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1206 - Concepts of
Physics
Wednesday, December 02nd 2009
Notes
• Assignment #8 due today !!!
• Assignment #9 due on Monday - for those
who would like to get points
• Please pick up your assignments ...
• This afternoon is a good time to do that
Notes
• I will be in meetings - starting tomorrow
(Thursday, December 03) until Tuesday,
December 8th
• Therefore it will be harder to just come by
my office.
• You can as always sent emails - it might be
evening (late) until I can answer them keep that in mind
Office hours
Note! Since meeting locations changed these changed too
• Thursday, December 03 -- 13:00 - 14:30
• Friday, December 04 -- not available
• Monday, December 07 -- 11:30 - 13:30
• Tuesday, December 08 -- not available
( I will be at Laurentian, so “appointments”
possible)
• Wednesday, December 09 -- 9:00 - 10:15
Notes
• Today - random exercises and repeats in
preparation for final exam
• Friday - similar to today
• Monday - the “Physics of the 2012 movie”
• Wednesday - your chance to ask last
minute questions and discussion of
solutions for assignment #9
• I might switch Monday and Wednesday ...
Conservation Laws
• Conservation laws are very important
• They are also useful for solving Problems
• We have talked about several of them:
Energy conservation
Linear momentum conservation
Angular momentum conservation
The principle of conservation of energy states that energy cannot be created or
destroyed, although it can be changed from one form to another. Thus in any
isolated or closed system, the sum of all forms of energy remains constant. The
energy of the system may be interconverted among many different forms—
mechanical, electrical, magnetic, thermal, chemical, nuclear, and so on—and as
time progresses, it tends to become less and less available; but within the limits of
small experimental uncertainty, no change in total amount of energy has been
observed in any situation in which it has been possible to ensure that energy has not
entered or left the system in the form of work or heat. For a system that is both
gaining and losing energy in the form of work and heat, as is true of any machine in
operation, the energy principle asserts that the net gain of energy is equal to the
total change of the system's internal energy.
J. P. Joule and others demonstrated the equivalence of heat and work by showing
experimentally that for every definite amount of work done against friction there
always appears a definite quantity of heat. The experiments usually were so
arranged that the heat generated was absorbed by a given quantity of water, and
it was observed that a given expenditure of mechanical energy always produced
the same rise of temperature in the water. The resulting numerical relation
between quantities of mechanical energy and heat is called the Joule equivalent,
or is also known as mechanical equivalent of heat.
In view of the principle of equivalence of mass and energy in the restricted theory of
relativity, the classical principle of conservation of energy must be regarded as a
special case of the principle of conservation of mass-energy. However, this more
general principle need be invoked only when dealing with certain nuclear phenomena
or when speeds comparable with the speed of light (3 × 108 m/s) are involved.
Example - your turn
Mass of the ball m = 10 kg, the height h is
0.20 m. Calculate potential and kinetic
energy for the blue, dark purple and light
purple position of the ball. Explain how you
made use of the principle of energy
conservation.
Example - your turn
Mass of the ball m = 10 kg, the height h is
0.20 m. Calculate potential and kinetic
energy for the blue, dark purple and light
purple position of the ball. Explain how you
made use of the principle of energy
conservation.
There is no rotation, therefore we only need to use the linear terms. There is also no
elastic potential energy. Gravitational potential energy is given as PE = m g h and
kinetic energy is defined as 1/2 m v2. Energy conservation tells us that the sum of
potential and kinetic energy at every point must be the same. At the blue position, the
ball is not moving, therefore all energy is in the potential energy, which we can
calculate with he given values: PE = m g h = 10 kg * 9.8 m/s2 * 0.20 m = 19.6 J
The dark purple ball has not potential energy - everything has been converted into
kinetic energy and back to potential energy for the light purple location.
Pool balls are a good example
inelastic means shape changes!!
Objects stick together
Astronomers observe the planet Mars as the Martians fight a nuclear war. The Martian
bombs are so powerful that they rip the planet into three separate pieces of liquified
rock, all having the same mass. If one fragment flies off with velocity components(all in
the center of mass frame) what is the magnitude of the third one's velocity?
v1x = 0
and
v2x = 10000 km/h
v1y = 10000 km/h
and
v2y = 0 km/h
In the center of mass frame, the planet initially had zero
momentum. After the explosion, the vector sum of the
momenta must still be zero. Vector addition can be done by
adding components, so
m v1x + m v2x + m v3x = 0
m v1y + m v2y + m v3y = 0
where we have used the same symbol m for all the terms, because the fragments all
have the same mass. The masses can be eliminated by dividing each equation by m,
and we find
v3x = v3y = - 10000 km/h
which gives a magnitude of
|v3| = sqrt(v3x2 + v3y2) = 14000 km/h
A 120 kg lineman moving west at 2 m/s tackles an 80 kg football
fullback moving east at 8 m/s. After the collision, both players move
east at 2 m/s. Draw a vector diagram in which the before- and aftercollision momenta of each player is represented by a momentum vector.
Label the magnitude of each momentum vector.
YOUR TURN .. .
A 120 kg lineman moving west at 2 m/s tackles an 80 kg football
fullback moving east at 8 m/s. After the collision, both players move
east at 2 m/s. Draw a vector diagram in which the before- and aftercollision momenta of each player is represented by a momentum vector.
Label the magnitude of each momentum vector.
Angular momentum is the tendency of a rotating object to keep rotating at the same
speed about the same axis of rotation. Angular momentum is measured with respect to
a certain axis of rotation.
Objects executing motion around a point possess a quantity called angular momentum. This is an important physical quantity
because all experimental evidence indicates that angular momentum is rigorously conserved in our Universe: it can be
transferred, but it cannot be created or destroyed. For the simple case of a small mass executing uniform circular motion
around a much larger mass (so that we can neglect the effect of the center of mass) the amount of angular momentum takes a
simple form. As the adjacent figure illustrates the magnitude of the angular momentum in this case is L = mvr, where L is the
angular momentum, m is the mass of the small object, v is the magnitude of its velocity, and r is the separation between the
objects.
Formation of a neutron star
A star rotates with a period of 30 days about an axis through its center. After the star
undergoes a supernova explosion, the stellar core, which had a radius of 1.0 x 104 km,
collapses into a neutron star of radius 3.0 km. Determine the period of rotation of the
neutron star. (Assume that during the collapse no external torque acts on it and it
remains spherical with the same relative mass distribution and its mass remains
constant. )
Formation of a neutron star
A star rotates with a period of 30 days about an axis through its center. After the star
undergoes a supernova explosion, the stellar core, which had a radius of 1.0 x 104 km,
collapses into a neutron star of radius 3.0 km. Determine the period of rotation of the
neutron star. (Assume that during the collapse no external torque acts on it and it
remains spherical with the same relative mass distribution and its mass remains
constant. )
The distribution of mass is symmetric, so the moment of inertia can be expressed as
i = k m r2, where k is a numerical constant. For a solid sphere k = 2/5. The star
undergoes a change in period, we will call the initial period Ti and the final one Tf. The
period is the time required for the star’s equator to make one complete revolution
around the axis of rotations. The star’s angular speed is given by ω = 2π/T. We can use
the conservation of angular momentum (L = iω).
Formation of a neutron star
A star rotates with a period of 30 days about an axis through its center. After the star
undergoes a supernova explosion, the stellar core, which had a radius of 1.0 x 104 km,
collapses into a neutron star of radius 3.0 km. Determine the period of rotation of the
neutron star. (Assume that during the collapse no external torque acts on it and it
remains spherical with the same relative mass distribution and its mass remains
constant. )
The distribution of mass is symmetric, so the moment of inertia can be expressed as
i = k m r2, where k is a numerical constant. For a solid sphere k = 2/5. The star
undergoes a change in period, we will call the initial period Ti and the final one Tf. The
period is the time required for the star’s equator to make one complete revolution
around the axis of rotations. The star’s angular speed is given by ω = 2π/T. We can use
the conservation of angular momentum (L = iω).
Conservation of angular momentum: Li = Lf --> iiωi = ifωf
use expression for ω: ii (2π/Ti) = if (2π/Tf)
use expression for i:
k m ri2 (2π/Ti) = k m rf2 (2π/Tf)
Simplify and solve for Tf: Tf = (rf/ri)2 Ti
All values on the right side are given:
Tf = (3.0 km/10000 km) (30 days) = 2.7 x 10-6 days = 0.23 s
Doppler Submarines
A submarine (sub A) travels through water at a speed of 8.00 m/s, emitting a sonar
wave at a frequency of 1400 Hz. The speed of sound in the water is 1533 m/s. A second
submarine (sub B) is located such that both submarines are traveling directly toward
each other. The second submarine is moving at 9.00 m/s.
(a) What frequency is detected by an observer riding on sub B as the subs approach
each other?
(b) The subs barely miss each other and pass. What frequency is detected by an
observer riding on sub B as the subs recede from each other?
(c) While the subs are approaching each other, some of the sound from sub A reflects
from sub B and returns to sub A. If this sound were to be detected by an observer on
sub A, what is its frequency?
Doppler Submarines
A submarine (sub A) travels through water at a speed of 8.00 m/s, emitting a sonar
wave at a frequency of 1400 Hz. The speed of sound in the water is 1533 m/s. A second
submarine (sub B) is located such that both submarines are traveling directly toward
each other. The second submarine is moving at 9.00 m/s.
(a) What frequency is detected by an observer riding on sub B as th subs approach
each other?
(b) The subs barely miss each other and pass. What frequency is detected by an
observer riding on sub B as the subs recede from each other?
(c) While the subs are approaching each other, some of the sound from sub A reflects
from sub B and returns to sub A. If this sound were to be detected by an observer on
sub A, what is its frequency?
As the title already say’s we have to use what we learned about the Doppler effect.
Both subs are moving, therefore we start with the general case formula and we need
to remember the sign conventions depending on direction of movement.
f’ = {(v + v(O))/(v - v(S))} f
Doppler Submarines
A submarine (sub A) travels through water at a speed of 8.00 m/s, emitting a sonar
wave at a frequency of 1400 Hz. The speed of sound in the water is 1533 m/s. A second
submarine (sub B) is located such that both submarines are traveling directly toward
each other. The second submarine is moving at 9.00 m/s.
(a) What frequency is detected by an observer riding on sub B as th subs approach
each other?
(b) The subs barely miss each other and pass. What frequency is detected by an
observer riding on sub B as the subs recede from each other?
(c) While the subs are approaching each other, some of the sound from sub A reflects
from sub B and returns to sub A. If this sound were to be detected by an observer on
sub A, what is its frequency?
As the title already say’s we have to use what we learned about the Doppler effect.
Both subs are moving, therefore we start with the general case formula and we need
to remember the sign conventions depending on direction of movement.
f’ = {(v + v(O))/(v - v(S))} f
(a) f’ = {(1533 m/s + 9.00 m/s)/(1533 m/s) - (8.00 m/s)} 1400 Hz = 1416 Hz
(b) same formula, but now the sign of the two speeds v(O) and v(S) is changed, so
f’ = {(1533 m/s + (- 9.00 m/s))/(1533 m/s) - (- 8.00 m/s)} 1400 Hz = 1385 Hz
(c) The sound of apparent frequency 1416 Hz for (a) is reflected from a moving
source (sub B) and then detected by a moving observer (sub A), so
f’ = {(1533 m/s + (+ 8.00 m/s))/(1533 m/s) - (+ 9.00 m/s)} 1400 Hz = 1432 Hz
The mistuned piano strings
Two identical piano strings of length 0.750 m are each tunes exactly to 440 Hz. The
tension in one of the strings is then increased by 1.0%. If they are now struck, what is
the beat frequency between the fundamentals of the two strings.
The mistuned piano strings
Two identical piano strings of length 0.750 m are each tunes exactly to 440 Hz. The
tension in one of the strings is then increased by 1.0%. If they are now struck, what is
the beat frequency between the fundamentals of the two strings.
A small change in tension on one string will cause a small change in frequency
compared to the original string. Therefore we expect to “hear” a beats effect.
We need to set up the ration of the fundamental frequencies for the two strings
using f = (v/2L) for each of them.
The mistuned piano strings
Two identical piano strings of length 0.750 m are each tunes exactly to 440 Hz. The
tension in one of the strings is then increased by 1.0%. If they are now struck, what is
the beat frequency between the fundamentals of the two strings.
A small change in tension on one string will cause a small change in frequency
compared to the original string. Therefore we expect to “hear” a beats effect.
We need to set up the ration of the fundamental frequencies for the two strings
using f = (v/2L) for each of them.
Set up ratio: f2/f1 = (v2/2L)/(v1/2L) = v2/v1
substitute the wave speeds with sqrt(T/M), where M mass per unit length, and T
tension we get:
f2/f1 = sqrt{(T2/M)/(T1/M)} = sqrt(T2/T1)
T2 = 1.010 T1 (1% increase)
Therefore f2/f1 = 1.005
So f1 = 1.005 f1 = 1.005 * 440 Hz = 442 Hz
The difference is 2 Hz
One block pushes another
Two blocks of masses m1 and m2 with m1 > m2
are placed in contact with each other on a
frictionless, horizontal surface. A constant
horizontal force F is applied to m1, as shown.
F
m1 m2
(a) Find the magnitude of the acceleration of the system
(b) Determine the magnitude of the contact force between the two blocks
(c) When the force is applied toward the left on m2, is the magnitude of the contact force
the same as it was when the force was applied toward the right on m1?
One block pushes another
Two blocks of masses m1 and m2 with m1 > m2
are placed in contact with each other on a
frictionless, horizontal surface. A constant
horizontal force F is applied to m1, as shown.
F
m1 m2
(a) Find the magnitude of the acceleration of the system
(b) Determine the magnitude of the contact force between the two blocks
(c) When the force is applied toward the left on m2, is the magnitude of the contact force
the same as it was when the force was applied toward the right on m1?
(a) “particle” under a net force. The two blocks are connected - so they can be treated
as one particle with mass m = m1 + m2. We have to apply Newton’s second law to
find the acceleration. The particle will move in +x direction.
ΣFx = F = (m1 + m2) ax therefore ax = F/(m1+m2)
One block pushes another
Two blocks of masses m1 and m2 with m1 > m2
are placed in contact with each other on a
frictionless, horizontal surface. A constant
horizontal force F is applied to m1, as shown.
F
m1 m2
(a) Find the magnitude of the acceleration of the system
(b) Determine the magnitude of the contact force between the two blocks
(c) When the force is applied toward the left on m2, is the magnitude of the contact force
the same as it was when the force was applied toward the right on m1?
(b) To find the contact force P we have to consider the blocks individually. First we draw a
free-body diagram for each block, where P12 is the contact force from m1 on m2.
FN1
F
m1
m1 g
FN2
P21
P12
m2
m2 g
Apply Newton’s second law to m2:
ΣFx = P12 = m2 ax
Substitute ax from (a)
P12 = {m2/(m1 + m2)} F
Apply Newton’s second law to m1:
ΣFx = F - P21 = F - P12 = m1 ax
P12 = {m2/(m1 + m2)} F
One block pushes another
Two blocks of masses m1 and m2 with m1 > m2
are placed in contact with each other on a
frictionless, horizontal surface. A constant
horizontal force F is applied to m1, as shown.
F
m1 m2
(a) Find the magnitude of the acceleration of the system
(b) Determine the magnitude of the contact force between the two blocks
(c) When the force is applied toward the left on m2, is the magnitude of the contact force
the same as it was when the force was applied toward the right on m1?
Both blocks are connected - therefore they must experience the same acceleration.
We have a net force acting on the “system” or “particle, so we will use Newtons second
law: F = m a for each component (horizontal and vertical)