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1206 - Concepts in
Physics
Wednesday, September 30th
NOTES
• Check out changes in assignment #2
The first and second exercise have gotten
more comments and #3 has been replaced
by and easier version.
• Additional tutorial (reminder):
THURSDAY - 16:30-18:00 F536 for all
first year students, so it doesn’t replace our
own tutorial on Monday’s
NOTES
• From now on, we will use the definitions
and notations given in the textbook as
much as possible, but be aware that this is
one particular choice these author’s made.
Other textbooks might use different
notations. You should always think if what
you are calculating makes sense.
• Especially the “word” alone does not
necessarily tell you if a quantity is a vector
or a scalar. These things get mixed up. Using
different words is often meant as a help
while learning.
Notes - Midterm
•
Topics for Midterm on November 2nd are: particle
kinematics, forces, Newton’s laws, velocity, acceleration,
circular motions, momentum, solids and fluids
•
Best preparation is to understand assignments and
example exercises during class. For more practice, there
are lots of examples in the textbook
•
We will use the last class before reading week
(Friday, Oct 23rd) to discuss assignment #4
•
OFFER: extra tutorial on Monday, October 25th (reading
week) at 10:30 am F055. We’ll go through some exercises
and you can ask questions.
Satellite
orbits
Reminder: circular motion
Uniform circular motion is the motion of an object traveling at a constant (uniform) speed on
a circular path.
Often it is more convenient to describe it by specifying the period of the motion, rather than the
speed. The period T is the time required to travel once around the circle - one complete
revolution. The speed v is the distance traveled (circumference of the circle) divided by the time T:
v = 2πr/T
where r is the radius of the circle.
If the radius is known the speed can be calculated from the period and vise versa.
The wheel of a car has a radius of r = 0.29 m and is being rotated at 830 revolutions minute.
Determine the speed at which the outer edge of the wheel is moving.
We need the period T first: Since the tire makes 830 revolutions in one minute, the number of
minutes required for a single revolution is: 1/(830 revolutions/min) = 1.2 x 10-3 min/revolution
Therefore, the period is T = 1.2 x 10-3 min or 0.072 s.
Now we calculate v = 2πr/T = 2π(0.29m)/0.072s = 25 m/s
circular motion has
constant speed (scalar), but the velocity
(vector) is constantly changing direction.
Changing velocity (magnitude or
direction) mean there is acceleration. This
particular acceleration is called
“centripetal acceleration” because is
points toward the centre of the circle.
The magnitude of the centripetal
acceleration ac depends on the speed v of
the object and the radius r of the circular
path. As always, acceleration is the change
in velocity Δv divided by the elapsed time
Δt.
a = Δv/Δt
The magnitude of ac is given by
ac = v2/r
The direction always points to the centre
of the circle.
Example:
The bobsled track at the 1994 Olympics in Lillehammer, Norway, contained turns with radii of 33 m
and 24 m. Find the centripetal acceleration at each turn for a speed of 34 m/s, a speed that was
achieved in the two-man event. Express the answers as multiples of g as well.
We can use the formula for centripetal acceleration:
ac = v2/r
Since r is in the denominator, we expect the acceleration
to be smaller when r is larger
Radius = 33 m
2
2
ac = (34m/s) : 33 m = 35 m/s = 3.6 g
Radius = 24 m
ac = (34m/s)2 : 24 m = 48 m/s2 = 4.9 g
Centripetal Force
We know from Newton’s second law, that whenever an object accelerates, there must be a net
force to create the acceleration. Thus, in uniform circular motion there must be a net force to
produce the centripetal acceleration. This force is given by the product of the objects mass and its
acceleration according to Newton’s second law, so in this case:
Fc = mv2/r
The centripetal force always points toward the center of the circle and continually changes direction
as the object moves.
A model airplane has a mass of 0.90 kg an moves at a constant speed
on a circle that is parallel to the ground. The path of the airplane and
its guideline lie in the same horizontal plane, because the weight of the
plane is balanced by the lift generated by its wings. Find the tension T
in the guideline (length = 17 m) for speeds of 19 m/s and 38 m/s.
The tension T in the guideline is the only force pulling the plane
inward , therefore it must be the centripetal force.
T = Fc = mv2/r
v = 19 m/s:
T= (0.90 kg)(19 m/s)2 / 17m = 19 N
v = 38 m/s:
T = (0.90 kg)(38 m/s)2 / 17m = 76 N
YOUR TURN:
In a circus, a man hangs upside down from a trapeze, legs bent over the bar and arms downward,
holding his partner. Is it harder for the man to hold his partner when the partner hangs straight
down and is stationary or when the partner is swinging through the straight-down position?
YOUR TURN:
In a circus, a man hangs upside down from a trapeze, legs bent over the bar and arms downward,
holding his partner. Is it harder for teh man to hold his partner when the partner hangs straight
down and is stationary or when the partner is swinging through the straight-down position?
When the man and his partner are stationary, the man’s arms
must support his partner’s weight. When the two are swinging,
however, the man’s arms must do an additional job. Then, the
partner is moving on a circular arc and has a centripetal
acceleration. To produce this acceleration.
Because of the additional pull, it is harder for the man to
hold his partner while swinging.
Centripetal force and safe driving
Compare the maximum speeds at which a car can safely negotiate an unbanked turn
(radius = 50.0 m) in dry weather (coefficient of static friction = 0.900) and icy weather (coefficient
of static friction = 0.100).
At the maximum speed, the maximum centripetal force acts on the tires, and static friction must
provide it. The magnitude of the maximum force of static friction is specified as follows:
fsmax = μsFN where μs is the coefficient of static friction and FN is the magnitude of the
normal force. We need to find the normal force, substitute it into the expression for the
maximum force of static friction and then equate the result to Fc. We expect the maximum
speed to be greater for dry road from experience.
Since the car does not accelerate in the vertical direction, the weight mg of the car is balanced by
the normal force, so FN = mg
Fc = μs FN = μs m g = mv2/r
Consequently: μs g = v2/r
v = sqrt(μs g r)
Note! The mass of the car has been eliminated - the max speed down not depend on the mass.
dry road (μs = 0.900)
icy road (μs = 0.100)
v=sqrt(0.900x9.8m/s2x50m) = 21.0 m/s
v=sqrt(0.100x9.8m/s2x50m) = 7.00 m/s
Excursion: Fundamental Forces
In nature there are two general types of forces, fundamental and nonfundamental. Fundamental forces
are the ones that are truly unique, in the sense that all other forces can be explained in terms of
them. Only three fundamental forces can describe nature today:
1. Gravitational force
2. Strong nuclear force
3. Electroweak force
We will discuss the gravitational force today. The strong nuclear force plays a primary role in the
stability of the nucleus. The electroweak force is a single force that manifests itself in two ways:
The electromagnetic force that electrically charged particles exert on one another and the socalled weak force that plays a role in the radioactive decays taking place for certain nuclei.
This understanding has involved over time and will continue to do so. Theoretical physicists today
are working on the so-called grand unification and hope to find just one force describing
everything. In the past (1860-1870) James Clerk Maxwell showed that the electric force and the
magnetic force can be explained as manifestations of a single electromagnetic force. Then in the
1970s (100 years later), Sheldon Glashow, Abdus Salam and Steven Weinberg presented a theory
that explains how the elctromagnetic force and the weak force are related to the electroweak
force. They received a Novel Prize in 1979 for this theory.
The gravitational force
Newton’s law of universal gravitation
We know, that objects fall downward because of gravity. The downward acceleration on earth is
described by using a value of g = 9.80 m/s. The acceleration due to gravity is like any other
acceleration, and Newton’s second law indicates that is must be caused by a net force.
NEWTON’S LAW OF UNIVERSAL GRAVITATION
Every particle in the universe exerts an attractive force on every other particle. A particle is a
piece of matter, small enough in size to be regarded as a mathematical point. For two particles that
have masses m1 and m2 and are separated by a distance r, the force that each exerts on the other is
directed along the line joining the particles and has a magnitude given by:
Where the symbol G denotes the universal
gravitational constant, whose value is found
experimentally to be G = 6.673 x 10-11 N m2/kg2
Some history
The constant G is called the universal gravitational constant, because it has the same value for
all pairs of particles anywhere in the universe, no matter how far apart they are.
The value for G was first measured in an experiment by the English scientist Henry Cavendish
(1731 - 1810)
To see the main features of Newton’s law of universal
gravitation, we look at two particles (p1, p2) with
masses m1 and m2, separated by a distance r.
+F, the gravitational force exerted on p1 by p2
-F, the gravitational force exerted on p2 by p1
The two forces have equal magnitudes and opposite
directions. They act on different bodies, causing them
to be mutually attracted. In fact, these forces are an
action -reaction pair, as required by Newton’s third law.
Example
Assume mass m1 = 12 kg and m2 = 25 kg with r = 1.2 m
Now put these number in the equation:
F = G (m1 * m2)/r2 =
(6.67 x 10-11 N m2/kg2) (12 kg)(25kg) / (1.2 m)2 =
1.4 x 10-8 N
In many cases, the gravitational force is very small - mainly because G is small. For example, you
exert a force of about 1N when pushing a doorbell.
YOUR TURN:
However, if one of the bodies has a large mass, like the earth (5.98 x 1024 kg), the gravitational
force can be large. Let’s look at the case of moon and earth (still assuming they are points). The
mass of the moon is: 7.36 x 1022 kg and the distance between earth and moon is 384400 km.
What is the gravitational force between the two “particles”?
Example
Assume mass m1 = 12 kg and m2 = 25 kg with r = 1.2 m
Now put these number in the equation:
F = G (m1 * m2)/r2 =
(6.67 x 10-11 N m2/kg2) (12 kg)(25kg) / (1.2 m)2 =
1.4 x 10-8 N
In many cases, the gravitational force is very small - mainly because G is small. For example, you
exert a force of about 1N when pushing a doorbell.
YOUR TURN:
However, if one of the bodies has a large mass, like the earth (5.98 x 1024 kg), the gravitational
force can be large. Let’s look at the case of moon and earth (still assuming they are points). The
mass of the moon is: 7.36 x 1022 kg and the distance between earth and moon is 385000 km.
What is the gravitational force between the two “particles”?
F = 6.67x10-11 N (m/kg)2 (5.98x1024 kg)(7.36x1022 kg)/(3.85x107)2 m
= (6.67 x 5.98 x 7.36/3.852) x 1021 N
= 1.98 x 1020 N
Weight
Definition of weight
The weight of an object on or above the earth is the gravitational fore that the earth exerts on the
object. The weight always acts downward, toward the center of the earth. On or above another
astronomical body, the weight is the gravitational force exerted on the object by that body.
SI unit of weight: Newton [N]
Using W for the magnitude of the weight, m for the mass of the object and ME for the mass of the
earth, if follows easily from the universal gravitation law:
W = G ME m/r2
The force acts even when the distance r is not equal to the radius of the earth. However, the
gravitational force becomes weaker as r increases, since r is in the denominator. The decrease is a
quadratic function. So twice the distance means a quarter of the force.
YOUR TURN:
The mass of the Hubble Space Telescope is 11600 kg. Determine the weight of the telescope when is
was resting on earth and as it is in its orbit 598 km above the earth surface. Use the center of the
earth as your reference point for the radii. The diameter of the earth is 12760 km.
YOUR TURN:
The mass of the Hubble Space Telescope is 11600 kg. Determine the weight of the telescope when is
was resting on earth and as it is in its orbit 598 km above the earth surface. Use the center of the
earth as your reference point for the radii. The diameter of the earth is 12760 km.
On the earth’s surface:
r = 6380 km = 6380000 m = 6.38 x 106 m
G = 6.67 x 10-11 N (m/kg)2
ME = 5.98 x 1024 kg
m2 = 11600 kg
W = G ME m/r2 = 1.14 x 105 N
In orbit - 598 km above the surface:
r = (6380 + 598) km = 6978000 m = 6.98 x 106 m
G = 6.67 x 10-11 N (m/kg)2
ME = 5.98 x 1024 kg
m2 = 11600 kg
W = G ME m/r2 = 0.95 x 105 N
As expected the weight is lower in orbit.
Satellites in circular orbits
There are many satellites in orbit about the
earth. The ones in circular orbits are examples
for uniform circular motion. Each satellite is kept
on its circular path by a centripetal force. The
gravitational pull of the earth provides the
centripetal force and acts like an invisible
guideline for the satellite.
There is only one speed that a satellite
can have if the satellite is to remain in
an orbit with a fixed radius.
The gravitational force is the only force acting on the satellite in the radial direction, it alone
provides the centripetal force. Using Newton’s law of gravitation, we can write:
Fc = G
m ME
r2
=
m v2
r
v = √GME/r
where G is the universal gravitational constant, ME the mass of the earth and r is the distance from
the center of the earth to the satellite.
Orbital speed of the Hubble space telescope
Let’s determine the speed of the Hubble Space Telescope
orbiting at a height of 598 km above the earth’s surface.
We already calculated the orbital radius in a previous slide
to be 6.98 x 106 m.
G = 6.67 x 10-11 N (m/kg)2
ME = 5.98 x 1024 kg
v = √GME/r = 7.56 x 103 m/s
look at units:
sqrt{N m2 kg-2 kg m-1} = sqrt{kg m s-2 m kg-1} = sqrt{(m/s)2}
look at 10n:
10-11 x 1024 x 10-6 = 107
(6.67 x 5.98)/6.98 = 5.71
look at numbers:
Just take sqrt(5.71x107) = 7.56 x 103
Example:
A motorcyclist is trying to leap across the canyon
v0 = 38 m/s
Once the cyclist leaves the cliff
no forces other than gravity act on
the cycle. We ignore air resistance.
vf = ?
h0 = 70 m
hf = 35 m
Find the speed with which the cycle
strikes the ground on the other
side.
No work is done by external nonconservative forces. So all we need to know is, that the total
mechanical energy is the same at the final and initial positions of the motorcycle.
We can write the principle of conservation of mechanical energy as follows:
KE0 + PE0 = KEf + PEf
1/2mv02 + mgh0 = 1/2mvf2 + mghf
The mass m appears in every term and can therefore be eliminated. Now, we are looking for vf
vf2 = v02 + 2g (hf - h0) = (38 m/s)2 + 2(9.8m/s2)(35m) = 2130 (m/s)2
vf = 46.2 m/s
YOUR TURN:
The tallest and fastest roller coaster in the world is now the Steel Dragon in Mie, Japan. The ride
includes a vertical drop of 93.5 m. The coaster has a speed of 3.0 m/s at the top of the drop. Neglect
friction and find the speed in the riders at the bottom.
This time, we will take nonconservative forces into account - like friction. The actual speed at the
bottom is 41.0 m/s. Assuming again, that the coaster has a speed of 3 m/s at the top, find the work
done by nonconservative forces on a 55.0 kg rider during the descent from a height h0 to a height hf,
where h0 - hf = 93.5 m.
Neglect friction. We can use the principle of conservation of mechanical energy to find the speed
of the riders at the bottom.
E0 = Ef
1/2mv02 + mgh0 = 1/2mvf2 + mghf
vf2 = v02 + 2g (hf - h0) = (3.0 m/s)2 + 2(9.8m/s2)(93.5m)
vf = 42.9 m/s
Since the speed at the top, the final speed, and the vertical drop are given, we can determine the
initial and final total mechanical energies of the rider. The work-energy theorem, Wnc = Ef - E0
Wnc = Ef - E0 = 1/2mvf2 + mghf - (1/2mv02 + mgh0)
Wnc = 1/2m(vf2 - v02) - mg(h0 - hf)
Wnc = 1/2(55.0 kg)[(41.0m/s)2-(3.0m/s)2]-(55.0 kg)(9.8 m/s2)(93.5 m)
Wnc = 27.5 kg 1672 (m/s)2-(50396.5) kg (m/s)2
Wnc = 45980 Nm - 50397 Nm
Wnc = - 4417 J