Transcript VT 2300

Final Physics Review
A COMPREHENSIVE GUIDE TO SURVIVING
THE END OF THE YEAR
Topics
 Linear Motion
 Newton’s Laws
 Projectile Motion
 Momentum and Impulse
 Friction
 Torque and Static Equilibrium
 Circular Motion and Centripetal Force
 Energy: Work, Kinetic, Potential, Conservation
 Electrostatics
Linear Motion
Linear Motion
 Position

Defined in terms of a frame of reference
Linear Motion
 Displacement
 Defined as the change in position
 xf
Δx = xf – xi
is final position, xi is initial position
Linear Motion
 Note: the displacement of an object
may not the distance the object travels

If you throw a ball in the air and it comes back to your hand,
its displacement is 0, but its total distance traveled is not
Linear Motion
 Average Speed
 Defined as total distance an object travels over time
Average speed =
𝒕𝒐𝒕𝒂𝒍 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆
𝒕𝒐𝒕𝒂𝒍 𝒕𝒊𝒎𝒆
Units = m/s
Linear motion
 Velocity
 Defined as displacement of an object over time
V=
Δ𝑥
Δ𝑡
xf – xi
=
tf – ti
Units = m/s
Note: Direction is the same as the direction of displacement
Linear Motion
 Acceleration
 Defined as change of velocity over time
a=
Δ𝑉
Δ𝑡
Units = m/s2
Linear Motion: Important Equations
 Constant Velocity
d= v t + di
Linear Motion: Important Equations
 Accelerated Motion
d = ½ a t2 + vi t + di
Linear Motion: Important Equations
 Other useful equations:
 No time? No problem!
V2 = vi2 + 2 a (∆ d)
Linear Motion: Important Equations
 Other useful equations:
 No time? No problem!
V2 = vi2 + 2 a (∆ d)
Newton’s
Laws of Motion
I. LAW OF INERTIA
II. F=MA
III. ACTION -REACTION
Newton’s Laws of Motion
 1st Law – Unless acted upon by an unbalanced
force, an object at rest will stay at rest, and an
object in motion will stay in motion at constant
velocity.
 2nd Law – Force equals mass times acceleration.
 3rd Law – In any direct interaction between
objects, for every action there is an equal and
opposite reaction.
1st Law
 Inertia is the
tendency of an
object to resist
changes in its
velocity:
whether in
motion or
motionless.
These pumpkins will not move unless acted on
by an unbalanced force.
2nd Law F net =ma.
Also stated as
a = F net / m
Acceleration is the result!
Net force on a mass is the cause!
2nd Law (F = m x a)
 How much force is needed to accelerate a 1400





kilogram car 2 meters per second/per second?
Write the formula
F=mxa
Fill in given numbers and units
F = 1400 kg x 2 meters per second/second
Solve for the unknown
 2800 kg-meters/second/second or
2800 N
Newton’s 2nd Law proves that different masses
accelerate to the earth at the same rate, but with
different forces.
 We know that objects
with different masses
accelerate to the
ground at the same
rate.
 However, because of
the 2nd Law we know
that they don’t hit the
ground with the same
force.
F = ma
F = ma
98 N = 10 kg x 9.8
m/s/s
9.8 N = 1 kg x 9.8
m/s/s
3rd Law
For every action, there is an
equal and opposite reaction.
 Consider the flying motion of birds. A bird flies by
use of its wings. The wings of a bird push air
downwards. In turn, the air reacts by pushing the
bird upwards.
 The size of the force on the air equals the size of
the force on the bird; the direction of the force on
the air (downwards) is opposite the direction of the
force on the bird (upwards).
 Action-reaction force pairs make it possible for
birds to fly.
Projectile Motion
Treat motion in x, y directions
independently
 Horizontal direction
 assume constant velocity, vx = constant
 x = vx t
Vertical direction
- freefall, accel due to gravity
- y = ½ a t2 + vi t + yi
- v = at + vi
Summary of concepts so far

Without air drag, projectile motion can be analyzed as a
combination of
Constant velocity in the x or horiz direction
Using specific c.v. equation for position

Accel motion, i.e. freefall in the y, vert dir
Using specific equations for position and velocity


Although motions are independent, time is common
Simplified for horiz launched projectiles…
 X direction
x = vx t
Y direction
y = 5t2
v = 10 t
Our equations are strictly x direction, strictly
y direction
 How can we figure out the x and y initial velocities of
a potato launched with some initial velocity at an
angle?
 Velocity in x dir
vx = ?
Initial vel in y dir
vy i = ?
Heeeey, SOHCAHTOA!
 What’s your vector, Victor?
 PhET computer activity
Assume that
Vel. of potato as
it leaves the cannon
V = 25 m/s
Then
vx = V [cos (θ angle)]
vy i = V [sin (θ angle)]
SOHCAHTOA
SOHCAHTOA
Friction and Equilibrium
Friction Forces
When two surfaces are in contact, friction forces oppose relative motion or
impending motion.
P
Friction forces are parallel to the surfaces in
contact and oppose motion or impending
motion.
Static Friction: No relative
motion.
Kinetic Friction: Relative motion.
Friction and the Normal Force
8N
4N
n 2N
n
12 N
4N
n
6N
The force required to overcome static or kinetic
friction is proportional to the normal force, n.
fs = msn
fk = mkn
Summary: Important Points to Consider
When Solving Friction Problems.
• The maximum force of static friction is
the force required to just start motion.
fs
n
P
f s  ms
W
Equilibrium exists at that instant:
Fx  0;
Fy  0
n
Summary: Important Points (Cont.)
• The force of kinetic friction is that force
required to maintain constant motion.
n
fk
P
W
f k  mk
n
• Equilibrium exists if speed is constant,
but fk does not get larger as the
speed is increased.
Fx  0;
Fy  0
Summary
Static Friction: No
relative motion.
fs ≤ msn
Kinetic Friction:
Relative motion.
fk = mkn
Procedure for solution of equilibrium
problems is the same for each case:
Fx  0 Fy  0
Momentum….
just another way to talk about
motion
and changes in motion
WHAT NEWTON WAS REALLY TALKING
ABOUT IN HIS FAMOUS PAPER…
Momentum
 The linear momentum of an object of mass m
moving with a velocity is defined as the product of
p
the mass and the velocity



v
SI Units
p aremkgvm / s
Vector quantity, the direction of the momentum is the same as
the velocity’s
Impulse causes change in momentum…
 In order to change the momentum of an object, a
force must be applied
 The rate of change of momentum of an object is
equal to the net force acting on it


p m(v f  v i )

 Fnet
t
t
Gives an alternative statement of Newton’s second law
Impulse cont.
 When a single, constant force acts on the object,
there is an impulse delivered to the object



I
is defined as the impulse
Vector
I quantity,
 F t the direction is the same as the direction of
the force
Impulse-Momentum Theorem
 The theorem states that the impulse acting on the
object is equal to the change in momentum of the
object


If the
force
F
t isnot
p constant,
mv use
mthe
v average force applied
f
i
CONSERVATION OF MOMENTUM!
 One of the great principles of physics!
 This explains the transfer of momentum in any
collision, explosion, interaction of objects moving
through space and time!
Σ p (before, initial) = Σ p (after, final)
 Add up the momenta of all objects before the
interaction
 Add up the momenta of all objects after the
interaction
 Before = after
Collisions!
 Momentum is conserved in any collision
 Inelastic collisions
 Kinetic energy is not conserved


Some of the kinetic energy is converted into other types of energy
such as heat, sound, work to permanently deform an object
Perfectly inelastic collisions occur when the objects stick
together

Not all of the KE is necessarily lost
More Types of Collisions
 Elastic collision
 both momentum and kinetic energy are conserved
 Actual collisions
 Most collisions fall between elastic and perfectly inelastic
collisions
More About Perfectly Inelastic Collisions
 When two objects
stick together after the
collision, they have
undergone a perfectly
inelastic collision
 Conservation of
momentum becomes
m 1v 1i  m 2 v 2i  (m 1  m 2 )v f
Sketches for Collision Problems
 Draw “before” and
“after” sketches
 Label each object
include the direction of
velocity
 keep track of subscripts

Sketches for Perfectly Inelastic Collisions
 The objects stick
together
 Include all the velocity
directions
 The “after” collision
combines the masses
Rotational mechanics
HOW CAN YOU CHANGE THE ROTATIONAL MOTION OF
AN OBJECT?
A change in rotational motion!!
 Torque causes a change in rotational motion
 All of Newton’s laws apply! Yay Newton!!
 If force is applied perpendicular to the lever arm,
then
 Torque
= force F x lever arm d
Since all of Newton’s laws apply…
 If the sum of the torques = 0,
 Then the object’s rotational motion will not change
 Wasn’t rotating….still not rotating
 (static equilibrium)
 Was rotating…continues at the same rate
Since all of Newton’s laws apply…
 If the sum of the torques is NOT = 0,
 Then the object’s rotational motion WILL change,
it will acclerate in an angular or rotational sense
 Wasn’t rotating….starts to rotate
 Was rotating…rate of rotation is changed
Activity from class.....Newton’s 1st law….
 Weighing an elephant (balancing a meterstick)
 Observations of balanced system
Heavier wts. closer to pivot pt
 Lighter wts. farther away from pivot pt

 Mathematical relationship
 Torque on left side = torque on right side
Fxd = Fxd
 Balanced torques…..sum of the torques = 0

Thoughts from class…
 Multiple weights? 2 on left, 1 on right??
 Add up torques
 (F1 x d1) + (F2 x d2) = ( F3 x d3)
Pivot NOT
at the CG?
 Can you balance the stick if the pivot is NOT at the
CG?
 The weight acting through the CG causes a torque
on the right side

Torque = wt x (dist between CG and pivot)
 Must place a weight on left side to create an equal
and opposite torque!
Circular Motion
Centripetal Acceleration
Centripetal Force
Energy
Gravitational Potential Energy
 Gravitational Potential Energy is the energy
associated with the relative position near the Earth’s
surface
 Objects interact with the
earth through the
gravitational force
 Actually the potential energy
is for the earth-object system
Kinetic Energy
 Energy associated with the motion of an object
 Scalar quantity with the same units as work
 Work is related to kinetic energy
1 2
KE  mv
2
Conservation of Energy, cont.
 Total mechanical energy TME is the sum of the
kinetic and potential energies in the system
Ei  E f

KE i  PE i  KE f  PE f
Another way to describe it –


PE + KE = TME = constant
Other types of potential energy functions can be added to
modify this equation
Work
 Provides a link between force and energy
 Impulse: change in momentum :: work: change in
energy


Force applied for a particular time causes a change in
momentum
Force applied for a particular distance causes a change in
energy
Work, cont.
W  (F c o s q ) x
 We are studying only
cases in which F and x
are parallel, q =zero
F is the magnitude of
the force
 Δ x is the magnitude of
the object’s
displacement
 q is the angle between

F and x
Work-Kinetic Energy Theorem
 When work is done by a net force on an object and
the only change in the object is its speed, the work
done is equal to the change in the object’s kinetic
energy

W net  K E f  K E i   K E
Speed will increase if work is positive
 Speed will decrease if work is negative

Work and Gravitational Potential Energy
 PE = mgh or mgy

W g rav it y  P E i  P E f
 Units of Potential
Energy are the same
as those of Work and
Kinetic Energy
Power
 Often also interested in the rate at which the
energy transfer takes place
 Power is defined as this rate of energy transfer

W

 Fv
t
 SI units are Watts (W)

J kg m 2
W  
s
s2
Practice Problem
 When Albert hits a 46-g golf ball with a club, the
ball picks up 43 J of kinetic energy. A constant
force of 2300 N is applied to the ball while the ball
and the club are in contact. Over what distance is
the club in contact with the ball?
W = F x d, so d = W/F =
43J/2300N = 0.019 m
electrostatics
MOVEMENT OF CHARGES
Rubber rod/rabbit fur
At start: both neutral
Process of charging: requires
friction
_ _
_ _
_
+ + +
+
+
Fur loses electrons : +
charged
Rod gains electrons: - charged
Objects End up oppositely
charged
Negative - rod/ neutral
electroscope
• No contact!
• Neither object loses or
gains electrons
• Electrons are free to move
_ _
_ _
_
++
+ +
_+ _+
_ +_+
_+ _+
_ +_+
_ _+ _+
___
+_+_
Want to get away from the
negative charge on the neg
rod, like charge
Move down the electroscope
toward the foils
Knob + charge
Foils – charge: repel and
angle outward
Positive +rod/ neutral
electroscope
• No contact!
• Neither object loses or gains
electrons
• Electrons are free to move
++
++
__+
_+_
_ _
+_+_
_+ _+
_ +_+
_+ _+
_ +_+
+ +
+ +
Electrons attracted to positive
rod, opposite charge
Move up the electroscope
toward the knob
Knob -- charge
Foils + charge: repel and
angle outward
Remove rod, foils return to
Now the electroscope is negatively charged…..
 What happens if we bring another charged object close to
it?

A. foils remain the same, same outward angle?
B. foils repel even more, bigger angle?
C. foils repel less, smaller angle?

Approach with another negatively charged object?




A B or C? Draw and explain.
Approach with a positively charged object?

A B or C? Draw and explain
Electric Current
Electric Current
 Whenever electric charges of like signs move,
an electric current is said to exist
 The current is the rate at which the charge
flows through this surface

Look at the charges flowing perpendicularly to a
surface of area A
Q
I 
t
 The SI unit of current is Ampere (A)

1 A = 1 coulomb/sec
Resistance
 In a conductor, the voltage applied across the ends of
the conductor is proportional to the current through
the conductor
 The constant of proportionality is the resistance of
the conductor
V
R
I
Resistance, cont
 Units of resistance are ohms (Ω)
 1Ω=1V/A
 Resistance in a circuit arises due to collisions
between the electrons carrying the current with the
fixed atoms inside the conductor
 Mostly depends on material, thickness and length of
wire
Ohm’s Law
 Experiments show that for many materials,
including most metals, the resistance remains
constant over a wide range of applied voltages or
currents
 This statement has become known as Ohm’s Law

ΔV = I R
 Ohm’s Law is an empirical relationship that is valid
only for certain materials

Materials that obey Ohm’s Law are said to be ohmic
Energy Transfer in the Circuit
 Consider the circuit
shown
 Imagine a quantity of
positive charge, Q,
moving around the
circuit from point A
back to point A
Energy Transfer in the Circuit, cont
 Point A is the reference point
 It is grounded and its potential is taken to be zero
 As the charge moves through the battery from A to B,
the potential energy of the system increases by QV

The chemical energy of the battery decreases by the same
amount
Electrical Energy and Power, cont
 The rate at which the energy is used is the power
Q

V  I V
t
 Electrical appliances are often described in terms
of power required or watts
 Power determines the brightness of a lamp

More voltage, more current or both
Chapter 35
 Electric Circuits
What makes a good circuit?
 Must use all conductors
 Must include voltage source
 Complete path
 From (+) to (-) end of voltage source
 No short circuits
 No breaks, gaps
 All electrical devices working
Observations about
Series circuits
 Same current throughout the single loop
 Voltage at battery = sum of voltage drops across the bulbs =
V1 +V2,etc

Add more bulbs, bulbs get dimmer, less power
 One bulb is unscrewed, all bulbs go out – disadvantage
 total resistance increases as more bulbs are added to the
circuit
 R total = R1 + R2 + etc.
 Advantage – uses less power, batteries last longer
Observations about
parallel circuits
 Same voltage throughout the circuit
 Voltage at battery = v1 = v2 = etc.
 Current at the battery=sum of currents in all branches
 I total = I1 + I2 + etc.
 One bulb unscrewed, rest stay on – advantage
 Add more bulbs, total resistance decreases
 Add more bulbs, total current increases
 Disadvantage – can overload the circuit causing overheating of wires

Can protect the circuit by including circuit breaker or fuse in the main branch
Resistors/resistances in circuits
 In series
 Total or equivalent R = R1 +R2 + etc
 total increases with more resistors
 In parallel
 Total or equivalent R


1 / R total = 1/R1 + 1/R2 + etc.
Total decreases with more resistors
Sample problem
 Three resistors are connected in series. If placed in a circuit with a 12-
volt power supply. Determine the equivalent resistance, the total circuit
current, and the voltage drop across and current at each resistor.
V 1 = I1 •
R1 V1 =
(0.31579
A) • (11 )
V1 =
3.47 V
V 2 = I2 •
R2 V2 =
(0.31579
A) • (7 )
V2 =
2.21 V
V 3 = I3 •
R3 V3 =
(0.31579
A) • (20 )
V3 =
6.32 V