *m*g*r
After simplification, we get:
2
v 0 = 5gr
An s w er : (a ) v B,min =
(b ) v 0,min =
1/2
(gr)
1/2
(5gr)
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Na me:_____KE Y________
L a st
F ir st
UNIVERSITY OF NEVADA, LAS VEGAS
DEP AR TMENTOF MEC H ANIC AL ENGINEER ING
MEG 207, Spring 2002, Final Examination
Closed Book, tw o pages of han dw ritten notes allow ed. Enter t he an swer for each question into the
space provided. Enter correct dimensional SI units where applicable.
1. (10 points) A point ma ss of 10 kg is tossed horizont a lly from 30 m a bove ground. The
mass land s on th e ground a t a distan ce of 40 m from a point on the ground directly below
the tossing point. Determine the initial velocity of the
mass.
v 0*t = x 1 = 40 (1)
2
y 1 = 0 = y 0 ---½
- *g*t
(2)
fr om (1): t = sq rt (2/g* y 0) = 2.47 s
inserting into (1) gives: v 0 = 40/t = 16.17 m/s
Answ er: v 0 =
16.17 m/s
2. (15 points) A point mass m is suspended from two wires AB an d CD. D etermine the
tension in the other (b)
wireimmediately
CD
before
AB
cut
(a)after
AB
is is
cut.
A, eR
B
mg
eθ
mg
A
o
o
(a ) La w of sines: A/sin (40 ) = mg /sin(120 )
Thus: A = 0.742*mg
(b) After AB is cut: Reaction B disa ppea rs. U sing pola r coordina tes:
Summing in radial dir: A-mg*cos(20 o) = m*r_ddot = 0, thus A = mg*cos(20 o) = 0.94*mg
(a ) Tension in Ca ble CD before cut =
(b) Tension in Cable CD after cut
0.742*mg
= 0.94*mg
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3. (20 points) A collar with ma ss 5 kg slides in th e vertica l plane a long t he curved rod
show n. It is a tta ched to a n elastic spring w ith un deformed length of 150 mm, and k = 600
N/m. The collar is released from rest at A.
Determine the collar's speed at B.
T1 + V1 = T2 + V2
2
½
k (x -x 0)
+
0.1meters
g
mgh = ½
mg*0.2
T1 = 0
2
mv 2
+
½
2
k(x 2-x 0)
0.05 met ers
inserting the da ta a nd solving for v 2 gives:
2
v 2 = k(0.12 ---- 0.052) + 2mg*0.2
2
2 2
v 2 = 0.9 + 3.924 = 4.824 m /s
vB
=
2.196
(
m/s
unit s)
4. (15 points) The 12-kg mass B is dropped with a horizontal velocity v 0 = 2.5 m/s ont o th e
30-kg luggage cart (µk = 0.5), which is initially at rest and can roll freely. Determine the
velocity of the luggage cart after mass B has reached
the sa me velocity a s th e car t.
An approximate solution can be found using the energy
principle. This neglects the energy loss due to friction
a s th e package slides on the cart.
2
T1 + V1 = T2 + V2
T1 = ½ m B v 1
U se dat um line at A. there is no cha nge in potential: V1 = V2 = 0
2
2
So we get: T1 = ½ m B v 1 = ½ (m B + m cart )*v 2
2
2
v 2 = (12*6.25)/42 = 1.78 m /s
Answ er (a ) v
=
1.336
2
(
m/s
unit s)
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5. (20 points) Center B of t he double pulley ha s a velocity of 0.6 m/s, a nd a n a ccelera tion
2
of 2.4 m/s , both directed downwa rd. R= 2m, r = 0.8 m. Determine the total acceration
vector of point D.
Point C is an instantaneous center.
ω = v B /r = -0.75 ra d/s clockw ise
2
α = a B /r = -3 r a d/s clockwise
a D = (0 --- 2.8* ω2)∗i + 2.8*α*j
a D = --- 1.575∗i -8.4*j
aD
--- 1.575∗i -8.4*j
=
(
m/s
unit s)
6. (20 points) The uniform rod AB with mass m an d length L is relased from rest at an
a ng le of Θ0 a s shown. Assu ming t ha t no sliding occu r s, det er mine (a ) t he a ng u la r
acceleration of the rod just after release. (b) the normal reaction and friction force at point
A just a fter release.
Pure rotation about A.
θ0
J
i
B
½ *L * α
L
g
Summing moments about A:
2
-1/2*L *sin (θ0)*mg = I B *α where IB = mL /3
mg
thus α = -3/2*g /L* sin (θ0)
(B ) Rea ctions a t A:
Forces in x-dir: F = m*a x = m*1/2*L*cos(θ0)*α
A
Forces i n y-dir : N --- mg = -m*1/2*L*sin (θ0)*α
2
or
ÿ
Rod
N=
N = m g[1- ¾ * sin (θ0) ]
=
α = -3/2*g /L* sin (θ0)
2
m g[1- ¾ * sin (θ0) ]
F = ¾ *mg* cos(θ0)* sin(θ0)
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Note: Exam results are up significantly in comparison to exam 2.
Name:_____KEY________
Last
First
UNLV, DEPARTMENT OF MECHANICAL ENGINEERING
MEG 207, Spring 2003, Third Test
Closed Book,
page in
ofall
handwritten
notes
allowed.
Enter
provided.
Enterone
SI units
answer spaces
with
brackets
( ). the answer for each question into the space
2
1. (25 points) A disk(mass m= 2kg, r= 0.1m Inertia: IG = mr /2.) is rigidly attached to massless rod AB
(Length R = 0.5m) and released from rest in the horizontal position. At the instant immediately following
the release, determine
(a) the angular acceleration of the rod.
(b) the acceleration, aB, of the point B.
1
0.
B
A
m
r=
V
R=0.5m
B
(fixed)
g
ANSWER
Newton: Sum of moments about A: using L = R. m B = m a ss a t B
m B *g*R= I A*α
I A = I G + m B *(R)2 = (0.01/2+ 0.25)* m B = 0.255*m B
Inserting into th e first equa tion gives:
m B g*R = 0.255*m B *α
after simplification and regrouping we get:
αRod = R*g /(0.255) = 0.5*9.81/0.255 = 19.2 ra d/s 2
Part (b):
2
A is center of rotation. Thus a B = R*α = 0.5∗α = 9.62 m/s
Answers: a) αRod = 19.2 ra d/s 2
2
b) aB = R*α = 0.5∗α = 9.62 m/s
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2. (25 points) A uniform rectangular crate of height = 0.5 m, width 0.2m and mass m = 200 kg is placed
onto a truck bed as shown. The crate cannot slide at point B, but can rotate about B. Determine the
maximum allowable acceleration of the truck so that the crate does not rotate.
[
0.2m
J
mg
Crate
g
i
m
.5
0
FB
atruck
B
N
N = mg
Moments about G: 0.25*FB - 0.1*mcrate*g = 0
(1)
Forces on crate in x-direction: FB = mcrate*atruck
(2)
Insert (2) into (1): 0.25* mcrate*atruck = 0.1*mcrate*g
atruck = g/2.5
2
Answer aTruck,max = 3.924
(
m/s
Units)
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3. (30 points) The spring (k = 1500 N/m) is initially uncompressed. The 3-kg block shown is released from
rest 4m above the spring and falls onto the spring. Using the energy method, determine
(a) the velocity at which the block hits the spring at y = 0 m.
(b) the amount of maximum spring
compression,
by
the
δ, caused
impacting block.
y
k
g
3
g
Answer
2
T2 – T 1 = - INT( [-mg]*dy = ½*m*v2
m
4
T1 = 0
2
2
2
v2 = 2gy =8g =78.48 m /s
(c)
2
(d) T
T –
=T
0 = ½*k*δ + m*g*δ
3
3
Spring is initially
uncompressed
2
δ
k = 1.5 kN/m
2
T =½*m*v
2
2
2
2
2
= ½*3*kg*78.48 m /s = ½*1500* δ + m*g*δ
We now solve for δ
(substitute d = δ a t r i gh t )
Answer
T2
2
750 ⋅d + 3 ⋅9.81 ⋅d
vimpact (y=0) =
8.86
(
m/s
Units)
δmax. compression =
0.377
(
m
Units)
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0
4. (20 Points) A point mass pendulum (mass m, Length L) is released from rest at an initial angle θ0 = 45
as shown.
(a) Determine the maximum velocity of the pendulum
(b) Determine the pendulum tension at θ = 0.
Given: Length L , mass m, g. No friction. (Use Energy method)
B
fixed
(a)
2
T2 – T 1 = ½*m*v2 = mgL(1- cos(θ0)
Rope length L
T1 = 0 (release from rest)
g
2
v2 = 2g L(1- cos(θ0)
θ0 = 450
m
1
h =L (1- cos(θ0)
θ=0
v2 =vmax
2
(b) Max. tension Force exists at (2): Sum of forces in radial dir. (inward towards center B):
2
T –mg = m* v2 /L
2
T = m(g+ v2 /L)
Answer: (a) vmax =
[ 2g L(1- cos(θ0)]1/2
(b) Max.Tension Force at θ = 0 =
2
m(g+ v2 /L)
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Name:________KEY______
Last
First
UNIVERSITY OF NEVADA, LAS VEGAS
DEPARTMENT OF MECHANICAL ENGINEERING
MEG 207, SPRING 2003, Final Examination
Closed Book, three pages of handwritten notes allowed. Enter the answer for each question into the space provided.
Enter correct dimensional SI units where applicable.
1.
(15 points) A player throws a ball horizontally from point A at an elevation 2m above ground.
The ball lands on the flat ground at point B, at a distance of 20 m from the origin.
Determine
(a) the initial velocity at which the ball was thrown
(b) the time elapsed between points A and B
y
A (x0,y0)
vo
g
yo = 2 m
x
horiz. distance
d = 20 m
B
The only acceleration is g in –j-direction. vo = vo*i. After integrating twice in i-direction:
xB = 20 = vo*t + 0 (xo = 0)
(1)
2
After integrating in j-direction: yB = 0 = 0 – ½*g*t + 2 (yo = 2)
2
From (2): t = 4/g
ÿ
Answer: vo = 31.3
t = 0.64
inserting into (1) gives: vo = 20/t = 20*(g/4)
(2)
1/2
= 31.3 m/s
m/s
s
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2.
g
k = 200 N/m
Mass =
F= 50 N
10 kg
no friction
2
x
unstretched
(15 points) A constant force F = 50 N is applied to mass
m=10 kg, which is initially at rest with the attached
spring (k = 200 N/m) unstretched. Determine
(a) the maximum velocity of the mass.
(b) the maximum stretch of the spring.
spring position
Spring is unstretched
at time of release
F ⋅x −
1
2
1
2
⋅k ⋅x
F
−
2
k ⋅x
⋅m⋅( v( x) )
2
max
= 1.25 (m/s)
2
The maximum velocity occurs
m ⋅ v ( x) ⋅
d
dx
when the derivative of the velocity is zero:
thus v
2
(a) T2 – T 1 = F*x - ½*k* x = ½*m*v2
v ( x)
= 0, x = F/k = 50/200 = 0.25,
2
2
(b) Both start and final velocities must be zero: T2 – T 1 = F*x - ½*k* x = 0
Two solutions: x1 = 0 (start point)
x2 : 2F - k* x = 0
x2 = 2F/k = ½ meters
(a) vmax
(b) xmax
=
=
J
1.12
0.5
m/s
m
VB/A
B
C
3. (20 points) Knowing that vΑ
= 5 m/s is constant to the left,
determine for the instant shown :
(a) the angular velocity of rod
BC
10 m
i
4m
o
(b) the
angular acceleration of
rod AB.
(a) Vector equation:
VB = VA + VB/A
o
o
Thus: RB/C*ωBC*j = -vA*i + RB/A*ωAB*(-sin30 i + -cos30 j) Components:
o
o
i: vA = RB/A*ωAB*sin30 ÿ
ωAB = vA /(RB/A*sin30 ) = - 5/(10*0.5) = -1 rad/s
o
o
j: RB/C*ωBC = RB/A*ωAB *cos30 ÿ ωBC = -RB/A*ωAB cos30 /RB/C = +10*0.866/4 = 2.165 rad/s;
Unit vectors
30
A
vA = 5 m/s
(b) aA = 0. only centripetal and angular accel terms exist. Vector eq. AB = AA + AB/A
2
2
o
o
o
o
-RB/C*ωBC *i + RB/C*αBCj = RB/A*ωAB *(-sin30 j + cos30 i) + RB/A*αAB*(-sin30 i - cos30 j)
2
o
o
j: RB/C*αBC = -RB/A*ωAB *sin30 - RB/A*αAB* cos30
2
2
o
o
i: -RB/C*ωBC = RB/A*ωAB * cos30 - RB/A*αAB*sin30
2
αAB = -(RB/C*ωBC
2
o
o
+ RB/A*ωAB * cos30 )/(RB/A*sin30 ) = - (4*4.69 + 10*1*0.866)/5 = - 5.48 rad/s
ωBC =
2.165 rad/s
αAB =
- 5.48 rad/s cw
2
2
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4. (15 points) The frictionless system of masses A (40 kg) and B (10 kg) is released from rest.
Determine:
(a) the acceleration of mass A.
(b) the tension in the cable
Mass A moves in i-direction, and B in j-direction
A
g
m = 40 kg
J
B
only.
Mass A moves in I-direction, and B in j-direction only.
Newton: A: ÿ Fx = T = m A ÿxÿ Constraint eq.: x = -y
B:
m = 10 kg
i
Unit vectors
ÿxÿ =
mB g
ma
+
mB
ÿ Fy
=
T
− mB
*g
=
m B ÿyÿ = −m B ÿxÿ
2
= 98.1/50 = 1.962 m/s
2
(b) Using first eq.: T = 40kg*1.962 m/s = 78.5 N
2
(a) aA =
1.962 m/s
(b) T =
78.5 N
5. (15 points) Gear D is stationary. Gear C has radius r C = 100 mm. Bar AB has a length of 200
mm. As Bar AB rotates counterclockwise at ωAB = 5 rad/s, determine
(a) The angular velocity of gear C
(b) The total acceleration vector (in Cartesian i-j
coordinates) of point P on gear C. Point P is
located 100 mm below B, normal to the line AB.
J
Ctr
i
Unit vectors
P
(a) The Contact of gear D with gear B is an inst. Center. We have: v B = 200*ωΑB = 100*ωC
= 2*5 = 10 rad/s
(b) AP = AB + AP/B aP =
ωC
ωC
Vector
=
aP =
10
2
- 200*5 *i +
2
100*10 *j
rad/s
-5 i +
10 j
2
m/s
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6. (20 points) The uniform rod AB with mass m is released from rest at an angle of 60 degrees as
shown. Assuming that no sliding occurs, determine (a) the angular acceleration of the rod just after
release. (b) the normal reaction and friction force at point B. Moment of Inertia for the rod, at center
2
of mass: IG = mL /12.
A
J
L
i
g
mg
60 0
B
We have a pure rotation about
fixed point B.
Parallel axis Theorem: IB = IG
2
2
+ m(L/2) =mL (1/12+1/4) =
2
mL /3
Newton about B:
1
L
ÿ M B = − 2 mg * 2
=
I B *α
=
α = -3g/(4L) (oriented in k-direction normal to the i-j plane)
(b) Reactions: Newton in x- and y- directions:
ÿyÿ = −
L
ÿ Fy
=
N
−
mg
ÿ Fx
=
F
=
mÿxÿ = mg * 0.433
α
4
= −3g
/ 16
=
ÿxÿ = −
L
2
mÿyÿ
α
* 0.866 = + g * 0.433(Crossprodu ct
>
0)
N = m(g – 3g/16) = 13mg/16
α
L/2
0.866*L
L/4
αRod =
3g/(4L)
Ν=
13mg/16
F = 0.433*mg
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