Download Example: I take an object of mass m = 1 kg and raise it to the height

Example: I take an object of mass m = 1 kg and raise
it to the height of 2 m above the ground. As I do it,
the velocity of the object is approximately constant.
What work did I do?
The net force was zero… There were two forces,
however: F upwards, and mg downwards.
1. The work done by the net force is zero, ∆KE=0
2. The work done by mg is WG= (-mg) d = - mgd = -19.6 J
3. The work done by my hand is (F=mg) WF = F d = mgd
=19.6 J
I worked against gravity and now the object has
something it did not have before…
y
What is it?
I now let it go… I do not do any work now.
y
It falls down. The only force is mg, the distance is –d,
The work done by gravity is WG = (-mg)(-d) = mgd = 19.6 J
Where did it go?
The object was accelerating (before reaching the floor),
Let’s calculate its final velocity:
v = gt
But we do not know t...
t=
2d
g
1 2
gt Since it passed d over t...
2
2d
⇒ v=g
= 2 gd
g
d=
Now let’s find KE = ½mv2 = mgd = WG !!!
The work done by the force gravity went to KE
One more time:
We did the positive work against gravity,
then the gravity did the same positive work, that
eventually went to the kinetic energy of the object.
In the beginning, the gravity was not “capable” of doing
that positive work!
The result of our positive work against the gravity was
changing a position of the object, so it became capable…
We call this capability a POTENTIAL ENERGY.
Still more…
We do the work against the gravity, the object’s
potential energy increases by the amount of this work.
Then the potential energy is converted into kinetic as a
result of work done by gravity.
1. Wext = ∆PE, 2. ∆PE + ∆KE = 0
In general,
Wnet, external except gravity = ∆KE + ∆PE
Conservation of mechanical energy:
In a closed system without losses (no friction, resistance,
etc.)
KE + PE = constant,
(KE + PE)
before
= (KE + PE) after
Simplified previous example:
You drop an object which then falls freely from height
d. What is its speed just before it hits the ground?
∆PE = PEafter − PEbefore = 0 − mgd = − mgd
∆KE = KEafter − KEbefore
1 2
1 2
= mv − 0 = mv
2
2
1 2
∆KE = −∆PE ⇒
mv = −(−mgd ) = mgd
2
v 2 = 2 gd
⇒ v = 2 gd
Notice, it’s the same result we got using Newton’s 2nd
law and kinematics!
Another example:
A sled slides from a hill of a height d. What is its
speed at the bottom of the hill?
∆PE = PEafter − PEbefore = 0 − mgd = − mgd
∆KE = KEafter − KEbefore
1 2
1 2
= mv − 0 = mv
2
2
1 2
∆KE = −∆PE ⇒
mv = −(−mgd ) = mgd
2
v 2 = 2 gd
⇒ v = 2 gd
Same result !
Back to collisions:
Elastic collisions are such collisions in which the KE is
conserved. If it is not conserved, the collision is called
inelastic.
An example of elastic collision: A point particle of
mass 1 kg and velocity of 2 m/s, hits a similar point
particle initially at rest. If the collision is elastic,
find the velocities of both particles after collision.
mv = mv1 + mv2
v = v1 + v 2
1 2 1
1
2
2
mv = mv1 + mv2
2
2
2
v 2 = v1 + v 2 + 2 v1v 2
They EXCHANGE velocities…
2
2
v 2 = v1 + v 2
2
v1 = 0
v1 = 0 ,
or
2
v2 = 0
v2 = v
If any similar objects collide elastically head on, they
exchange their velocities.
Another example: two carts of equal mass have equal
opposite velocities, they elastically collide head on. Find
their velocities after collision.
Initial momentum = (mv)1 + (mv) 2 = 0
After collision : 0 = mv1 + mv2 ⇒ v1 = −v2
⎛1
⎞
KE before = 2⎜ mv 2 ⎟ = mv 2
⎝2
⎠
1
1
1
1
2
2
2
2
KE after = mv1 + mv2 = mv1 + mv1
2
2
2
2
2
= mv1
Therefore : v1 = −v,
v2 = v
Last example: A particle of mass 2 kg and velocity of 1
m/s hits another particle of 2 kg initially at rest, head on.
If the first particle’s velocity after collision is 0.2 m/s,
Find the velocity of the second particle.
m1v0 = m1v1 + m2 v2
2 ⋅1 = 2 ⋅ 0.2 + 2 ⋅ v2
v2 = 1 − 0.2 = 0.8 m/s
What happened to the kinetic energy?
1
KE before = 2 ⋅12 = 1 J
2
1
1
2
KE after = 2 ⋅ 0.2 + 2 ⋅ 0.82 = 0.04 + 0.64 = 0.68 J
2
2
∆KE = 0.68 − 1 = −0.32 J
Where did the energy go?
Other forms of energy:
• Internal energy – we will discuss it very soon
• Chemical energy (energy stored in a compound, for
example wood)
• Electrical energy
• Elastic potential energy
Nuclear energy
• Nukilar
A spring gun:
Potential elastic energy of the spring
Kinetic energy of the projectile
Potential energy of the projectile
Kinetic energy of the projectile
Sound waves and “heat” (internal energy)
A climber on the rope…
The work done by his muscles
Potential energy in the g field
Kinetic energy and heat
Sound, heat, arms damage
CONSERVATION OF ENERGY:
“Energy is never created or destroyed, it is just
converted from one form into another.”
“ The total energy in a isolated system is constant.”
The only problem with this is that sometimes it is
difficult to follow the energy conversions similar to
follow the forces. If a car hits a tree, for example, it
is not easy to tell how much energy went to damaging a
car, and how much into damaging a tree, and so on…
QUESTION:
Think of these examples. What type of energy is
converted into what different type of energy ?
• Car Engine
TYPES Of ENERGY:
• Elevator
1. Chemical energy
• Battery
2. Gravitational potential
energy
• Generator
• Hydroelectric power
plant
3. Electrical energy
• You on your Bicycle
5. Elastic potential energy
4. Kinetic energy
6. Nuclear Energy
POWER
When you run up a flight of stairs or walk it up
slowly, the work you do is the same.
Why are you more exhausted when you run than
when you walk ?
It depends on the rate at which you do work.
This rate is called power:
P = W/∆t
Why Power?
ƒ Power = rate of work done by a force
ƒ Power – average or instantaneous – depending of ∆t.
d
W Fd
P=
=
=F
= Fv
∆t ∆t
∆t
ƒ Instantaneous power is equal to the product of the
force by the instantaneous velocity in the direction of
the force.
ƒ The work done by the force is related to the energy –
alternatively, if the object “does work” and loses
(expends energy), the dissipated power is equal to the
rate of loss of energy
W ∆E
ƒ As a result of doing work,
=
=
P
the object loses energy
∆t
∆t
What do you need?
ƒ To achieve something one needs enough force, power, or
energy?
ƒ If you want to break something, raise something, move
something – you need to exert a needed force. You
cannot “have” it, you can either able or not to exert it.
ƒ If you are running, you are exerting force all the time –
to move and support yourself (~mg), but you are moving
with a velocity v, and dissipate power ~mgv. If you want
to increase the speed of your motion, you need to
increase this dissipation of power, but you do not “have”
it, all you have is energy!
ƒ You need to “HAVE ENERGY”, not force, or power. You
cannot store neither force nor power.
UNITS:
Power: 1 J/s = 1 Watt = 1 W
Example:
Light bulbs: 60 W means the light bulb uses 60
Joules of energy per second.
Is that a lot ?
When you pay your electric bill, what unit is the
‘amount of electricity’ you used measured ?
kWh = kiloWatt-hours = 1000 Watt x hour
What kind of unit is that ? What does it measure ?
Energy !
1 kWh = 1000 W x h = 1000 J/s x 1 hour
= 1000 J/s x 3600 s = 3.6 x 106 J
QUESTION:
How long would you have to leave a 60W light bulb on to
use 1 kWh of energy ?
P = E/t ⇒ t = E/P = 1000 W×h/ 60 W = 16.7 hours
EXAMPLE: Horse Power
What does 100 hp mean?
It’s the maximum power an engine can provide.
1 hp = 0.746 kW ⇒ 100 hp = 74.6 kW
How much average power is needed to accelerate a
3000 kg SUV from 0 to 60 mph in 10 seconds ?
60 mph = 26.8 m/s
Work = gain in kinetic energy = ½ m v2 = 1.08 x 106 J
Power = Work/Time = 1.08 x 106 J/10 s = 108,000 W
= 108 kW = 144 hp
When an ice skater spins, and she brings her arms closer
to her body. The rate of her rotation increases.
WHY ?
1. She is reducing air resistance.
2. She is adding energy by forcefully
bringing her arms in.
What is the determining factor? Is there another physical
quantity?
ƒ Yes there is, and you are right if you have guessed that it
is related to momentum – it is an angular momentum
ƒ It is not, however, as intuitively simple as the linear
momentum…
Angular momentum – momentum about an axis
ƒ L = r × mv
where v is only a component of the
velocity directed tangentially
ƒ Just imagine that an object is fixed to an axis and
there is a “supply” of the centripetal force if needed
due to, say, an elastic force, and the only motion that is
allowed to the object is rotation around this axis. Then
L=rmv is exactly equal to the angular momentum of this
object with respect to this axis in a frame of
stationary axis. It won’t change if this frame (with the
axis) is moving uniformly, because the other component
of the velocity won’t be “rotational around this axis.”
ƒ The angular momentum is conserved in absence of
external forces that change rotation about the axis…
tangential forces.
Central forces – angular momentum conservation
ƒ If in a system of two particles, the forces are directed
along the straight line connecting these particles, the
angular momentum with respect to any axis is conserved.
Such forces are called central.
ƒ Example: Because the force of gravity is central, the
angular momenta of the planets and the Sun are
conserved.
ƒ Consequences: Kepler’s laws of planetary motion
ƒ First Kepler’s law: trajectories of all planets (comets,
etc.) are ellipses with the Sun in one of its foci. Each of
such ellipses is a closed curve in one plane.
ƒ Second Kepler’s law: While moving along its trajectory
any planet (comet, etc.) “sweeps” the same area in a
given time interval. [The further the object from the
Sun along its trajectory, the smaller its speed is.]
Note, the Third Kepler’s law is equivalent to the Newton’s law
of gravity
Kepler’s second law example:
Satellite around earth. Moves on elliptical orbit.
BO is smaller
the speed is
larger
O
AO is larger
the speed is
smaller
Gravitational force is central
Solar
system
up to
Jupiter
Another example:
I’m spinning a tennis ball on a rope. Without disturbing
the system in any other way, I shorten the length of the
rope. What happens ?
It will speed up because the angular momentum is conserved,
since the Force is central !!!
Li = m vi ri = m vf rf = Lf
So, if rf < ri, then vf > vi to keep L constant.
What about energy conservation?
Since it is now going faster isn’t the kinetic energy
increased?
The kinetic energy is indeed increased, because we did work
on the system.
Work = F d ≈ (m v2/r) d
This is energy we added to the system, increasing its kinetic
energy.
How to change angular momentum?
ƒ To change the angular momentum, an external tangential
force is needed. (Imagine that you want to spin a bicycle’s
wheel with you hand.)
ƒ It turns out to be convenient to introduce another
physical quantity to formalize such procedure, the torque
ƒ Torque – product of the force by the distance between
its line of action and an axis
τ = r×F
ƒ Then it turns out that the rate of change of angular
momentum is equal to the net external torque. In absence
of the net external torque, the angular momentum is
conserved.
Looks and sounds familiar,
does not it?
∆(r × mv)
=τ = r× F
∆t
Statics in the inertial frame
ƒ Under what conditions the system will remain static or
in a state of the uniform motion?
ƒ Particle? – First Newton’s law: Zero net external force
ƒ A system of particles:
Static
ƒ Zero net external force
ƒ Zero net external torque
Non-static
Lever
ƒ Levers are used since the times of ancient Egypt to amplify
applied forces
ƒ Imagine yourself using this lever: you push down and the load
goes up. You can stop your motion at any position. Then the
lever bar is in equilibrium.
ƒ There are three forces acting on the lever bar – the load
pushes down with its weight, you pushing down with your
force, and the support reaction compensates them, so that
their sum is zero.
ƒ Two forces supply torques to the bar with respect to the
support: the load’s weight (counter-clockwise), and your force
(clockwise). They also have to compensate each other:
Fload rload = Fw rw
rload
where rload and rw are lever arms of
the load and a worker respectively.
rworker
Golden rule of mechanics
The expression that we
derived from statics leads
geometrically to the similar
expression:
Fload dload = Fworker dworker
known as the golden rule of
mechanics
ƒ It expresses the rule of the lever – that you can raise a
heavy load with a smaller force
ƒ At the same time, the displacement through which this
force acts is larger than the displacement of the load by
the same factor
ƒ The work done by the worker is the same as the work done
by the reaction of the lever raising the load!
We finished the first part of the course,
mechanics
Kinematics – description of motion
Dynamics – reasons of motion and its changes
Newton’s laws applied to point-like objects
The main quantity is momentum, mv
First applications, gravity, examples of forces
Work done by force
Energy, mechanical energy, kinetic energy, power
Angular momentum, torque
“First principles of mechanics”