Diapositiva 1

```PHYSICS FOR ENGINEERS.
B-EXAMS 2006-2007.
B1
B2
B3
B4
1
PROBLEM 1
07B1Eng
HOME EXAM B1
(a)
You have a spring whose natural length is L0 = 10 cm -figure (a)-.
When a mass M = 250 g is hung on the spring, its length increases
by L = 40 cm –figure (b)-. Finally, the hanging mass oscillates after
the spring is stretched A = 10 cm and then released –figure (c)-.
a)
b)
c)
d)
(c)
L0
What is the constant of the spring? 0.5 p
Find the period of the oscillation
0.5 p
Find the position of the mass 6.98 s after the oscillations start. 1.0 p
Find the period of the oscillation if you had hung the same 2.0 p
mass M from two identical springs like this one disposed in
a paralell way.
L
 AB
PROBLEM 2
A
M
B
The mechanism depicted in the figure is
composed by two rigid rods, each of lenght L.
Rod AB girates around toggle A, whereas rod
BC has a joint in point B and its end C slides
on the floor. You know the lenght L, the
angular velocity of rod AB, AB, and the angles
1 and 2. Find:
a) The velocity of point B, vB, and the angle
formed by vB with the horizontal. 2.0 p
(b)
M
2
L

k
A
L

j

i
1
b) The angular velocity of rod BC, BC, and the velocity of point C, vC. 4.0 p
C
2
y
07B1_Eng
PROBLEM 1
HOME EXAM B1
(a)
You have a spring whose natural length is L0 = 10 cm -figure (a)-.
When a mass M = 250 g is hung on the spring, its length increases
by L = 40 cm –figure (b)-. Finally, the hanging mass oscillates after
the spring is stretched A = 10 cm and then released –figure (c)-.
a)
b)
c)
d)
(b)
(c)
L0
What is the constant of the spring? 0.5 p
Find the period of the oscillation
0.5 p
Find the position of the mass 6.98 s after the oscillations start. 1.0 p
Find the period of the oscillation if you had hung the same 2.0 p
mass M from two identical springs like this one disposed in
a paralell way.
L
y  A
y  A
A
M
F
From Hooke’s law:
a)
F k L
From Newton’s 2nd law: F  Mg
k
Mg 0.25  9.8

 6.125 N/m
0.40
L
M
Mg
b)
T  2
M
0.25
 2
 1.27 s
k
6.125
We choose t = 0 when y = A
That implies  = 0
c)
A  0.10 m
y  A cos t   
y  0.1cos4.95 t 

y  0.1cos4.95  6.98  0.01 m
k
6.125

M
0.25
y  A
3
y
PROBLEM 1 (SOLUTION CONTINUED)
From Newton’s 2nd law:
From Hooke’s law:
d)
F1  F2  Mg  Ma
F1  k y
d2y
M 2  2k y  Mg
dt
F2  k y
(The springs
are identical)
d2y
2k


yg
dt 2
M
Equation of the oscillation driven by two identical springs:
d 2 y 2k

yg
dt 2 M
y
where
d2y
Let us write the equation as
2 y  g
2
dt
F1
F2
The solution of this equation is
ma
The period is
M
T' 
2

 2
y

2k
M
M
g  A cos t   
2k
M
0.25
 2
 0.90 s
2k
2  6.125
The set of two identical springs disposed in a parallel way
(each constant = k) behaves as a single spring of constant 2 k.
Mg
4
PROBLEM 2
The mechanism depicted in the figure is composed by two rigid rods, each of lenght L. Rod AB
girates around toggle A, whereas rod BC has a joint in point B and its end C slides on the floor.
You know the lenghtBL, the angular velocity of rod AB, AB, and the angles 1 and 2. Find:
a) The velocity of point B, vB, and the angle formed by vB with the horizontal. 2.0 p
b) The angular velocity of rod BC, BC, and the velocity of point C, vC. 4.0 p
 AB
B
2
L

k
A
1
L

j

i
C
5
PROBLEM 2 (solution continued)
a) Find: The velocity of point B, vB, and the angle formed by vB with the horizontal.
Rod AB
From dot product definition:


 


i sin 1  j cos1  i  i sin 1  j cos1 i cos 




vB  v A   AB  rB / A
 




vB  0   AB  k  L i cos 1  j sin 1 





j
cos


i
1
1


i sin 1  j cos 1 i
i sin 
cos   
See that
 k  i   j
 k  j  i

 AB   AB



vB   AB L i sin 1  j cos 1 
 AB
B

rB / A
A

vA  0
L
cos   sin 1  cos90  1 
  90  1

i


vB

rB / A


 L i cos 1  j sin 1 
 

k


i sin 1  j cos 1

j

k

i
1
6
b) Find: The angular velocity of rod BC, BC, and the velocity of point C, vC.




vC  vB  BC  rC / B

i



vC i   AB L i sin 1  j cos 1   0
L sin 
We know:



vB   AB L i sin 1  j cos 1

Vector rC / B


Let’s call   1   2  90
Although we don’t know their
values, we can write for angular
velocity and velocity of point C:
BC and vC are the
quantities to find.
vC  BC L cos    AB L sin 1
BC L sin    AB L cos 1
B

2
L
1

k
 BC   AB

vB
 BC
L
90  1

j
A
0
This vectorial equation can be splitted into
two equations, one for each component:
 AB



rC / B  L i sin   j cos  

 BC





vC i   AB L i sin 1  j cos 1   i  BC L cos   j  BC L sin 



rC / B  L i sin 1   2  90  j cos1   2  90
BC  BC k


vC  vC i

k

j
0
 L cos 
1  2  90

i

rC / B
cos 1
cos 1
  AB
sin 
sin 1   2  90
 cos 

vC   AB L
cos 1  sin 1 
 sin 

 cos1   2  90

vC   AB L
cos 1  sin 1 
 sin 1   2  90

C

vC
C end slides on the floor
7
HOME EXAM B2 (2007)
PROBLEM 1
MODEL A
A rigid thin rod of lenght L = 1.80 m, mass M = 6 kg is articulated on
a toggle (point O in the figure). The rod is kept tilted by an steel
towline attached to the wall. The angles between the towline, the rod
and the wall are 1 = 60&ordm; and 2 = 50&ordm; respectively. A counterweight m
= 4 kg hangs from the other end of the rod.
a) Draw the free body diagram for the rod (2 p).
b) Find the tension on the towline and the rectangular
components of the reaction in the point O (2 p).
2
1
L
m
O
M
PROBLEM 2
The picture shows a disc of radious 3R, with four circular holes, each of radious R, lying on the indicated
positions. The surface density of the disc is  (kg&middot;m-2). The disc moves on the floor. Answer the following
questions for the numerical values given below:
a) Calculate the Izz moment of inertia of the disc, where Z is the
perpendicular axis which passes through its simmetry center (not
shown in the figure). (2 p)
b) The initial angular velocity of the disc when it takes contact with
the floor is 0 (clockwise direction), meanwhile the initial linear
velocity of its center of mass is zero. The dynamics friction
coefficient is . Make a FBD taking into account the external forces
acting on the disc, and find the time it takes the disc to roll without
slipping, the linear velocity of its center of mass and the angular
velocity of the solid at the moment in which rolling without slipping
begins. (4 p)
Find the numerical results for the above questions, being R = 14,7 cm,  = 50 kg&middot;m-2, 0 = 0,60 rad/s,  =8 0.15
HOME EXAM B2 (2007)
PROBLEM 1 Solution
MODEL A
A rigid thin rod of lenght L = 1.80 m, mass M = 6 kg is articulated on
a toggle (point O in the figure). The rod is kept tilted by an steel
towline attached to the wall. The angles between the towline, the rod
and the wall are 1 = 60&ordm; and 2 = 50&ordm; respectively. A counterweight m
= 4 kg hangs from the other end of the rod.
a) Draw the free body diagram for the rod (2 p).
b) Find the tension on the towline and the rectangular
components of the reaction in the point O (2 p).

FBD
2
 O   Mg
2
T
180  1
90   2
Y
1
X
L

Rx
180  

O
180  
mg
Mg
Ry
  180  1  2

2
1
L
m
O
M
L
sin 180     mgL sin 180     TL sin 180  1   0
2
sin   M

T
  m g
sin 1  2

Fx  Rx  T cos90   2   0

Rx 
sin  sin  2  M

  m g
sin 1  2

Fy  R y  T sin 90   2   Mg  mg  0
Ry  M  mg 
sin  cos  2  M

  m g
sin 1  2

T  74.4 N; Rx  57.0 N;
Ry  50.2 N
9
PROBLEM 2 Solution (continued)
Moment of inertia of a disc ( surface density, a radious)
1
1
2
( Mass )  Radious     a 4
2
2
In our problem, we have two different types of disc: 1. A solid disc of radious a = 3R and surface density  = .
About to a normal axis passing through its simmetry center (Izz)
I1zz 
3R
1
81
  3R 4    R 4
2
2
All of them are simmetrically disposed around, being 2R the distance between
each center and the center of the body.
R
2R
I zz 
2R
Momentum of inertia of every
hole about to the Z axis passing
through its simmetry center
2R
1
I 2 zz     R 4
2
We have now to calculate the momentum of inertia of every hole about to the
Z axis passing through the center of the body. We apply Steiner theorem:

Z axis normal to the
plane, not shown
1
81
  3R 4    R 4
2
2
9
2
 I 2 zz     R 2 2 R      R 4
2
I1zz 
Z axis
I 2' zz

9
2
I 2' zz  I 2 zz     R 2 2 R      R 4
2


9
45
 81
I zz  I 2 zz  4 I 2' zz    4    R 4 
  R4
2
2
2
45
I zz 
  R4
2
10
PROBLEM 2 Solution (continued)
Initial picture of the problem
 
 




vC 0  0  k  3R  j   30 R  i 

vCM 0  0

0  k
This means that initially point C moves towards the
left, therefore the friction force goes to the right.
2nd

Newton’s law:



Fx  FR  m aCM
aCM   g


 m g i  m aCM i
Rotation of the disk:



Y
CM
X

 3 m g R 

k
I CM



3R  j   m g i  I CM 
The moment of inertia ICM is the same IZZ previously found,
because our disc is a flat figure, so the angular acceleration
can be written as:
3 m g R

Where the mass is given by

C

FR
Initial situation

 (t )
Angular velocity decreases at the same time
that velocity of the center of mass increases
I CM
m    3R   4   R 2  5  R 2
and the moment of inertia is I CM


vC 0
 CM  3R  j  FR  I CM 


3R  j 
Z
3 5  R 2 g R 2  g


3R
45 / 2  R 4
2
45
 I zz 
  R4
2
 2 g 

k
3R
Anticlockwise


FR

t &gt; 0, but before rolling without 11
slipping
PROBLEM 2 Resolution
Translation &amp; Rotation equations
Rolling condition
 g t f  3R0 
vCM (t )  aCM t   g t
 (t )  0   t  0 

vCM (t f )   (t f )  3R
2 g
3R t f
3R
tf 
 (t )
R 0
g
When rolling without slipping begins, we have
vCM (t f )  aCM t f   g t f  R 0
 (t f )  0   t f  0 
2 g
t
3R
2 g
1
t f  0
3R
3
vCM (t )
45
  R4
2
2 g

3R
I zz 


FR

t &gt; 0, but before rolling without slipping
R = 14,7 cm,  = 50 kg&middot;m-2, 0 = 0,60 rad/s,  = 0.15
IZZ = 1.65 kg&middot;m2
 = 6.67 rad&middot;s-2 tf = 0.060 s vCM(tf) = 8.82&middot;10-2 m/s (tf) = 0.20 rad/s
12
07B3_Eng HOME EXAM B3
PROBLEM 1 (4 p)
A hollow sphere (inner radius R1 = 10 cm, outer radius R2 = 20 cm) is made of material of density 0 =
0.80 g&middot;cm-3 and is floating in a liquid of density L = 1.60 g&middot;cm-3. The interior is now filled with material
of density m so that the sphere just floats completely submerged. (a) Find the volume fraction of the
floating hollow sphere under the liquid surface level before filling its interior. (b) Find density m.
PROBLEM 2 (6 p)
A cylindrical-shaped tank is used in a farm as water reservoir. The area of the base of the tank is 10 m2.
A faucet on the top lets go in 0.60 kg of water per second, meanwhile the outcoming flux across the
sink lying on the bottom pours 0,50y kg/s outside, where y means the height of the liquid above the flat
bottom of the tank. The tank has also a spillway 1 m above the bottom. Assuming that at the beginning
of the day the tank contains 100 liters, then we open at the same time the incoming faucet and the sink,
find:
a) How long does the surface of the water take to reach the spillway (in case it reaches).
b) In case it would not have any spillway, find the maximum height the surface of the water could reach.
c) Plot the height of the surface of the water versus time, and mark in your plot the values you obtained as
results for the paragraphs a) and b).
13
PROBLEM 1 SOLUTION
A hollow sphere (inner radius R1, outer radius R2) is made of material of density 0 and is floating
in a liquid of density L. The interior is now filled with material of density m so that the sphere
just floats completely submerged. (a) Find the volume fraction of the floating hollow sphere under
the liquid surface level before filling its interior. (b) Find density m.
(a) Volume and mass of the hollow sphere:
E
0

4
M 0    0 R23  R13
3
4
V0   R23
3
R1
R2
M0g
L
VL
From Archimedes’ Principle, the sphere is buoyed up by
a force E equal to the weight of the displaced fluid. As
the sphere floats, E should be equal to its weight M0g.
E   L VL g
Setting E equal to the weight, we find VL
4 
VL   0 R23  R13
3 L



VL is the volume of the sphere’s submerged portion
4
E   L VL g  M 0 g    0 R23  R13  g
3
Submerged fraction


0  R13 
VL 0 R23  R13
1 - 3 


3
V0  L
R2
 L  R2 
14
PROBLEM 1 SOLUTION
(b) When the interior of the sphere is filled with material of density m, the sphere just floats
completely submerged.
Now the sphere is buoyed up by a force E’ equal to the
weight of the displaced fluid, which coincides with the
E'
weight of an amount of fluid whose volume is the same
0
that the sphere’s volume.
m
E '   L V0 g  M 0  M ' g
R1
M0g  M '

R2
L
4
4
M '    L R23    0 R23  R13
3
3
The mass of the material filling the hollow is
Setting equal both expressions for M’ we have
M '   L V0  M 0

4
4
M '    L -  0  R23    0 R13
3
3
4
M '    m R13
3
R23
 m   0   L -  0  3
R1
0 (g/cm3) = 0,80
L (g/cm3) = 1,60
Numerical
solutions
R1 (m) = 0,10
R2 (m) = 0,20
VL/V0 = 0,44
m (g/cm3) = 7,20
15
PROBLEM 2 SOLUTION
A cylindrical-shaped tank is used in a farm as water reservoir. The area of the base of the tank is 10 m2.
A faucet on the top lets go in 0.60 kg of water per second, meanwhile the outcoming flux across the
sink lying on the bottom pours 0,50y kg/s outside, where y means the height of the liquid above the flat
bottom of the tank. The tank has also a spillway 1 m above the bottom. Assuming that at the beginning
of the day the tank contains 100 liters, then we open at the same time the incoming faucet and the sink,
find:
a) How long it takes the surface of the water to reach the spillway (in case it reaches).
b) In case it would not have any spillway, find the maximum height the surface of the water could reach.
c) Plot the height of the surface of the water versus time, and mark in your plot the values you obtained as
results for the paragraphs a) and b).
m 1
Spillway
Continuity equation:
dm
 m 1  m 2
dt
m(t )   V (t )   S y (t )
h
V (t )
y (t )
0.60
V0
y0
S
  density of water
dm
 C1  C2 y (t )
dt
2
m
V0
S
dm
dV (t )
dy

S
dt
dt
dt
0.50
 1  C1 (kg/s)
m
y0 
S
dy
 C1  C2 y
dt
 2  C2 y (C2  kg/(m  s))
m
16
PROBLEM 2 SOLUTION
y

dy
1

dt
C1  C2 y  S
dy
S
 C1  C2 y
dt
t
dy
1

C1  C2 y  S
y0
1

LnC1  C2 y 
C2
y
y0
1

t
S
t

 C 
C C
y  1   1  y0  exp   2 t 
C2  C2
 S 

m 1
1
du
C2
 C 
C1  C2 y  C1  C2 y0  exp   2 t 
 S 
t y h
 C1

  h 
C
S


Ln  2
C2
 C1

  y0 
 C2


 C  C
C1  C1
   y0  exp   2 t   1
t  C2  C2
  S  C2

Numerical values
Spillway
V 0 (litros) =
h
V (t )
100
V 0 (m ) =
0,1
S (m2) =
 (kg/m3) =
10
1000
0,6
C 2 (kg/m.s)= 0,5
y 0 (m) =
0,01
C 1/C 2 (m) = 1,2
C 1/C 2-y 0 = 1,19
V0
S

du
1

LnC1  C2 y 
u
C2
ym&aacute;x  lim
C 1 (kg/s) =
y0
dy
1

C1  C2 y
C2
u  C1  C2 y  dy  
a) Find how long takes the
surface of the water to reach the
spillway (in case it reaches).
3
y (t )

0
 C  C2 y 
C
   2 t
Ln 1
S
 C1  C2 y0 
0
b) In case it would not have any
spillway, find the maximum
height the surface of the
watercould reach.

dt
2
m
C 2/ S =
0,00005
h (m) =
1
y  1.2 1.19 exp  0.00005 t 
t y h  
1.2  1  35668 s
1
Ln
0.00005 1.2  0.01

 C  C
C1  C1
   y0  exp   2 t   1  1.20 m
t  C2  C2
  S  C2

ym&aacute;x  lim
17
PROBLEM 2 SOLUTION
c) Plotting: level of water versus time
y

 C 
C1  C1
   y0  exp   2 t 
C2  C2
 S 

y  1.2 1.19 exp  0.00005 t 
1,4
y (m)
Maximum height in case there
weren’t any spillway
1,2
1.20 m
1,0
y=h=1m
0,8
0,6
The level of the water
reaches the spillway
0,4
t = 35668 s
0,2
0,0
t (s)
0,0
4
2,0x10
4
4,0x10
4
6,0x10
4
8,0x10
5
1,0x10
5
1,2x10
5
1,4x10
18
07B4
HOME EXAM B3
Only spanish version available
Un gas ideal de coeficiente adiab&aacute;tico  = 1.4 ejecuta un ciclo de potencia formado por las siguientes etapas:
1→2. El gas se expande politr&oacute;picamente (&iacute;ndice de politrop&iacute;a k1 = 1.35) desde las condiciones V1 = 1 litro,
P1 = 7.87 bar, hasta que su volumen se duplica.
2→3. El gas se enfr&iacute;a a volumen constante, hasta que su temperatura es T3 = 280 K.
3→1. El gas se comprime politr&oacute;picamente hasta restituir las condiciones iniciales (sea k2 el &iacute;ndice de
politrop&iacute;a de este proceso, que deber&aacute; determinarse).
Se supone que todas las etapas son reversibles. La masa de gas es n = 0.20 moles, y la constante universal de
los gases es R = 8,314 J/(K&middot;mol). Se pide:
A) Calcular las coordenadas de presi&oacute;n y temperatura en todos los puntos notables del ciclo (2 p).
B) Determinar el &iacute;ndice k2 y representar gr&aacute;ficamente el ciclo en un diagrama de Clapeyron (P-V) (2 p).
C) Calcular el trabajo asociado con cada una de las etapas del ciclo, discutiendo su signo (2 p).
D) Calcular el calor asociado con cada una de las etapas del ciclo, discutiendo su signo (2 p).
E) Determinar el rendimiento del ciclo (1 p).
F) Calcular la variaci&oacute;n de entrop&iacute;a de cada una de las etapas del ciclo (2 p).
19
EXAMEN DE CASA B4 (2007)
input data MODELO A
R (J/K.mol)
8,314
n (mol)
0,2
V1 (l)
1
C&aacute;lculo entrop&iacute;as
P1 (bar)
7,87
Volumen auxiliar
Qpolitr&oacute;pico 
1 /  1
 p f V f 

V 
 pV 
i
i


2
V(k1) (m3)
0,00109
T3 (K)
280
V(k2) (m3)
0,00371
k1

1,35
1,4
S12 (J/K)
V2 (l)
V1 (m3)
0,001
P1 (Pa)
787000
V2 (m3)
0,002
k1
S23 (J/K)
Sciclo (J/K)
0,000
cv (J/K.mol)
Polit 12
Isoc 23
Polit 31
Qin, Wout
Rend =
439,68
0,136
8
1
piVi  p f V f
W politr&oacute;pico 
k 1
nRTi  T f 

k 1
7
k1
cV 
6
R
  1
W
4
P1V1k2 4,2073464
Q
3
V (m3)
0,001
0,002
0,002
W (J)
484,38
0,00
-424,41
Rendimiento  
59,97
P (Pa)
787000
308734
232792
U (J)
-423,83
-379,71
803,54
2 Q
k2
W
3 V (litros )
2
1,0
1,2
1,4
1,6
Q politr&oacute;pico ,
1.
Ti  T f
Ti  T f
W politr&oacute;pico
1,8
2,0
Q politropico k   k  
W politropico
0,00
+
-
2.
Ti  T f
Ti  T f
+
+
-
59.97
Wout W polit12  W polit31

 0.136

Qin Q polit12  Q polit31 439.68
Incremento de entrop&iacute;a (politr&oacute;picas)
V
S   nR ln i
V
 1
Q
20,785
T (K)
473,3
371,3
280,0
Q (J)
60,55
-379,71
379,13
 k 

 nRT f  Ti 

 k  1  1
P1V1k1 70,141449
1,7573
1
2
3
1,030 Politropica
0,002
k2
k 1
p f V f  piVi
p ( bar )
-1,174 Isocora
S31 (J/K)

5
1,35
V3 (m3)
0,144 Politropica
piVi  p f V f
Incremento de entrop&iacute;a (isoc&oacute;rica)
1 /  1
 p f V f 

donde V  
 pV 
 i i 
S 

QV
T
Tf


Ti
 Tf
ncv dT
R
n
ln 
T
  1  Ti



20
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