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Chapter 5 Circular Motion; Gravitation
Chapter 5
Circular Motion; Gravitation
• Kinematics of Uniform Circular Motion
• Dynamics of Uniform Circular Motion
• Highway Curves, Banked and Unbanked
• Non-uniform Circular Motion
• Centrifugation
Will be covered
after chapter 7
• Centrifugation
• (5.6) Newton’s Law of Universal Gravitation
• (5.7) Gravity Near the Earth’s Surface
Recalling Recalling LastLast LectureLectureRecalling Recalling LastLast LectureLecture
MidtermMidterm
Date: October 16, 1pm – 2:15pm. Location: CL-110.
Remind:
- There are four problems. One of them will be selected between problems 52 and
87 found in assignment 5. The midterm includes everything covered until last
Thursday, including friction and inclines (chapters 1 to 4).
- You ARE allowed to use a calculator. An equation sheet will be provided (you can
find it at http://ilc2.phys.uregina.ca/~barbi/academic/phys109/phys109.html )
- Solutions of the assignments can be found on the webpage listed above. You can
also find a link to old midterms from this webpage.
- I will be exceptionally available for discussions, questions, etc between 14:30 and
15:00h today, and between 13:00 and 15:00h tomorrow.
- I strongly recommend the Wednesday’s tutorial session: an old midterm will be
solved.
- Remember, your marked assignments are available for you. Feel free to come
anytime to my office (LB-212) if you have failed to pick it up in classroom.
FrictionFriction and and InclinedInclined PlanesPlanes
Three forces can ALWAYS be identified acting on an object moving on an inclined
place:
• Gravity (vertical);
• Friction (along the surface);
• Normal (perpendicular to the surface).friend 1
friend 2
you
(4.5)
(4.6)
Problem 4-41 (textbook A 15.0-kg box is released on a 32º incline and accelerates
down the incline at 0.30 m/s2. Find the friction force impeding its motion. What is the
coefficient of kinetic friction?
Tension in a Flexible CordTension in a Flexible Cord
= 0.30 m/s2
32o
= 0.30 m/s2
Problem 4.41 (textbook):
A free-body diagram for the box is shown.
Write Newton’s 2nd law for each direction:
Notice that the sum in the y direction is 0, since
y
x
θ
θ
mgr
NFr
frFr
fr
N
sin
cos 0
x x
y y
F mg F ma
F F mg ma
θ
θ
= − =
= − = =
∑∑
Notice that the sum in the y direction is 0, since
there is no motion (and hence no acceleration) in
the y direction. Solve for the force of friction.
Now solve for the coefficient of kinetic friction. Note that the expression for the
normal force comes from the y direction force equation above.
mg
( ) ( )( )fr
2 o 2
fr
sin
sin 15.0 kg 9.80m s sin 32 0.30m s 73.40 N 73 N
x
x
mg F ma
F mg ma
θ
θ
− = →
= − = − = ≈
( )( )( )fr
fr N 2 o
73.40 Ncos 0.59
cos 15.0 kg 9.80m s cos 32k k k
FF F mg
mgµ µ θ µ
θ= = → = = =
Problem 4-65 (textbook A bicyclist of mass 65 kg (including the bicycle) can coast
down a 6.0º hill at a steady speed of 6.0 Km/h because of air resistance. How much
force must be applied to climb the hill at the same speed and same air resistance?
Tension in a Flexible CordTension in a Flexible Cord
Problem 4.65 (textbook):
Consider a free-body diagram for the cyclist coasting downhill at a constant speed.
Here we call Ffr the friction due to air resistance (and not sliding or static friction).
Since there is no acceleration, the net force in each direction must be zero. Write
Newton’s 2nd law for the x direction.
This establishes the size of the air friction force at
6.0 km/h, and so can be used in the next part.
fr frsin 0 sin
xF mg F F mgθ θ= − = → =∑ y
θ
NFr
frFr
θ6.0 km/h, and so can be used in the next part.
Now consider a free-body diagram for the cyclist
climbing the hill. Fp is the force pushing the cyclist
uphill. Again, write Newton’s 2nd law for the x direction,
with a net force of 0.
xθ
mgr
θ
y
x
θθ
mgr
NFr
frFr
PFr
fr Psin 0
xF F mg Fθ= + − = →∑
( )( )( )P fr
2 o 2
sin 2 sin
2 65 kg 9.8 m s sin 6.0 1.3 10 N
F F mg mgθ θ= + =
= = ×
Newton’s Newton’s Law of Universal GravitationLaw of Universal Gravitation
If the force of gravity is being exerted on objects on Earth, what is the origin of that
force?
Newton’s realization was that the forceNewton’s realization was that the force
must come from the Earth.
He further realized that this force must be
what keeps the Moon in its orbit.
Newton’s Newton’s Law of Universal GravitationLaw of Universal Gravitation
More than that: the Earth exerts a downward force on you, and you exert an
upward force on the Earth. This follows from Newton’s third law.
When there is such a difference in masses, the reaction force is undetectable, but for
bodies more equal in mass it can be significant.
Therefore, the gravitational force must be proportional to both masses.
Newton’s Newton’s Law of Universal GravitationLaw of Universal Gravitation
By observing planetary orbits, Newton also concluded that the gravitational force
must decrease as the inverse of the square of the distance between the masses.
Summary: The gravitational force is directly proportional to the masses and inversely
proportional to the square of their distance.
Or, as stated in Newton’s law of universal gravitation:
“ Every particle in the universe attracts every other particle with a force that is
proportional to the products of their masses and inversely proportional to the
square of the distance between them. This forces acts along the line joining the
two particles.”
Newton’s Newton’s Law of Universal GravitationLaw of Universal Gravitation
The magnitude of this force is given by:
Where:
(4.7)
Newton’s Newton’s Law of Universal GravitationLaw of Universal Gravitation
We know the relation between force and acceleration: . But what is a in
?
If m2 is the object exerting a force on m1 , then the gravitational acceleration felt
by object 1 can be identified as:
(4.8)⇒ ⇒
The opposite is also true: object 2 will feel an acceleration
due to the gravitational force applied by object 1.
You can then write for the force acting on m1 and for the
force acting on m2 .
Gravity Near the Earth’s SurfaceGravity Near the Earth’s Surface
An object of mass m ON the surface of the Earth will feel a force given by:
Where,
mE = mass of the Earth ;
rE = 6.38 x 106 m = radius of the Earth.
g has been measured and it is known to be 9.80 m/s2.
(4.9)
g has been measured and it is known to be 9.80 m/s2.
In fact, the value of g can be considered constant at any position near the Earth’s
surface (this is what we have been assuming without much discussion so far).
Note that knowing G, rE and g, we can calculate the mass of the Earth. From eq. 4.9:
More accurate calculations lead to mE = 5.974 x 1024 Kg
Newton’s Newton’s Law of Universal GravitationLaw of Universal Gravitation
Problem 5-31 (textbook): A hypothetical planet has a radius 1.5 times that of Earth,
but has the same mass. What is the acceleration due to gravity near its surface?
Solution:
The acceleration due to gravity at any location on or above the surface of a planet is
given by
2
planet Planetg G M r=
where r is the distance from the center of the planet to the location in question.
( )
2
2Planet Earth Earth
planet Earth22 2 2 2 2
EarthEarth
1 1 9.8 m s4.4 m s
1.5 1.5 1.51.5
M M Mg G G G g
r RR= = = = = =
Newton’s Newton’s Law of Universal GravitationLaw of Universal Gravitation
Problem 5-34 (textbook): Calculate the effective value of g, the acceleration of
gravity, at (a) 3200 m, and (b) 3200 km, above the Earth’s surface.
Solution:
The acceleration due to gravity at any location at or above the surface of a planet is
given by
where r is the distance from the center of the planet to the location in question. For
this problem,
(a)
2
p lanet P lanetg G M r=
24
P lanet E arth5 .97 10 kgM M= = ×
(a)
(b)
6
Earth3200 m 6.38 10 m 3200 mr R= + = × +
( ) ( )( )
24
11 2 2 2Earth
22 6
5.97 10 kg6.67 10 N m kg 9.77 m s
6.38 10 m 3200 m
Mg G
r
−×
= = × =× +
�
6 6 6
Earth3200 km 6.38 10 m 3.20 10 m 9.58 10 mr R= + = × + × = ×
( ) ( )( )
24
11 2 2 2Earth
22 6
5.97 10 kg6.67 10 N m kg 4.34 m s
9.58 10 m
Mg G
r
−×
= = × =×
�
