Download Second Mid-Term Exam Solution

Water Resources, Hydrologic and Environmental Sciences
Civil Engineering Department
Fort Collins, CO 80523-1372
(970) 491-7621
CE261 ENGINEERING MECHANICS: DYNAMICS
Fall 2002
SECOND MIDTERM EXAM
Thursday, October 24, 2002 – 5-6:30 PM Hammond Auditorium Room 120 Engineering Building
You may not consult any books, notes, or inanimate references. You may not consult with another
person. You may not copy another student’s solutions.
PROBLEM 1 (20 points)
The block shown is observed to have a
velocity v1 = 20 ft/sec as it passes point
A and a velocity v2 = 10 ft/sec as it
passes point B on the incline. Calculate
the coefficient of kinetic friction µk
between the block and the incline if x =
30 ft and θ = 15°.
PROBLEM 2 (20 points)
The ballistic pendulum is a simple
device to measure projectile velocity v
by observing the maximum angle θ to
which the box of sand with embedded
projectile swings. Calculate the angle θ
if the 2-oz projectile is fired horizontally
into the suspended 50-lb box of sand
with a velocity v = 2000 ft/sec. Also find
the percentage of energy lost during the
impact.
1
Jorge A. Ramirez
PROBLEM 3 (30 points)
The two spheres of equal mass m are
able to slide along the horizontal rod. If
they are initially latched in position a
distance r from the rotating axis with the
assembly rotating freely with an angular
velocity wo, determine the new angular
velocity w after the spheres are released
and finally assume positions at the end
of the rod at a radial distance of 2r. Also
find the fraction n of the initial kinetic
energy of the system that is lost. Neglect
the small mass of the rod and shaft.
PROBLEM 4 (30 points)
The bungee jumper, an 80-kg man, falls
from the bridge at A with the bungee
cord secured to his ankles. He falls 20 m
before the 17-m length of elastic bungee
cord begins to stretch. The 3 m of rope
above the elastic cord has no appreciable
stretch. The man is observed to drop a
total of 44 m before being projected
upward. Neglect any energy loss and
calculate (a) the stiffness k of the bungee
cord (increase in tension per meter of
elongation), (b) the maximum velocity
vmax of the man during the fall, and (c)
his maximum acceleration amax. Treat the
man as a particle located at the end of
the bungee cord and treat the bungee
cord as a linear spring.
2
Jorge A. Ramirez
Water Resources, Hydrologic and Environmental Sciences
Civil Engineering Department
Fort Collins, CO 80523-1372
(970) 491-7621
CE261 ENGINEERING MECHANICS: DYNAMICS
Fall 2002
1/5
GRAVITATION
F =G
b)
m1 m2
r2
as
a
Variable acceleration as
function of velocity, a(v):
a
t
v = vo + ∫ a (τ )dτ
0
me
R2
R2
g = go
( R + h) 2
g =G
2/2
Variable acceleration
function of time, a(t):
t
s = s o + ∫ v(τ )dτ
0
c)
RECTILINEAR MOTION
v
t = to +
ds
= s
v=
dt
d 2s
dv
= v = 2 = s
a=
dt
dt
vdv = ads
sds = sds
dv
∫ a (v )
vo
v
s = so +
vdv
∫ a (v )
vo
d)
where s is displacement, v is velocity, a
is acceleration, all of which are, in
general, functions of time, t.
Variable acceleration as
function of displacement, a(s):
a
s
v 2 = vo2 + 2 ∫ a ( s )ds
so
a)
Constant acceleration
2/3
PLANE CURVILINEAR MOTION
t
v = v o + ∫ a (τ )dτ = v o + at
v=
dr
= r
dt
a=
dv
= v
dt
0
s
v 2 = v o2 + 2 ∫ a ( s )ds = vo2 + 2a ( s − s o )
so
t
s = s o + ∫ v(τ )dτ =s o + vo t +
0
1 2
at
2
2/4
RECTANGULAR COORDINATES
r = xi + yj
v = r = xi + y j
a = v = r = xi + yj
where the initial conditions at time to = 0
are so, and vo.
1
Jorge A. Ramirez
v 2 = v x2 + v y2
Circular motion
v = v x2 + v y2
tan θ =
v = rθ
a n = v 2 / r = rθ 2 = vθ
a = v = rθ
vy
vx
t
a 2 = a x2 + a y2
2/6
POLAR COORDINATES (r-θ)
a = a x2 + a y2
r = re r
e = θe
r
Projectile motion
e θ = −θe r
v = re + rθe
v x = (v x ) o
r
x = x o + (v x ) o t
a = (r − rθ )e r + (rθ + 2rθ)e θ
v r = r
1 2
gt
2
v y2 = (v y ) o2 − 2 g ( y − y 0 )
y = y o + (v y ) o t −
vθ = rθ
v = v r2 + vθ2
Trajectory equation:
a r = r − rθ 2
a = rθ + 2rθ
gx 2
sec 2 θ
2u 2
θ
a = a r2 + aθ2
corresponding to an initial position xo =
0; yo = 0; initial velocity of magnitude u
and making an angle θ with the
horizontal.
2/5
2/7
SPACE CURVILINEAR MOTION
a)
Rectangular coordinates
R = xi + yj + zk
= xi + yj + zk
v=R
NORMAL AND TANGENTIAL
COORDINATES (n-t)
= xi + yj + zk
a = v = R
v = ve t = ρβe t
b)
a=
v2
an =
ρ
ρ
Cylindrical coordinates (r-θ-z)
R = re r + zk
v = re r + rθe θ + zk
a = (r − rθ 2 )e r + (rθ + 2rθ)e θ + zk
e n + ve t
v2
θ
2
v y = (v y ) o − gt
y = x tan θ −
θ
= ρβ 2 = vβ
at = v = s
a = a n2 + at2
2
Jorge A. Ramirez
•
Spherical coordinates (R-θ-φ)
c)
Rectangular coordinates:
∑F
∑F
v = v R e R + vθ e θ + vφ e φ
v R = R
vθ = Rθ cos φ
v = Rφ
x
= ma x
y
= ma y
a = axi + a y j
φ
a = a R e R + aθ e θ + aφ e φ
a = a x2 + a y2
− Rφ 2 − Rθ 2 cos 2 φ
aR = R
•
cos φ d
( ( R 2θ)) − 2 Rθφ sin φ
R dt
1 d
aφ = ( ( R 2φ)) + Rθ 2 sin φ cos φ
R dt
aθ =
2/8
Normal
and
coordinates:
∑F
∑F
RELATIVE MOTION (Translating
Axes)
rA = rB + rA / B
r A = rB + r A / B
•
n
= ma n
t
= ma t
Polar coordinates:
∑F
∑ Fθ
v A = v B + v A/ B
rA = rB + rA / B
Tangential
r
= ma r
= maθ
a A = aB + a A/ B
3/6
Law of Sines
a)
a
b
c
=
=
sin A sin B sin C
•
dU = F ⋅ dr = F cos αds
a 2 = b 2 + c 2 − 2bc cos A
a)
dU = Ft ds
U 1− 2 = ∫ F ⋅ dr = ∫ ( Fx dx + Fy dy + Fz dz )
KINETICS OF PARTICLES - Force,
Mass, and Acceleration
s2
U = ∫ Ft ds
s1
Newton’s Second Law
•
ΣF = ma
3/5
Calculation of Work
dU = F ⋅ dr
Law of Cosines
3/2
KINETICS OF PARTICLES - WORK
AND ENERGY
Work and Curvilinear Motion
PLANE CURVILINEAR MOTION
3
Jorge A. Ramirez
2
s2
1
s1
V g = mgh
U 1− 2 = ∫ F ⋅ dr = ∫ Ft ds
2
∆V g = mg (h2 − h1 ) = mg∆h
mgR 2
Vg = −
r
1 1
∆V g = mgR 2 ( − )
r1 r2
2
U 1− 2 = ∫ F ⋅ dr = ∫ ma ⋅ dr
1
1
v2
2
1
U 1− 2 = ∫ ma ⋅ dr = ∫ mvdv = m(v 22 − v12 )
2
1
v1
U 1− 2 =
•
•
1
m(v 22 − v12 )
2
x
1
Ve = ∫ kxdx = kx 2
2
0
Principle of Work and Kinetic
Energy
T=
∆V e =
1 2
mv
2
•
U 1− 2 = T2 − T1 = ∆T
•
1
k ( x 22 − x12 )
2
Work-Energy Equation
U 1'− 2 = ∆T + ∆V g + ∆Ve
Power
P = F⋅v
em =
•
Elastic Potential Energy (Springs)
3/9
Pout
Pin
•
KINETICS OF PARTICLES –
LINEAR IMPULSE AND LINEAR
MOMENTUM
Linear momentum:
Work done by Gravitational Forces
G = mv
U 1− 2 = −mg ( y 2 − y1 )
U 1− 2 = mgR E2 (
d
∑ F = mv = dt (mv) = G
1 1
− )
r2 r1
∑F
∑F
∑F
y
= G x
= G
z
= G z
x
•
Work done by Springs
x
x
2
2
1
U 1− 2 = − ∫ Fdx = − ∫ kxdx = − k ( x 22 − x12 )
2
x1
x1
3/7
•
•
POTENTIAL ENERGY
Linear Impulse and Linear
Momentum
t2
∫ Fdt = mv
Gravitational Potential Energy
y
2
− mv 1 = G 2 − G 1 = ∆G
t1
4
Jorge A. Ramirez
t2
t2
∫ ∑ F dt = (mv
x
∫ ΣM
) − (mv x )1
x 2
t2
∫ ∑ F dt = (mv
y
) − (mv y )1
t2
∫ Σ( M
y 2
t1
z
dt = (H o ) 2 − (H o )1 = ∆H o
o
) x dt = ( H o ) x 2 − ( H o ) x1
t1
t2
∫ ∑ F dt = (mv
o
t1
t1
z
) 2 − (mv z )1
= m[( yv z − zv y ) 2 − ( yv z − zv y )1 ]
t1
•
•
Conservation of Linear Momentum
∆G = 0 or G 1 = G 2
IMPULSE
3/10 ANGULAR
ANGULAR MOMENTUM
Plane motion applications
t2
∫ ΣM
o
dt = ( H o ) 2 − ( H o ) 1
t1
AND
t2
∫ ΣFr sin θdt = mv d
2
•
2
− mv1 d 1
t1
Angular momentum
H o = r × mv = ( H o ) x i + ( H o ) x j + ( H o ) x k
i
j
k
Ho = m x
y
z
vx
vy
vz
( H o ) x = m( yv z − zv y )
( H o ) y = m( zv x − xv z )
( H o ) z = m( xv y − yv x )
•
Rate of change of angular
momentum
ΣM o = r × ΣF = r × mv = H
o
Σ( M o ) x = ( H o ) x
Σ( M ) = ( H )
o
y
o
y
Σ( M o ) z = ( H o ) z
•
Principle of Angular Impulse and
Angular Momentum
5
Jorge A. Ramirez