Eastern Mediterranean University
Faculty of Engineering
Department of Mechanical Engineering
Rigid Body Dynamics (MENG233)
Instructor:
Assistant Professor Dr. Mostafa Ranjbar
Assistant:
Sadegh Mazloomi
Homework #2
Sunbmitted By:
Name:
ST ID
Submission Deadline:
22th April 2015
Q1-
A sphere is fired downwards into a medium with an initial
speed of 27 m>s. If it experiences a deceleration of
a = ( -6t) m>s2, where t is in seconds, determine the
distance traveled before it stops.
SOLUTION
Velocity: v0 = 27 m>s at t0 = 0 s. Applying Eq. 12–2, we have
A+TB
dv = adt
v
L27
t
dv =
L0
- 6tdt
v = A 27 - 3t2 B m>s
(1)
At v = 0, from Eq. (1)
0 = 27 - 3t2
t = 3.00 s
teaching
Web)
or
Distance Traveled: s0 = 0 m at t0 = 0 s. Using the result v = 27 - 3t2 and applying
Eq. 12–1, we have
A+TB
Dissemination
Wide
copyright
States
World
permitted.
A 27 - 3t2 B dt
the
not
instructors
is
L0
learning.
L0
t
ds =
of
s
in
ds = vdt
(2)
student
the
by
and
use
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on
s = A 27t - t3 B m
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assessing
s = 27(3.00) - 3.003 = 54.0 m
of
protected
the
(including
for
work
At t = 3.00 s, from Eq. (2)
Ans.
Q2-
It is observed that the time for the ball to strike the ground
at B is 2.5 s. Determine the speed vA and angle uA at which
the ball was thrown.
vA
uA
A
1.2 m
50 m
SOLUTION
Coordinate System: The x–y coordinate system will be set so that its origin coincides
with point A.
x-Motion: Here, (vA)x = vA cos uA, xA = 0, xB = 50 m, and t = 2.5 s. Thus,
+ B
A:
xB = xA + (vA)xt
50 = 0 + vA cos uA(2.5)
vA cos uA = 20
(1)
y-Motion: Here, (vA)y = vA sin uA,
= -9.81 m>s2. Thus,
yB = -1.2 m,
ay = -g
and
or
yA = 0 ,
Web)
laws
1
a t2
2 y
teaching
yB = yA + (vA)y t +
in
A+cB
(2)
learning.
is
of
the
not
instructors
States
World
vA sin uA = 11.7825
by
and
use
United
on
Solving Eqs. (1) and (2) yields
work
student
the
vA = 23.2 m>s
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the
(including
for
uA = 30.5°
Ans.
permitted.
Dissemination
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copyright
1
(- 9.81) A 2.52 B
2
-1.2 = 0 + vA sin uA (2.5) +
B
sale
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destroy
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and
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courses
part
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is
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provided
integrity
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and
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assessing
is
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of
protected
the
(including
for
work
student
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and
use
United
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learning.
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of
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permitted.
Dissemination
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in
teaching
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laws
or
Q3-
Determine the velocity of t he log if t he truck at C pulls
the cable 4 m/s to the right.
B
SOLUTION
2sB + (sB - sC) = l
3sB - sC = l
3VB - VC = 0
Since Vc= - 4 then
3VB = - 4
VB = - 1.33 ft = 1.33 ft :
Ans.
C
Q4At the instant shown, cars A and B are traveling at speeds
of 25 m/s and 30 m/s, respectively. If B is increasing its
speed by 5 m>s2, while A maintains a constant speed,
determine the velocity and acceleration of B with respect
to A. Car B moves along a curve having a radius of
curvature of 300 m.
vB = 30 m/s
B
A
vA = 25 m/s
SOLUTION
vB = - 30 cos 30°i + 30 sin 30°j = {-25.98i + 15j} m/s
vA = {-25i } m/s
vB>A = nB - nA
= (-25.98i + 15j) - (-25i) = {-0.98i + 15j} m/s
vB>A = 20.982 + 152 = 15.03 m/s
(aB)t = NT2
or
v2# 02
=
= 3NT2
r
Ans.
teaching
Web)
(aB)n =
= .°
laws
u = tan - 1
Ans.
Wide
copyright
in
aB = ( cos 60° - cos 30°)i + ( sin 60° + sin 30°)j
permitted.
Dissemination
NT
instructors
States
World
= {i + j}
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learning.
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aA = 0
student
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aB>A = aB - aA
the
(including
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= \J K^- 0 = {JK}
Ans.
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5.1
=119°
u = tan - 1
-2.83
work
this
assessing
is
work
solely
of
protected
aB>A = 2()2 + ()2 =NT2
30°
Q5Block A weighs 10 N and block B weighs 3 N. If B is
moving downward with a velocity 1vB21 = 3 m >s at t = 0,
determine the velocity of A when t = 1 s . The coefficient of
kinetic friction between the horizontal plane and block A is
mA = 0.15. (Use impulse and momentum concept)
A
SOLUTION
sA + 2sB = l
(vB)1
vA = -2vB
mv1 + ©
10
10
(v )
(2)(3) - T(1) + 0.15(10)(1) =
9.81
9.81 A 2
F dt = mv2
-(vA)2
b
2
in
a
or
3
3
(3) + 3(1) - 2T(1) =
9.81
9.81
Web)
L
laws
mv1 + ©
Dissemination
Wide
copyright
- T - 1.019(vA)2 = 4.616
not
instructors
States
World
permitted.
- 2T + 0.153 1vA22 = - 3.917
and
use
United
on
learning.
is
of
the
T = 1.50 N
by
(vA)2 = - 6.00 m>s = 6.00 m>s :
the
(including
for
work
student
the
Ans.
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assessing
is
integrity
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is
This
destroy
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1 + T2
F dt = mv2
teaching
-
L
sale
+ 2
1;
3 m /s
B
Q6-
A 750-mm-long spring is compressed and confined by the
plate P, which can slide freely along the vertical 600-mm-long
rods. The 40-kg block is given a speed of v = 5 m>s when it is
h = 2 m above the plate. Determine how far the plate moves
downwards when the block momentarily stops after striking
it. Neglect the mass of the plate.
v ⫽ 5 m/s
A
h⫽2m
SOLUTION
Potential Energy: With reference to the datum set in Fig. a, the gravitational potential
energy of the block at positions (1) and (2) are A Vg B 1 = mgh1 = 40(9.81)(0) = 0
P
and A Vg B 2 = mgh2 = 40(9.81) C - (2 + y) D = C - 392.4(2 + y) D , respectively. The
compression of the spring when the block is at positions (1) and (2) are
s1 = (0.75 - 0.6) = 0.15 m and s2 = s1 + y = (0.15 + y) m. Thus, the initial and
final elastic potential energy of the spring are
1
2
1
2
1
2
1
2
k ⫽ 25 kN/m
A Ve B 1 = ks21 = (25)(103)(0.152) = 281.25 J
laws
or
A Ve B 2 = ks22 = (25)(103)(0.15 + y)2
in
teaching
Web)
Conservation of Energy:
Dissemination
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T1 + V1 = T2 + V2
provided
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assessing
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solely
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the
(including
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student
by
1
(40)(52) + (0 + 281.25) = 0 + [-392.4(2 + y)] +
2
1
(25)(10
( 3)(0.15 + y)2
2
12500y2 + 3357.6y - 1284.8 = 0
and
use
United
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learning.
is
of
the
not
instructors
States
World
permitted.
1
1
mv21 + c A Vg B 1 + A Ve B 1 d = mv22 + c A Vg B 2 + A Ve B 2 d
2
2
part
the
e
Solving for the positive root of the above equation,
any
courses
Ans.
destroy
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and
will
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y = 0.2133 m = 213 mm
600 mm
Q7-