Phy 2053 Announcements
Exam 2 on Thursday, 8:20 – 10:10 pm
1.
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More Phy 2053 Announcements
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20 questions, multiple choice
There is only one correct answer to the question “In order to
receive credit for this problem, you must correctly code
(“bubble in”) your UFID and your 5-digit test number…”
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Æ I have correctly bubbled my UFID number and 5-digit test code.
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Please get there at least 10 minutes early, and preferably 20
minutes
New protocols - stay in your seat until your exam is
collected. Raise your hands; proctors will come to collect it.
If you want to avoid the long wait at the end, either (a) come
early and sit at the front or (b) finish early
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Please circle your answer on the exam
Room assignments for Exam 2
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BRY130: A-ELU
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FLG220: EMM-HER
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FLG230: HEW-LAI
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FLG260: LAM-MON
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FLG270: MOO-P
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LIT121: R-SAW
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MAEB211:SCH-T
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MAT18: U-Z
You be allowed one handwritten formula sheet
(both sides), 8 ½” x 11” paper
Important concepts from the
beginning of the course
Review – Exam 2
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Exam 2 will cover from Chapter
5.4 through Chapter 8
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Eg, Kepler’s Laws
Physics builds on itself!
1D Kinematics
Newton’s Laws: Σ F = ma
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Concepts and material that you
have learned from the beginning of
the course will be needed
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No testing of material that was not
covered in class
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Work: W ≡ (F cos θ)Δx
(Translational) Kinetic Energy:
2
Work-energy theorem: KEt = (1 / 2) mv
Potential energy
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Hooke’s Law: F = - k x
1 2
Spring potential energy: PEs = kx
Power
P = Work/time = FΔx / t
Average Power P = F v
PEg = mgy
Chapter 6
Springs
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Gravitational potential energy:
Conservation of mechanical energy
Chapter 5
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Equilibrium, free-body diagrams, relationship
between friction and normal force
2
Linear momentum p = mv
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Relationship between force and momentum: F = Δ p/Δ t
r
r r
Impulse: I = Δp = FΔt
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When force is not constant, use average force
Conservation of momentum
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Always conserved, but conservation most useful in
isolated systems involving collisions (no external forces)
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the total momentum before the collision will equal the total
momentum after the collision
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1
Chapter 7
Chapter 6
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Types of collisions
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Head-on, 1D collisions
Inelastic collisions
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Mechanical energy is not conserved in the collision
Perfectly inelastic collisions occur when the objects stick
together after the collision
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Rotational kinematics:
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„ ω = Δθ/Δt, α = Δω/Δt
relationship between angular and linear quantities
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Final velocity vf is the same for both objects
Example is the famous ballistic pendulum
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Elastic collision
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Angle θ, angular velocity ω, angular
acceleration α
Displacement: Δs = Δθr
Velocity: vt = ω r
Acceleration: at = α r
both momentum and kinetic energy are conserved
For one dimensional head-on collisions:
v1i + v1f = v2i + v2f
No questions on
2D collisions (and glancing collisions):
Rocket propulsion
„ x-direction:
x
x
x
x
„ y-direction:
y
y
y
y
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Chapter 7
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v2
ac = t
r
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magnitude given by
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Relationship to angular velocity:
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F=G
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force that keeps an object following a circular
path
FC = maC
„ Eg, tension in a string, gravity,
friction
Banked curves:
Loop-the-loop: v top = gR
Total acceleration:
g=G
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Chapter 8
Torque, τ = r F
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Torque and equilibrium
^
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r
Second condition: Στ = 0
r
ΣF = 0 or
r
r
ΣFx = 0 and ΣFy = 0
Pick coordinate system and draw free-body diagram
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Apply
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Pick axis to sum torques and apply
r
Στ = 0
Torque and angular acceleration:
No questions on
escape velocity
Στ = Iα
2
i i
I = Σmr = MR2
Moment of inertia I:
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Rotational dynamics – Newton’s 2nd Law for rotation
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Rotational kinetic energy:
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Angular momentum:
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Solving rotational equilibrium problems:
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MEm
r
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τ = r F sin θ
First condition:
ME
r2
General expression for gravitational
potential energy
PE = −G
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m1m 2
r2
Acceleration due to gravity varies with
distance from the center of the earth
a = a 2t + aC2
Chapter 8
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Newton’s universal law of gravitation
aC = ω 2 r
Centripetal force
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Chapter 7
Centripetal acceleration
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Table 8-1 lists moments of inertia for different objects
Bucket problem
KE =
1 2
Iω
2
Mechanical conservation laws still apply!
L=Iω
Conservation of angular momentum (isolated system):
The angular momentum of a system is conserved when
the net external torque acting on the systems is zero.
Στ = 0, Li = Lf or Iiω i = If ω f
2
Newton’s Second Law for Rotation-Example
Draw free body diagrams of
each object
„ Only the cylinder is rotating,
so apply Στ = I α
„ The bucket is falling, but not
rotating, so apply ΣF = ma
„ Remember a = α r and
solve the resulting equations
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