UNIT 3: CIRCULAR MOTION, WORK AND ENERGY
Unit 3A Circular Motion – Acceleration with Changes in Direction
Acceleration has been defined as the rate of change of velocity.
Velocity has been defined as a vector with a magnitude (speed) and a
direction.
So far, we have considered acceleration only for objects slowing
down or speeding up, ie., changes in the speed component of velocity. But
objects can be accelerating without changing speeds. When objects are
going at a constant speed but continually change their direction, they are
accelerating as well. The result of this acceleration is circular motion.
Period T and Frequency f
The time it takes for one complete revolution of circular motion is
called the period and given a special symbol T. The inverse of this time
period T is the frequency f, which gives the number of revolutions in one
second. The units for frequency are Hertz Hz, where 1 Hz = 1
revolution/second.
f 
1
T
Example 1: A 30 cm vinyl record revolves on a turntable that "turns" a
record around and around at 33 1/3 RPM (revolutions per
minute)
Example:
above?
What is the period and frequency of the 33 1/3 RPM record
Since 33 1/3 RPM refers to 33 1/3 revolutions per minute
33
1 rev  1min 
rev
 0.55Hz =f

  0.555
3 min  60s 
s
T
Example:
1
1

 1.8s
f 0.555Hz
(per 1 revolution)
Old vinyl records were separated into two categories: albums
or LP’s (for long play) which played at 33 1/3 RPM, and
“singles” with one song on an A side and another song on the
B side, playing at 45 RPM.
a) What is the period of a single revolving on a turntable?
For one revolution, the needle covers the distance around the perimeter of
the record's circle, d = 2  R, where R is the radius of the circle.
For circular motion, the formula for speed v of an object at the perimeter of
the circle is a special application of the constant speed formula v = d/t
v
2 R
T
Going back to the record question: (45RPM for a single)
If the diameter of a single is 13.0 cm, what is the speed of the turntable
needle as it revolves at the outer tip of the record?
Here v represents the constant speed of an object undergoing circular
motion, its direction always at right angles to a line through the centre of the
circle or tangential to its speed at any point.
Acceleration in Circular Motion: Centripetal Acceleration
If the speed is not changing in circular motion, how does an object
accelerate in circular motion? To understand circular motion acceleration,
let us freeze the motion of an object in circular motion at several instants in
its circular path.
Two things to note: 1) while the speed v is constant, the direction of v
is constantly changing, and 2) the direction of v is always perpendicular (at
right angles) to the line from the center of the circle (tangential to the circle).
Following Newton's First Law (inertia), the object at any instant will want to
keep going at its constant velocity, which includes its constant speed and
constant direction. It keeps going at its constant speed, but it is constantly
changing direction, bending towards the circular path, so it is accelerating.
If we could draw a vector indicating which direction v is changing,
the arrow would point directly to the center of the circle.
This then is the direction of the acceleration: the direction of
acceleration in circular motion is always directed towards the center of the
circle. The term for this acceleration is the centripetal ("centre-seeking")
acceleration, ac, given by the formula
where v refers to the constant speed and r refers to the radius of the circular
orbit.
Note that this formula only gives the magnitude of the centripetal
acceleration as a function of the speed v. The direction of the acceleration
is always towards the center of the circle. We can extend this formula by
substituting the speed formula v = 2R/T to get
ac = (2R/T)2
R
= 4 2 R
T2
Example 2: Calculate the magnitude and direction of the acceleration of the
turntable needle at the outside of the 33 1/3 RPM (R=15 cm)
record.
Derivation of Centripetal Acceleration Formula (*optional)
Consider a velocity vector v1 originating from a point P1 on a circle of
radius R. The vector is of magnitude v and in the direction perpendicular to
the line from the centre of the circle C. Now consider another point P2 on
the circle a distance from P1. A line from C to P2 will also be of magnitude
R and set at an angle  from the line from C to P2.
We can draw a new vector v1 (in dashed lines above) originating from P2,
parallel to the original v1 vector. By similar angles this new vector v1 will
be set at an angle  = 90o –  from the line from C to P2.
At point P2 we can insert a second velocity vector v2, again of magnitude v
and now at right angles to the line from C to P2. To get a change in velocity
vector v, we can add the negative of vector v1 tail-to-tip to the second
vector v2. The resultant is v = v2 + - v1.
If we denote the displacement from P1 to P2 as the change in displacement
d, we get a triangle C-P1-P2. By similar angles, the angle between v1 and
v2 is also . Since v1 and v2 are of the same magnitude v and angle 
between them, and the lines C to P and C to P are both of the same
magnitude R with the same angle  between them, we have two similar
triangles. This allows us to state:
v = d
v
R
For small angle , we can approximate the displacement d = v t.
Substituting in the above, and rearranging, we get
v = v t
v
R
v= v (v)
t
R
Since by definition, acceleration is the change in velocity over the change in
time, then
ac = v 2
R
Centripetal Force
If there is acceleration, then, by Newton's 2nd Law (F=ma), there
must be a net force accompanying centripetal acceleration in circular
motion. This net force is directed towards the center of the circle like the
acceleration, and appropriately named the centripetal force.
centripetal force can be caused by;
 the tension of the string or rope when an object is whirled around in a
circle on a string
 the friction of the wheels on the road when a car is rounding an
unbanked curve
 the electric force of attraction between an electron circling a proton.
 gravity toward a large central body
Without the force – tension, friction or electrostatic - the object would
continue going in its straight path. The centripetal force causes the
centripetal acceleration, in this case a change in direction of v towards the
center of the circle. The constant change in the direction of v causes the
object to follow circular motion, instead of a straight line path.
Centripetal force is the net force.
The formula for centripetal force, Fc, derives directly from centripetal
acceleration and Newton's 2nd Law:
Fc  mac  m
v2
R
Again, we can substitute the formula for v in circular motion to get the
expanded formula for centripetal force
2
 4 2 R 2 
 2 R 




T 2  4m 2 R
T 


Fc  m
m

R
R
T2
Example 3: Calculate the centripetal force on the needle of the record
player in the above 33 1/3 RPM record example if the needle
weighs 1.5 grams.
Horizontal Circular Motion
On the surface of the earth, when an object is undergoing circular
motion along the horizontal (flat), since the force of gravity is perpendicular
to the centripetal force, there is no interaction- the net force is the
centripetal force causing circular motion.
Example 4: A 10 kg hammer is swung horizontally around and around on a
2.0 meter radius chain (including arm length) at a constant
speed of 20 m/s.
1) What is the centripetal acceleration of the hammer?
2) What is the tension on the chain?
3) What horizontal direction will the hammer travel when the
chain is released?
a) outward because of the centrifugal force
b) inward because of the centripetal force
c) in a straight line along the present v direction because of its inertia
d) continue along its circular path as it was doing before
4) Assuming the hammer is released from the horizontal, at a
height of 1.5 m,how far will the hammer travel horizontally before it hits the
ground?
Centrifugal Force
Centripetal force is a real force. The tension in a string whirling horizontally, the
force of the track on the roller coaster with or against gravity, or the friction between the
road and a car as it rounds a curve: these are all examples of centripetal force.
Unfortunately, we do not feel centripetal force as a normal push or pull on us, like
other forces. For example, what do you feel as you round a bend? Imagine you are
driving in a car and you are turning right around a corner. The centripetal force is
directed to the right. Yet, you feel yourself pushed to the left. Why? Likewise, in a
merry-go-round, or like the amusement park ride where you stand back against a wall
that circles around and around, until the floor drops off beneath your feet. The centripetal
force is directed inward, yet you find yourself being pushed outward, to the point where
you find yourself pinned against the outer wall and defying gravity in the amusement
park ride. Why is the direction of the force we feel opposite to the direction of the
centripetal force?
Imagine you are standing in an elevator, accelerating upward. What do you feel?
While the tension on the cable lifting the elevator is directed upward, you feel as if you
are being pushed down the elevator floor. This is very much like the feeling you get in
circular motion. Pretend, that instead of accelerating upward, the elevator is instead
circling horizontally at the end of a large rope. The elevator is still accelerating
"upward", but now it is a centripetal acceleration - and the centripetal force/tension on the
rope causing this elevator's acceleration is directed towards the centre of a circle. Inside
the elevator it would still feel like you were being pushed downward, to the floor. This is
very similar to the case of the amusement park ride where you feel as if you are being
pushed outward. The floor of the elevator corresponds to the outside wall of the circle.
In actuality, what you are feeling is called a pseudo-force (pseudo = false). In
both cases above, what is happening is that you are moving at constant velocity and your
inertia makes you want to stay moving at whatever speed and direction you're moving.
The elevator and the object, on the other hand, are accelerating. The elevator is
accelerating "upwards" and you - wanting to stay still - you feel as if you are being
pushed "downward", but it is the elevator that is actually pushing upwards at you.
Likewise in the circling object, inside the object, you want to keep moving in a straightline path. The object, though, is accelerating inward. The outer wall of the object is
pushing inward at you, not you pushing outward at all. The feeling of pushing outward
opposite to centripetal force in circular motion seems real, though, and it is given a
special name, called the centrifugal force. Centrifugal force is a pseudo-force.
(Newton’s 3rd Law)
Simulating Gravity
http://www.wired.com/wiredscience/2013/05/gravity-in-the-elysium-spacestation/
The force of gravity holds us to the surface of the earth. In space travel there
is no gravity, but we can simulate gravity through circular motion.
Speculation has space stations in the future built as giant wheels, revolving
around its own central axis. If the space station was revolving at the
appropriate rate, the centrifugal force would mimic the force of gravity on
the outside rim of the wheel.
Example 7: Let us put the radius of a circular space station at 50 meters.
What would be the speed needed in meters per second and
revolutions per minute to mimic the force of gravity here on the
surface of the earth? (hint: start with ac = g).
(22 m/s or 4.2
rpm)
Inside the revolving space station, an interesting situation arises in
terms of the forces. To an observer inside the revolving space station, the
situation is identical to the surface of the earth: she feels a force of gravity
downward, balanced by a normal force the floor of the space station is
exerting back at her. She feels like she is not moving at all, like standing on
earth, where the forces are balanced.
Newton- The Great Thinker (*optional)
There is a story about Newton’s discovery of the universal law of gravitation
while sitting in his garden one day. Back in Newton’s early days at Cambridge
university, the Black Plague (bubonic plague carried by rats) swept over Europe, wiping
out a large percentage of the population. In 1466, the university was forced to shut down
and Newton went home, back to his mother’s farm in Lincolnshire. It was during this
time that he formulated most of his important contributions to mathematics and physics
including the binomial theorem, differential calculus, vector addition, the laws of motion,
centripetal acceleration, optics, and universal gravitation. Newton was not even 22 at the
time.
Newton’s biographer writes in his memoirs that an apple falling from a tree
provided Newton his inspiration for the law of universal gravitation. As Newton thought
about the apple and its fall, he began to see the relationship between the similarity
between the moon and the apple’s motions. Might not the motion of the moon and the
apple be caused from the same thing?
“ I began to think of gravity extending to the orb of the moon, and … from
Kepler's rule I deduced that the forces which keep the Planets in their orbs must be
reciprocally as the squares of their distances from the centres about which they revolve:
and thereby compared to the force required to keep the moon in her orb with the force of
gravity at the surface of the earth, and found them to answer pretty nearly. All this was in
the two plague years of 1665 and 1666, for in those days I was in the prime of my age for
invention, and minded mathematics and philosophy more than at any time since…”
We can attempt to recreate Newton’s derivation of the force relationships between
the apple on earth and the moon as an example of the use of proportions rather than
equations to solve a physics problem. Knowing only the radius of the earth (RE = 6.37 x
108 m) and approximating the orbital radius of the moon as 60 times RE, Newton was
able to arrive at a very close approximation to the period of the moon, verifying the
inverse square relationship.
For an object in circular motion due to the force of gravity, we start with the
centripetal force Fc (force causing circular motion- see next unit) being due to the
gravitational force Fg.
Fg  Fc
v2
m
R
 2 R 
 T 
m 
R
4 m 2 R

T2
4 m 2 R

K R3

2
substituting (v 
2 R
)
T
substituting (T 2  K R 3 Kepler ' s Law)
4 m 2
K R2
 1 
 cons tan t  2 
R 
1
Fg  2
R
This inverse square relation gives an approximation to the force of gravity on the moon
relative to the force of gravity on an apple on the surface of the earth. If the distance R to
the moon is approximately 60 times the radius of the earth, then the force of gravity on
the moon is approximately 1/60 squared or 1/3600 as compared to the apple on earth.
Again, starting with the first principle that the gravitational force is the centripetal force:
Fg  Fc
v2
m g m
R
 2 R 


T 
g
R
2
4 R
 2
T
2
or ,
4 2 R
g
This equation allows Newton to estimate the period T of the moon’s orbit.
Substituting the new radius of the moon as 60 times the radius on earth, and the
gravitational field on the moon as 1/60 squared times the acceleration due to gravity on
earth, we can obtain a very close approximation of the period of the moon’s orbit.
4 2 R
Tmoon 2 
g
T2 

4  2 (60 RE )
60 g E
4  2 (60 6.37 106 m )

1
m
[9.81 2 ]
3600
s
 1 h  1 day 
Tmoon  2.353110 6 s 


 3600 s  24 h 
 27.2 days
The true value for the moon’s orbit is 27.3 days. It should be remembered that
Newton was performing these calculations in the days without computers or calculators.
Newton was a great thinker but rather eccentric, and often mean and vindictive.
He did not bother to publish many of his works or ideas, often until much later or when
he was goaded to by friends like Edmund Halley (of Halley’s comet). Yet when other
scientists or mathematicians claimed credit for a discovery, he was merciless in
denouncing them, as with his lifelong attacks of Robert Hooke (of Hooke’s Law and
microscope fame) and Gottfried Liebniz (co-discoverer, with Newton, of calculus). In
his later years he became quite famous and was appointed Warden of the Royal Mint, a
lucrative and important post. One of his duties there was to chase down counterfeiters
and Newton managed to convict many prominent ones who were then tried, executed by
hanging then drawn and quartered. He is reported to have died a virgin.
Kepler's Laws of Planetary Motion
Kepler's 1st Law states that planetary orbits are ellipses.
Kepler's 2nd Law states that equal areas subtended by a string from the
sun around a planet's orbital path will correspond to equal periods of
time. This means that planets move at different speeds along different parts
of the elliptical orbit.
Kepler's 3rd Law, which states the mathematical relationship between
the periods of orbit, T, and the radius of orbit, R, is given by:
Satellites in Orbit
Newton applied his ideas of forces to orbiting bodies. He said that an object
in orbit must have a centripetal force that is equal to the force causing it to
orbit (force of gravity).
Fc = Fg
The orbiting body must maintain a particular speed that would make the
forces equate. If the speed varies, then the orbiting body would not maintain
its same orbital radius. (This explains Kepler’s law of changing speed and
radius for a planet at different distances from the sun in it’s elliptical orbit.)
Geosynchronous /Geostationary Satellites
Lastly, satellites can be positioned so that they follow the
rotation of the earth (T = 24 hours). These geosynchronous or
geostationary satellites are fixed hovering over one point of the earth,
which is advantageous for instance in tv satellites. Because of its fixed
position above a point on the earth, your tv satellite dish need not move
while tuned to one satellite station. Like pearls on a string, a large number
of tv satellites are positioned above the equator to allow for the greatest
range covering the surface of the earth.
Example 16: Calculate the height above the earth necessary to plant a
satellite in geosynchronous orbit above Ecuador.
G forces
Pilots and astronauts have to withstand tremendous g-forces
sometimes in their line of work. G-forces refers to multiples of the
acceleration due to gravity, g. For example, 3 G's refers to 3 times the
acceleration due to gravity, and your weight would then be 3 times the
normal weight you would feel if you were standing on the surface of the
earth.
To test and train pilots ability to withstand G-forces, a training
machine using circular motion creates artificial g-forces. The pilot is
seated on the outer edge of the circle, facing inwards and the entire
apparatus starts revolving.
Using our formulae, we can quite readily calculate the speed needed for any
particular G-force. For instance, to calculate the speed needed to produce
1G of force (FN = 1xFg ):
Since Fc = FN ( =__xFg )
mv2 = 1x m g
R
v = ( R g )1/2
Exanple 9: Calculate the speed needed to revolve a 5.0 m radius G-force
trainer to produce 6.0 G's of force.
(17 m/s)
Simulated Weightlessness
Conversely, for space travel, astronauts must get used to weightlessness. To
train future astronauts for weightlessness, a special aircraft is flown in a
circular arc close enough to earth’s surface that g=9.81m/s2) which
simulates weightlessness and provides a brief training time in each arc. Fg
on the astronaut is balanced by the inertial force(centrifugal) that is created
by Fc. (see workbook page 264)
(this airplane: dubbed the “Vomit Comet” was used for the "space"
sequences in the movie, Apollo 13).
Example 10: What circular radius must a weightlessness(FN =0) simulator
aircraft with a speed of 1000 km/h have to fly? [Fc (centrifugal)
is balanced with Fg]
(7.87 km)
Vertical Circular Motion
On the surface of the earth, when an object is undergoing circular
motion along the vertical, because the force of gravity is acting along the
vertical, we have to add the force of gravity Fg to any tension T or FN, their
vector sum being the net or centripetal force Fc.
Example 5: A 1000 N man rides a roller coaster with a vertical loop of
radius 10 m. A radar gun clocks the roller coaster at a constant
speed of 35.66 km/h around the loop.
1) Sketch a free-body diagram of the roller coaster at the top of the
loop.
What is the apparent weight of the man if he stands on a Newton
scale?
At the top, since the apparent weight and gravity are in the same direction as
the net force Fc, then
Fc =
2) Sketch a free-body diagram of the roller coaster at the bottom of the loop.
What is the apparent weight of the man if he stands on a Newton
scale?
At the bottom, since the tension is the same direction as the (centripetal) net
force Fc, but the force of gravity is in the opposite (negative) direction, then
Fc =
Example 6: What is the tension at the top in a 1.0 m string with a 10 g
eraser circling vertically at 10.0 m/s?
What speed would T = 0?
What would happen to the object if you swing it at a speed below that
speed?
Banked Curves (*optional)
Friction between the road and car tires is what supplies the centripetal
force when cars round a curve. If the friction is not great enough, the cars
would continue in their straight line paths according to their current inertia.
This is a real problem road engineers must worry about, especially here in
Canada when the roads get icy. Besides getting cars to slow down before
the curve (reducing the centripetal force needed), road engineers can use
gravity to their favour and bank curves sloping down towards the center of
the giant circle that the curve represents.
From the above, the formula for banked curves is:
Example 11: The Indianapolis Motor Speedway is a 2.5 mile oval track,
home to the world famous Indianapolis 500, where cars can
reach speeds around 240 mph (400 km/h). Supposing turn No.
3 is banked 9.20 to the horizontal to handle the tremendous
speed. Assuming there is no friction and no slowing down in
the turn, what is the radius of the circle turn No. 3 inscribes at
400 km/h?
(7.8 km)
Unit 3B- Work and Energy
Energy is Work
Historically, the concept of energy arose from the industrial revolution and the
study of machines doing work for us (mechanical energy). Energy is defined strictly in
physics as the ability to do work.
Energy = Work
E
= W
The unit for energy and work is the same: the Joule (J).
The formula for work, like all previous formulae so far, derives from a graph:
here the force-displacement graph. Work is always equal to the area under the forcedisplacement (F vs d) graph. For a constant or average force, the area under the F vs d
graph is the area of a rectangle of length F and width d.
W = area of a rectangle
= length x width
= Fconstant x d
This gives the general formula for work with an average or constant force.
W = Fd
work with a constant or average force
It is important to realize that work – and energy – although derived from vectors
like force and displacement is a scalar quantity (no directions). The units for work or
energy is the Joule, equivalent to the force of 1 Newton across a distance of 1 metre.
1J = 1Nm
Usually, the force under consideration is the man-made or applied force, T (FT).
Work Needs a Distance
Example 1.
I hold a 100 g coffee cup in the air without moving. How much work am I
doing on the coffee cup?
Start with a free-body diagram. The work being done is from the applied
force T opposing the force of gravity Fg. By Newton’s 2nd Law:
But since displacement is zero (no motion)
W=Txd
= Fg x 0
=0
In general, if there is no displacement, the net work is zero.
Work Done Against a Force (of Gravity)
Example 2:
I lift a 100 g coffee cup in the air 1.00 m at constant velocity. How much
work am I doing on the coffee cup? Again, by Newton’s 2nd Law:
Note that the same forces for the coffee cup at rest apply for the constant
velocity case as well. The difference is that the constant velocity case has
a displacement which then yields the work done.
W
=Txd
= Fg x d
= (mg) h
= (0.100 kg) (9.81 m/s2) 1.00m
= 0.981 J
(almost 1 joule of work! Phew!)
Work Done to Accelerate
Example 3:
I move this 100 g coffee cup from rest a distance of 1.00 m across this
table with an average force of 1.00 N.
a) What is the work being done on the coffee cup?
Note that in this case there is no work being done to overcome a force like
the force of gravity, only work being done to accelerate the object.
W =
=
=
(1.00 J)
b) Assuming the table is frictionless, what would be the final speed of the
coffee cup after 1.00 m?
(4.47 m/s)
In general, there are two basic types of work:
1) Work done to overcome a force like the force of gravity Fg
2) Work done to accelerate an object (F).
Which work, if any or both, applies to any particular problem depends on what
the relevant force - usually the applied force FA – does.
Work Done by a Force at an Angle to the Displacement
Remembering that work is a product of force and displacement – both vectors the force and displacement must be in the same direction to count for work. Since the
component of force in the same direction as the displacement is given as F cos, the
general formula for work includes the angle  between the force and displacement.
W = F d cos
Example 4:
A 20.0 kg wagon is pulled with a force of 40.0 N by its handle at an angle
of 30.00 to the horizontal. Calculate the work done in pulling it 10.0 m..
(346 J)
For the case when force and displacement are in the same direction,  = 0o and cos = 1,
and we get back the original equation W = F d.
Example 5:
If the wagon above starts from rest, what is its speed after 10.0 m?
(Remember that the acceleration is due to the net force.)
(5.88 m/s)
Work Done Against Friction
The usual cases of motion are assumed ideal – to not involve friction. If friction is
considered, its direction opposite to motion means it is usually subtracted from the forces
in the direction of motion.
The work done by brakes on a vehicle is an example of a force in the opposite direction
of the displacement, eg the brakes of a car. This produces a decrease in energy or a
negative work (W= -F d).
Example 6:
A 1200 kg car is traveling down the road at 60.0 km/h.
a) If it takes 33.0 m to stop the car, calculate the work done by the brakes
in stopping the car.
(-1.67 x 105 J)
b) What is the average force of friction applied by the brakes during the
car’s stopping?
(-5.05 x 103 N)
Example 7:
A 20.0 kg wagon is pulled with a force of 40.0 N by its handle at an angle
of 30.00 to the horizontal against a force of friction of 20.0 N.
a) Determine the net force in the direction of motion.
(14.6 N)
a) Calculate the work done in pulling the cart 10.0 m.
(346 J)
b) If the wagon starts from rest, what is its speed after 10.0 m?
(Remember that the acceleration is due to the net force.)
(3.83 m/s)
Pg 294 do # 1-3
Work on an Incline
For an incline, the forces parallel to the surface determine the motion of the
object up or down the incline. This means the force of gravity resolved into
its parallel component Fg// is added or subtracted to the applied force FA
(and/or friction Ff as need be).
Example 8: A girl pushes a 20.0 kg box, initially at rest, up a frictionless
10.0 m long ramp inclined 30.00 to the horizontal. If the final
velocity of the box at the top of the ramp is 1.00 m/s, what was
the work done on the box by the girl?
Start by drawing the free-body diagram to determine the forces
acting on the box.
Determine, by Newton’s 2nd Law, the applied force FA. Here,
work is being done by the applied force FA to overcome the
force of gravity Fg and also accelerate the object with a net
force F.
Power
Power, P, is the rate of doing work, the work or energy per second or
time.
P = W = E= Fd = F vave
t
t t
The unit for power is the Watt (W), unfortunately the same symbol as
for work. To differentiate between Watt W and work W, two bars can be
placed above the Watt symbol W.
1 Watt = 1 Joule
1 second
1W= 1J
1s
Example 9: A 50.0 kg student runs up three flight of stairs at constant
velocity, a vertical distance of 10.0 m. The first time it takes
her 9.81 s. The second time it takes her 19.62 s. What was the
power she exerted for both times up the stairs? (Hint: first find
the work done)
Note that for both times up the stairs, she is doing the same
work (against the force of gravity). But because it takes her
half as long for the first time as the second time, she is exerting
twice the power the first time as for the second time up the
stairs.
Mechanical Energy
Energy has many forms: heat, light, sound, chemical, electrical,
nuclear. Any one form readily converts to any other. So, for instance,
chemical energy can be converted to electrical energy, nuclear energy to
heat energy, and so on.
Historically, the concept of energy arose from the Industrial
Revolution with the idea of machines able to do work for us. All forms of
energy can transform into the work of machines, or have the potential to be
transformed into the work that machines can do for us. Notably, one of the
first machines in the Industrial Revolution was the steam engine, used in the
coal mines of England to hoist water out of the mine. There, heat in the
form of steam (from boiling water) was used to power a giant pump,
replacing the work being done by horses drawing load after load of water out
from underground in the mine (from which came the term “horsepower” to
rate machines).
The first steam engines were very inefficient, burning up a lot of coal
(luckily there was lots nearby). It was James Watt who developed a vastly
improved steam engine, one much more efficient that generated a lot more
work for any given amount of heat. Watt’s version of the steam engine
essentially powered the beginning of the Industrial Revolution – used from
factories to trains.
For physicists then, the important forms of energy – historically –
were mechanical energy (energy of machines). Mechanical energy has two
parts:
1) Kinetic energy Ek or energy of motion,
Eg. Water moving or being moved
2) Potential energy Ep, the energy of position or state with the
potential to produce motion (or more correctly cause changes in
motion).
Eg. A boulder perched on top of a hill, a stretched spring, a
chemical bond.
In any isolated system where energy is not allowed to come in or out
the total mechanical energy ET is the sum of the individual kinetic and
potential energies.
ET= Ek + Ep
The law of conservation of energy states that energy is neither
created nor destroyed, or the total energy of any isolated system is a
constant value. In a perfect (frictionless) system, the total mechanical
energy is a constant value.
Example 10: Describe the mechanical energy transformations in a pendulum
brought up to a height of 0.100 metres above its lowest point,
and then released
a) in a frictionless system
b) in a system with friction.
Gravitational Energy Potential Ep
The potential energy of a system is related to the amount of work
required to put that system into its position or state (from a position where
the energy potential is zero).
Example 11: A 100 g coffee cup is lifted 1.00 m straight up at constant
velocity. What is the potential energy of the coffee cup?
Ep
=W
=Td
= Fg d
= (mg) h
= (0.100 kg) (9.81 m/s2) (1.00 m)
= 0.981 J
The potential energy from gravity is equal to the work done against
gravity. The general formula for potential energy due to gravity is
Ep = m g h
Note that the energy potential due to gravity is zero when the height is
zero. The general position of zero potential energy is the point where there
is no further potential to move, here the ground level zero. However,
potential energy can be made relative to any given position.The change in
potential energy (from gravity) exactly equals the work done from that
position against the force of gravity.
Example 12: A 250 g pendulum of length 0.500 m is initially at rest 1.00 m
from the floor. The pendulum bob is pulled sideways a distance
of 0.300 m. What is the potential energy of the pendulum bob
in its raised position?
Kinetic Energy Ek
Kinetic energy is energy of motion. Any object with mass m and
speed v has kinetic energy Ek given by
Ek = ½ mv2
Example 13: Which has more kinetic energy, a 100 g baseball thrown at 150
km/h, or a 10.0 g bullet speeding at 1500 km/h?
Like potential energy, the formula for kinetic energy is derived from
the work done, in this case the change in kinetic energy exactly equals the
work done to accelerate an object.
Example 14: Calculate the work done to accelerate a 5.0 g bullet to a speed
of 1100 km/h from a gun with a 10.0 cm barrel.
W = Fgun d
Elastic Potential energy and Hooke’s Law:
Elastic potential can be stored in any object which when deformed by a force
soon returns to its original shape undamaged. The work that was done to
change the object’s shape is returned by the object as work. There are many
common examples of elastic potential energy – diving boards, trampolines,
sling shot, springs in a car, etc.
A British scientist, Robert Hooke, stated that ‘The amount of deformation of
an elastic object is proportional to the force applied to deform it’. Up to a
certain point the amount of stretch or compression (x) in a spring varies
directly as the force (F).
Fnet = kx Since the spring responds in an opposite direction to the applied
force it is often written as Fr = -kx and called a ‘restoring force’. Fnet = -Fr
The slope of the graph is called the spring constant (k) in N/m. Beyond a
certain applied force, the graph will no longer be linear and damage can be
done to the spring by applying that amount of force. This is called the nonelastic region and at the end of this region is the breaking point for the
spring. Because force is a vector quantity we can give different signs for a
stretch than a compression. In either case ‘k’ is constant and +.
Note: As the x gets larger, the force gets larger. This is not a constant
force!! And therefore the acceleration of a spring is not constant either!
Eg. A spring which has a force constant of 15.0 N/m has a 1.50 x 102 g
mass attached to it. How far will the spring stretch?
When a spring is compressed/stretched, the work done on the spring to
compress/extend it is stored in the spring as POTENTIAL ENERGY. The
amount of potential energy is given by:
Ep = ½ kx2 (since the force changes, you can not use the work formula
unless average force is given – show formula!)
The Law of Conservation of Mechanical Energy
The sum of potential and kinetic energy is called the mechanical
energy.
Total Mechanical Energy = Energy Potential + Energy Kinetic
ET
=
EP
+
EK
Energy forms are readily converted into other energy forms. The first
law of thermodynamics states that energy is conserved in these energy
transformations. This means that the total amount of energy will always
remain the same before and after any energy conversion, or any form of
energy will convert completely to any other form of energy.
In ideal (frictionless) system this means that work can convert
completely (100%) to mechanical energy, or that one form of mechanical
energy like potential energy can convert completely to another form like
kinetic energy, and vice versa.
Example 19:
An archer draws back a bowstring with an average force
of 500 N to a distance of 20.0 cm from equilibrium, outline the
energy conversions. What is the maximum speed an 80.0 g
arrow could be shot from this drawn bowstring?
In an isolated frictionless system where energy cannot go in or out,
the total amount of energy is the total mechanical energy, which stays a
constant value. This means that in an isolated system, any system’s loss in
potential energy is equal to its gain in kinetic energy (or a system’s gain in
potential energy is equal to its loss in kinetic energy). If an isolated system
starts with only potential energy, it can convert any part or all of that
potential energy to kinetic energy. At the end it may have converted
anywhere from 0 to 100 per cent of that potential energy to kinetic energy.
Irrespective of the conversions, the total energy – here the total mechanical
energy – remains a constant.
ET(position 1) = ET(position 2) or 0(energy added)= Ek + Ep
Example 20: A coffee cup falls off a1.00 m high table. Assuming no friction,
complete the following table of the mechanical energy of the
coffee cup as it is
a) at the table top just before it starts falling
b) at the middle of its fall
c) at the bottom just before it hits the ground.
Note that the change in energy potential is exactly equal to the opposite
change in kinetic energy.
The law of conservation of energy allows the calculation of a speed at
any point along the energy conversion, knowing only the initial energy of a
system. By equating the total mechanical energy from any one point to any
other point, once the height h is known the speed v can be determined. Note
that mass m is not required as it cancels out from both sides of the equation.
E Top =
E Middle
= E Bottom
mgh = (mgh + ½ mv2)Middle = (½ mv2)Bottom
Example 21. Calculate the maximum speed of the coffee cup just before it
hits the floor.
E Top = E Bottom
Example 22. Calculate the speed of the coffee cup at the midpoint of its
drop to the floor.
Example 23: A weightlifter lifts a 250 kg barbell to a height of 2.00 m from
the floor.
When he releases it, what is the maximum speed before it hits
the ground?
Example 24: A 100 g coffee cup is lifted 1.00 m straight up from the floor at
constant velocity, then dropped. Assuming no friction,
a) What is the work done on the coffee cup?
b) What is the potential energy of the coffee cup at the top?
c) What is the kinetic energy at the bottom just before it hits the
ground?
d) What is the maximum final speed just before it hits the
ground?
e) What is the speed of the coffee cup at the middle of its fall?
Roller Coaster Problem:
Example 20 Initially the roller coaster is at rest at A (see diagram below).
It starts
rolling down the track. Assuming a frictionless surface, what is
the speed of the roller coaster at B, C and D:
Pendulum:
Falling object (no resistance):
What would the graph look like if there is friction?
t
The Work-Energy Theorem
The equations for kinetic and potential energy derive from Newtonian
mechanics and the work done in either accelerating (kinetic) or overcoming
the force of gravity (potential). By first considering the (changes in)
potential or kinetic energies, most motion problems can often be solved
much simpler than by Newtonian mechanics and its consideration of forces
across distances.
To analyze a motion problem from an energy standpoint, first
consider: are there any changes in kinetic energy Ek from changes in speed,
or potential energy Ep from changes in height or position? If so, then the
total work WT (Fd) involved will be a sum of these two types of energies.
WT= Ek + Ep (ideal systems-all energy is accounted for)
Let us work through a few examples from above to illustrate this point.
Example 15. What is the work done in holding a 100 g coffee cup
stationary on a table?
(Here there is no change in speed or position)
(0)
Example 16.
What is the work done in lifting the 100 g coffee cup
straight up from the table 1.00 m at constant velocity?
(Here there is a change in position)
Example 17. What is the work done in moving a 100 g coffee cup straight
across a frictionless table 1.00 m from rest to a final speed of
1.00 m/s?
(Here there is a change in speed)
Example 18: Calculate the work done to lift a 100 g coffee cup 1.00 m
straight up from the table while it accelerates to1.00 m/s from
rest. (Here there is both)
Work can be added to the system by an applied force, or taken away from
the system by friction. Work theorem is useful in both cases. Signs are used
to show an increase or decrease in the work. The change in energies (Ek,
Ep) need to be final – initial to reflect the increase or decrease with the
appropriate sign.
Eg. Pg 308 #1,2
Lab 7- Conservation of Energy – Friction Down an Incline
Objective:
To use the law of conservation of energy to determine what percentage of
energy is lost to friction by a cart rolling down an incline.
Background:
The law of conservation of mechanical energy predicts that for ideal systems,
potential energy can be completely (100%) converted to kinetic energy. For a cart rolling
down a ramp, this means the potential energy of the cart at rest at the top of a frictionless
incline will be exactly equal to the final kinetic energy of the cart at the bottom of the
incline.
Ep, top = Ek, bottom
Real systems on earth include friction. As a result, some energy will be lost to
friction. By the law of conservation of energy, the total energy of the system should still
remain constant. This means that the potential energy at the top is not converted 100% to
kinetic energy, but that some percentage of energy will be lost to friction.
Eg.
Ep, top = Ek, bottom + Wfriction
100% = 80% + 20%
(20% of the energy is lost to friction here)
Rearranging, we can determine the energy lost to friction by taking the
potential energy and subtracting from it the kinetic energy at the bottom. The potential
energy is calculated from the mass m and height g. The kinetic energy is calculated from
the mass m and speed v at the bottom. The difference is the energy lost to friction Wf or
Efriction:
Wf = Ep top - Ek bottom
WT= Ek + Ep (Theorem)
Wfriction = Ep, top - Ek, bottom
= mgh - ½ mv2
This energy lost to friction is usually in the form of heat or sound. Since the total
energy of the system to begin with is the energy potential at the top of the ramp
(Ep,top=100%), to calculate the percentage of energy lost to friction, we divide the energy
lost to friction Efriction (Wf) by this total energy and multiply by 100.
% Energy lost = Efriction x 100
Ep, top
Materials:
Ticker tape timer and tape (2.0 m)
80-100 cm ramp or incline
Cart (paper clip or tape to attach ticker tape to end)
metre stick
Lab Write-up:
Design an experiment using the above materials to determine the
percent of energy lost to friction on an incline. Perform the
experiment collecting your own values (work in groups of two) and
determining the percent of energy lost to friction. Your write-up
should include sufficient detail and information for the experiment
to be reproducable by one of your classmates.