x - Studentportalen

Mekanik HI 2014
Andreas Lindblad
F2
Föreläsningsplan
Tema
Kapitel
F1 Kinematik i linjär- och cirkulär-rörelse
1,2,3 samt
9.1-9.3
F2 Newtons lagar
4,5
F3 Arbete & Kinetisk Energi
6,7
F4 Impuls & Rörelsemängdsmoment
8
F5 Rotationsrörelse och tröghetsmoment
9
F6 Dynamik i rotationsrörelse, vridmoment 10
F7 Rep. rotationsrörelse
F8 Svängningsrörelse
14 + utdelat
F9 Svängningsrörelse 2
14 + utdelat
Resten av föreläsningarna: repetition och problemlösning
Kinematik
vx
• Ekvationerna till höger gäller
för rätlinjig rörelse med en
konstant acceleration ax.
x
=
x
=
2
vx
=
x0
=
v0x + ax t
a x t2
x0 + v0x t +
2
2
v0x + 2ax (x x0 )
✓
◆
v0x + vx
t
2
th
x
of
be
t = 0 the bird is at the origin. (a) Calculate the position and acceleration vectors of the bird as functions of time. (b) What is the
bird’s altitude (y-coordinate) as it flies over x = 0 for the first time
after t = 0?
3.47 ... CP A test rocket is Figure P3.47
launched by accelerating it
along a 200.0-m incline at
2
1.25 m>s starting from rest at
m
0
.
200
point A (Fig. P3.47). The
incline rises at 35.0° above
35.0°
the horizontal, and at the
A
instant the rocket leaves it, its
engines turn off and it is subject only to gravity (air resistance can be ignored). Find (a) the
maximum height above the ground that the rocket reaches, and (b)
the greatest horizontal range of the rocket beyond point A.
3.48 . Martian Athletics. In the long jump, an athlete launches
herself at an angle above the ground and lands at the same height,
trying to travel the greatest horizontal distance. Suppose that on
e
e
g
m
n
h
x
Problems
99
value of t is the velocity of the plane perpendicular to its
acceleration?
3.46 .. CALC A bird flies in the xy-plane with a velocity vector given
S
by v ! 1a - bt22ın " gt≥n, with a = 2.4 m>s, b = 1.6 m>s3, and
g = 4.0 m>s2. The positive y-direction is vertically upward. At
t = 0 the bird is at the origin. (a) Calculate the position and acceleration vectors of the bird as functions of time. (b) What is the
bird’s altitude (y-coordinate) as it flies over x = 0 for the first time
after t = 0?
3.47 ... CP A test rocket is Figure P3.47
launched by accelerating it
along a 200.0-m incline at
1.25 m>s2 starting from rest at
.0 m
0
0
2
point A (Fig. P3.47). The
incline rises at 35.0° above
35.0°
the horizontal, and at the
A
instant the rocket leaves it, its
engines turn off and it is subject only to gravity (air resistance can be ignored). Find (a) the
maximum height above the ground that the rocket reaches, and (b)
the greatest horizontal range of the rocket beyond point A.
3.48 . Martian Athletics. In the long jump, an athlete launches
e
e
g
m
n
h
x
Problems
99
value of t is the velocity of the plane perpendicular to its
acceleration?
3.46 .. CALC A bird flies in the xy-plane with a velocity vector given
S
by v ! 1a - bt22ın " gt≥n, with a = 2.4 m>s, b = 1.6 m>s3, and
g = 4.0 m>s2. The positive y-direction is vertically upward. At
t = 0 the bird is at the origin. (a) Calculate the position and acceleration vectors of the bird as functions of time. (b) What is the
bird’s altitude (y-coordinate) as it flies over x = 0 for the first time
after t = 0?
3.47 ... CP A test rocket is Figure P3.47
launched by accelerating it
along a 200.0-m incline at
1.25 m>s2 starting from rest at
.0 m
0
0
2
point A (Fig. P3.47). The
incline rises at 35.0° above
35.0°
the horizontal, and at the
A
instant the rocket leaves it, its
engines turn off and it is subject only to gravity (air resistance can be ignored). Find (a) the
maximum height above the ground that the rocket reaches, and (b)
the greatest horizontal range of the rocket beyond point A.
3.48 . Martian Athletics. In the long jump, an athlete launches
2.00 m>s. What angular velocity is required to achieve these values?
9.27 . Electric Drill. According to the shop manual, when
drilling a 12.7-mm-diameter hole in wood, plastic, or aluminum,
a drill should have a speed of 1250 rev>min. For a 12.7-mmdiameter drill bit turning at a constant 1250 rev>min, find (a) the
maximum linear speed of any part of the bit and (b) the maximum
radial acceleration of any part of the bit.
9.28 . At t = 3.00 s a point on the rim of a 0.200-m-radius wheel
has a tangential speed of 50.0 m>s as the wheel slows down with
a tangential acceleration of constant magnitude 10.0 m>s2.
(a) Calculate the wheel’s constant angular acceleration. (b) Calculate the angular velocities at t = 3.00 s and t = 0. (c) Through
what angle did the wheel turn between t = 0 and t = 3.00 s?
(d) At what time will the radial acceleration equal g?
R is cut
remaining h
E9.34a). (a
of inertia o
axis perpe
through po
your answe
the same a
plete disk o
would be t
of a quarte
radius R a
dicular to
through po
Redan de gamla grekerna: Aristoteles fysik
Aristoteles
384-322 f. kr.
Huvudprinciperna i Aristoteles fysik är:!
1. Naturliga platser: varje element vill vara på olika platser
relativt jordens mitt, som också är centrum för universum.!
2. Gravitation/Svävning: för att nå denna position så dras objekt
uppåt eller nedåt.!
3. Rätlinjig rörelse: rörelse som beror av denna kraft är längs en
rak linje och vid en konstant hastighet.!
4. densitet-hastighet-förhållande: hastigheten är omvänt
proportionell mot densiteten hos ett objekt.!
5. Vakuum är omöjligt: rörelse i vakuum är oändligt snabb.!
6. Heltäckande eter: alla platser i rymden är fyllda med materia.!
7. Oändligt universum: rymden kan inte ha en gräns.!
8. Continuumteori: mellan atomer finns vakuum, och därför kan
inte materia vara atomisk.!
9. Kvintessens: objekt ovan jord är inte skapad av jordlig
materia.!
10.Oföränderligt och evigt kosmos: solen och planeterna är
perfekta sfärer, som inte förändras.!
11.Cirkulär rörelse: planeter rör sig i en perfekt cirkulär rörelse.
Newtons lagar
Tröghetslagen: Då en kropp befinner sig i vila eller likformig
rörelse är summan av alla krafter som verkar på den noll.
Newtons lagar
Accelerationslagen: Om summan av alla yttre krafter inte är lika med
noll så är summan lika med accelerationen gånger massan.
Enhet: 1 N=1 kgm/s2
Newtons lagar
Reaktionslagen: Om en kropp A verkar med en kraft på en annan
kropp B, då finns det en kraft verkandes på A från B så att:
Newtons gravitationslag
G= 6.674×10−11 N m2 kg−2
5 st mekaniska krafter
Normalkraft
“Snörkraft”
Friktion
Tyngdkraft
in the
agram
rce on
r force
te the
ecraft.
while
craft is
omical
nd the
astroastro-
4.50 . A loaded elevator with very worn cables has a total mass
of 2200 kg, and the cables can withstand a maximum tension of
28,000 N. (a) Draw the free-body force diagram for the elevator. In
terms of the forces on your diagram, what is the net force on the
elevator? Apply Newton’s second law to the elevator and find the
maximum upward acceleration for the elevator if the cables are not
to break. (b) What would be the answer to part (a) if the elevator
were on the moon, where g = 1.62 m>s2?
4.51 .. CP Jumping to the Ground. A 75.0-kg man steps off a
platform 3.10 m above the ground. He keeps his legs straight as he
falls, but at the moment his feet touch the ground his knees begin to
bend, and, treated as a particle, he moves an additional 0.60 m
before coming to rest. (a) What is his speed at the instant his feet
touch the ground? (b) Treating him as a particle, what is his acceleration (magnitude and direction) as he slows down, if the acceleration is assumed to be constant? (c) Draw his free-body diagram
(see Section 4.6). In terms of the forces on the diagram, what is the
net force on him? Use Newton’s laws and the results of part (b) to
calculate the average force his feet exert on the ground while he
slows down. Express this force in newtons and also as a multiple
of his weight.
...
5.100 .. Accelerometer. The system shown in Fig. P5.100 can
be used to measure the acceleration of the system. An observer riding on the platform measures the angle u that the thread supporting
the light ball makes with the vertical. There is no friction anywhere. (a) How is u related to the acceleration of the system? (b) If
m 1 = 250 kg and m 2 = 1250 kg, what is u? (c) If you can vary
m 1 and m 2, what is the largest angle u you could achieve? Explain
how you need to adjust m 1 and m 2 to do this.
Figure P5.100
Ball
u
Platform (m2)
Horizontal surface
m1
the bus?
5.105 .
Problem
firm hol
over a
attached
(Fig. P5
sees the
the rope
climbs, d
or remai
climbs,
monkey
increase,
monkey
What ha
the mon
is fallin
ground,
bananas
5.106 ..
speed of
own an
ow the
s final
ad with
s slide
ate the
nd the
e road.
ange if
initial
riction
hat has
much
m> s in
from a
x = 0. Use the work–energy theorem to find the speed of the sled
at (a) x = 8.0 m and (b) x = 12.0 m. You can ignore friction
between the sled and the surface of the pond.
6.36 . A 2.0-kg box and a 3.0-kg box on a perfectly smooth horizontal floor have a spring of force constant 250 N>m compressed
between them. If the initial compression of the spring is 6.0 cm,
find the acceleration of each box the instant after they are released.
Be sure to include free-body diagrams of each box as part of your
solution.
6.37 .. A 6.0-kg box moving at 3.0 m>s on a horizontal, frictionless surface runs into a light spring of force constant 75 N>cm.
Use the work–energy theorem to find the maximum compression
of the spring.
6.38 .. Leg Presses. As part of your daily workout, you lie on
your back and push with your feet against a platform attached to
two stiff springs arranged side by side so that they are parallel to
each other. When you push the platform, you compress the springs.
You do 80.0 J of work when you compress the springs 0.200 m
from their uncompressed length. (a) What magnitude of force must
you apply to hold the platform in this position? (b) How much
additional work must you do to move the platform 0.200 m
$10 for each
limit, would
h 80.0 cm is
white rat with
mouse with
eleased from
e to hold on,
cal position?
ength 2.50 m,
t. The potato
port, with the
speed of the
he tension in
for a batted
at an angle of 60.0° above the horizontal. Use energy conservation
to find the ball’s greatest height above the ground.
7.65 .. In a truck-loading station at a post office, a small 0.200-kg
package is released from rest at point A on a track that is onequarter of a circle with radius 1.60 m (Fig. P7.65). The size of the
package is much less than 1.60 m, so the package can be treated as
a particle. It slides down the track and reaches point B with a speed
of 4.80 m>s. From point B, it slides on a level surface a distance of
Figure P7.65
A
R 5 1.60 m
m 5 0.200 kg
238
3.00 m
CHAPTER 7 Potential Energy and Energy Conservation
C
B
3.00 m to point C, where it comes to rest. (a) What is the coefficient of kinetic friction on the horizontal surface? (b) How much
work is done on the package by friction as it slides down the circular arc from A to B?
7.66 ... A truck with mass m has a brake failure while going
down an icy mountain road of constant downward slope angle a
from rest. (a
the motion
mum compr
7.71 .. An
vertical spri
is compress