Download Physics unit 06 REVIEW Name___C. ANSWERS__________

Physics unit 06 REVIEW
Circular Motion and Gravitation
v2
ac =
r
m v2
F c=
r
F g=G
Name___C. ANSWERS__________
m1 m2
r2
m
g=G 2
r
G=6.67·10–11 Nm2/kg2
mE=5.98·1024 kg
rE=6.38·106 m
TRUE / FALSE
1.
2.
3.
4.
5.
__F__ The only reason riders on a tilting rotating platform of a carnival ride don't fall out is
because they are strapped in.
__F__ When an object travels in a circle at constant speed, the acceleration of the object is zero.
__F__ The gravitational force of Earth on an object does not depend upon the object’s mass.
__T__ Any force that causes an object to move in a circular path is called a centripetal force.
__T__ The gravitational force on a ball decreases as the ball moves farther from Earth.
MULTIPLE CHOICE
6.
What term describes a change in the speed velocity of an object in circular motion?
a. tangential speed
b. centrifugal acceleration
c. centripetal acceleration
d. centripetal force
7.
When a car makes a sharp left turn, what causes the passengers to move toward the right side of
the car?
a. centripetal acceleration
b. centripetal force
c. centrifugal force
d. inertia
8.
Two small masses that are 10.0 cm apart attract each other with a force of 10.0 N. When they are
5.0 cm apart, these masses will attract each other with what force?
a. 5.0 N
b. 20.0 N
c. 2.5 N
d. 40.0 N
9.
What is the centripetal force that holds planets in orbit?
a. inertia
b. gravitational force
c. planetary force
d. friction
10. The force that Earth exerts on the moon
a. is greater than the force the moon exerts on the Earth.
b. is less than the force the moon exerts on Earth.
c. is equal in magnitude to the force the moon exerts on Earth.
11. How does the gravitational force between two objects change if the distance between the objects
doubles?
a. The force decreases by a factor of 4
b. The force decreases by a factor of 2
c. The force increases by a factor of 2
d. The force increases by a factor of 4
12. The gravitational force between two masses is 36 N. What is the gravitational force if the
distance between them is tripled?
a. 4.0 N
b. 18 N
c. 9.0 N
d. 27 N
13. According to Newton's law of universal gravitation, the force of attraction between any two
masses is directly proportional to the
a. velocity of the two masses
b. distance between the masses
c. product of the two masses
d. sum of the two masses
14. As the distance between two bodies increases, the force of attraction between the bodies
a. increases
b. decreases
c. remains the same
15. The centripetal force on an object in circular motion is
a. in the same direction as the tangential speed.
b. in the same direction as the centripetal acceleration.
c. in the direction opposite the tangential speed.
d. in the direction opposite the centripetal acceleration.
CONCEPTUAL QUESTIONS
16. Discuss the following statement: “A satellite is continually in free fall.”
The satellite is moving in an orbit (circular motion) under influence of only one force –
gravity. Therefore, it is in exactly the same situation as a free falling object. In fact, orbital
motion is an ultimate case of a horizontally launched projectile.
17. Earth exerts a 1.0 N gravitational force on an apple. Does the apple accelerate toward Earth, or
does Earth accelerate towards the apple? Explain your answer.
If the gravitational force is the only force involved (in other words, the apple is not attached
to the tree), then both: the apple is accelerating towards Earth and Earth is accelerating
towards the apple. They are both accelerating because there is a gravity force of attraction
acting on both.
18. Why is centripetal force known as a net force? Give three different examples of objects in circular
motion and the force(s) providing the centripetal force.
Centripetal force must be unbalanced by other forces. If it would be balanced by another
force, the net force would be 0, and an object would follow the Newton's 1 st law (moving at
the same speed in a straight line; not changing velocity). If there is a non-zero net force
acting on an object in the direction perpendicular to its velocity, the object will move in a
circle; and this net force will be the centripetal force.
1) Solar system orbiting the supermassive black hole in the center of our galaxy:
gravity force is the centripetal force
2) A car follows a turn on a highway: friction force is the centripetal force
3) A softball is moving in circle being held by the player right before the pitch:
pushing/holding force from the player's hand is the centripetal force
19. Assume two masses, ma and mb, separated by a distance, d, have a gravitational force, F acting
between them. Fill in the values for gravitational force (in terms of multiples of F) when the
masses and distance change according to the table.
Distance
Mass A
Mass B
Force
d
ma
2mb
2F
d
2ma
mb
2F
d
2ma
2mb
4F
2d
ma
mb
F/4
d/2
ma
mb
4F
d/3
3ma
mb
27F
CALCULATIONS
20. A car on a roller coaster loaded with passengers has a mass of 2000 kg. At the lowest point of the
track, the radius of curvature of the track is 24 m and the roller car has a speed of 17 m/s. Find
the centripetal acceleration of the roller coaster car at the lowest point on the track and the net
force on the roller car at this point.
2
m 2 289 m
( 17 )
v2
s
s2
m
ac= =
=
=12 2
r
24 m
24 m
s
The net force is FN – Fg, and it is the centripetal force:
FN
Fg
2
m
(17 )
2
v
s
m
Fc =m a c =m =2000 kg
=2000 kg⋅12 2 =24000 N
r
24 m
s
21. What is the speed of a 95 kg fullback who makes a turn on a football field and experiences a
centripetal force of 637 N as he moves in a circle that has a radius of 12.0 m.
v2
Fc =m a c =m
r
Therefore:
v 2=
Fc r
m
and
v=
√ √
Fc r
637⋅12
m
=
=9.0
m
95
s
22. A 0.45 kg ball attached to the end of a 1.3 m cord and swung in a horizontal circle. What is the
maximum speed the ball can go if the breaking tension of the string is 75N.
The maximum centripetal force possible is 75 N. So, using the approach from #21:
v=
√ √
Fc r
75⋅1.3
m
=
=15
m
0.45
s
23. Two trucks with equal mass are attracted to each other with a gravitational force of 6.7 Х 10 -4 N.
The trucks are separated by a distance of 3.0 m. What is the mass of one of the trucks?
Fg =G
m=
√
m⋅m
m2
=G
r2
r2
m 2=
Therefore,
√
Fg r
G
2
and
√
Fg r 2
6.7⋅10− 4⋅32
10−4⋅32
=
=
= √ 107⋅32 =3000 √ 10=9500 kg
−11
−11
G
6.7⋅10
10
24. Two large spheres are suspended close to each other. Their centers are 2.2 m apart. One sphere
weighs 920 N. The other sphere has a weight of 149 N. What is the gravitational force between
them?
Fg =G
m1⋅m2
… we can use this formula if we know the masses. Since we have the weights
r2
(weight is a gravity force acting on an object due to gravity of Earth), we can calculate the
masses using weight=mg, or m=weight/g:
m1 = 920 N / 9.8 N/kg = 94 kg
Then
m1⋅m2
Fg =G
r
2
=6.67⋅10−11
and
m2 = 149 N / 9.8 N/kg = 15 kg
94⋅15
=1.9⋅10−8 N
2
2.2
(Note: 2 significant figures in the answer)
25. What is the force of gravity acting on a 2.00 x 103 kg spacecraft when it orbits at a distance equal
to two Earth radii from the Earth’s center?
Fg =G
m1⋅m2
r
2
= Fg =G
m1⋅m2
2
( 2 r E)
−11
=6.67⋅10
3
24
2.00⋅10 ⋅5.98⋅10
= 4.9⋅103 N
6 2
( 2⋅6.38⋅10 )
26. What must be the speed of a satellite moving in a circular orbit about the Earth at an altitude of
3600 km?
First, the distance between the satellite and the center of Earth is:
r = rE + altitude = 6380 km + 3600 km = 9980 km = 9.98·106 m
Fg=Fc, so, if the satellite has a mass of m,
√
√
G
m⋅m E
r
2
=m
24
mE
m
−11 5.98⋅10
v= G
= 6.67⋅10
= √ 4.00⋅107=6320
6
r
s
9.98⋅10
v2
. So,
r
v 2=G
mE
r
and