Final Exam
Monday, May 3, 5:00 to 7:00 pm
McMaster Hall room 2002
You may bring in 1 standard sheet of paper (8.5” x
11.0”) to the Final Exam. -- both sides ok
Constants and some math formulas provided.
Review
Definitions
Scalar: length, speed, mass, energy, potential, pressure,
temperature, entropy, power, moment of inertia
Vector: displacement, velocity, force, momentum, torque
Classical Mechanics
Newton’s Laws of Motion
Conservation Laws (Total Energy, Momentum), collisions
Fluids, waves, equilibrium
Thermodynamics
Thermodynamics Laws
For constant acceleration
Five quantities involved: (x−x0), v0, v, t, a
Equations
Involves
missing
v = v0 + a t
v, v0, a, t
(x-x0)
x - x0 = v0 t + ½ a t2
(x−x0), v0, a, t
v
v2 – v02 = 2a ( x – x0 )
(x−x0), v0, a, v
t
x – x0 = ½ ( v0 + v ) t
(x−x0), v0, v, t
a
x – x0 = v t – ½ a t2
(x−x0), v, a, t
v0
Above equations only apply for a = constant
Chapter 3: Vectors
To describe motions in 2- or 3-dimensions, we need vectors
A vector quantity has both a magnitude and a
direction. e.g., acceleration, velocity, displacement,
force, torque, and momentum.
To describe 3-D motion, we will dissect it into 3
one-dimensional motions. Each 1-D will be a
component in the larger 3-D vector.
Add vectors by components
r
a =
a x î
r
b =
b x î
r r r
r = a + b = ( a x + b x ) î
+
a y ˆj
+
a z k̂
+
b y ˆj
+
b z k̂
+
( a y + b y ) ˆj +
a x = a cos θ
a y = a sin θ
a=
a 2x + a 2y
tan θ =
ay
ax
( a z + b z ) k̂
Multiplication of Vectors
• Multiply a vector by a scalar: b = s a
– Magnitude of b: s times the magnitude of a
– Direction of b : same as a if s > 0,
opposite of a if s < 0
• Multiply a vector by a vector
– Scalar product (tells you the projection of a onto b.)
• results in a scalar
– Vector product (tells you the area subtended by a and b.)
• results in another vector perpendicular to both a and b.
Forms of energy
Kinetic energy: the energy associated with motion.
Potential energy: the energy associated with position
or state. (Chapter 08)
Heat: thermodynamic quanitity related to entropy.
Conservation of Energy is one of Nature’s fundamental
laws that is not violated.
That means the grand total of all forms of energy in a
given system is (and was, and will be) a constant.
Etotal = K1 + U1 + Wnc = K2 + U2
= ½ mv12 + mgy1 + ½ kx12 = ½ mv22 + mgy2 + ½ kx22
Law of Conservation of Energy
Count up the initial energy in all of its forms.
E i = K i + U i + E thermali + E internali
Count up the final energy in all of its forms.
E f = K f + U f + E thermalf + E internalf
These two must be equal.
E i = K i + U i + E thermali + E internali
= E f = K f + U f + E thermalf + E internalf
Work-Kinetic Energy Theorem
Work is energy transferred to or from an object
by means of a force acting on the object
v v
W = ∫ F ⋅ ds
The change in the kinetic energy of a particle is equal the net
work done on the particle
∆K = K f − K i = Wnet
1
1
2
= mv f − mvi2
2
2
final kinetic energy = initial kinetic energy + net work
1
1
2
2
K f = mv f = K i + Wnet = mvi + Wnet
2
2
Force Potential Energy Relationship
F(x) = −dU(x)/dx
Thus, the force in the x-directions the negative derivative of the
potential energy! The same holds true for y- and z-directions.
∂U( x , y, z)
F( x ) = −
∂x
∂U( x , y, z)
F( y) = −
∂y
∂U( x , y, z)
F(z) = −
∂z
v  ∂U( x , y, z)   ∂U( x , y, z)   ∂U( x , y, z) 
 ĵ +  −
F = −
 î +  −
 k̂
∂x
∂y
∂z

 

 
Potential Energy Curve
We know:
Therefore:
∆U(x) = −W = −F(x) ∆x
F(x) = −dU(x)/dx
Now integrate along the displacement:
v v
dU
W = ∫ F ⋅ dx = − ∫
dx
dx
v v
dU
∫ F ⋅ dx = K f −K i = − ∫ dx = −(U f − Ui ) = Ui − U f
dx
v v
dU
∫ F ⋅ dx = − ∫ dx dx = K f − K i = U i − U f
Rearrange terms: K f + U f = K i + U i
Center of mass for a system of n particles:
n
n
n
1
1
1
y c.m . = ∑ m i y i
z c. m . = ∑ m i z i
x c. m . = ∑ m i x i
M i =1
M i =1
M i =1
n
M is just the total mass of the system M = ∑ m i
1 n
v
v
Using vectors, we have: rc.m. = ∑ m i ri
M i =1
i =1
Continuous mass distributions:
x c. m .
1
= ∫ x dm
M
y c.m .
1
= ∫ y dm
M
z c.m.
1
= ∫ z dm
M
Linear Momentum
The linear momentum of a
particle is defined as
v
v
p = mv
Newton’s second law in terms of momentum
v
v
dp d
dv v dm
v v
v
= (mv) = m + v
= ma = Fnet
dt dt
dt
dt
Newton’s Second Law
n v
v
v
v
Fnet = Ftotal = ∑ Fi = ma
i =1
Inelastic Collisions
The mechanical energy is not conserved in inelastic
collisions.
(Ei > Ef) or (Ei < Ef)
But, .... linear momentum is conserved!
Inelastic Collisions
How much mechanical energy was lost in the collision?
v

1
(m 1 + m 2 )V f2 = 1 (m 1 + m 2 ) m 1
2
2
 m1 + m 2
E lost = E i − E f =
E lost
1
1
m 1 v 12i − 
2
2
1
=
2
2
 2
 2
1
m 12
 v 1 i = 
 v 1 i
2  m1 + m 2 

 2
m 12
1  m 1m 2  2
 v 1 i = 
 v 1 i
m1 + m 2 
2  m1 + m 2 
 m 1m 2

 m1 + m 2
 2
 v 1 i

⇒
Elastic Collisions
Perfectly elastic collision: Linear momentum is conserved
v
v
v
v
v
m 1 v 1 i + m 2 v 2 i = m 1 v 1 f + m 2 v 2 f (= M v c . m . )
Perfectly elastic collision: Mechanical energy is conserved
1
1
v2
v
E i = m 1 v 1 i + m 2 v 22 i
2
2
=
1
1
v2
v
m 1 v 1 f + m 2 v 22 f = E f
2
2
Rotation with constant angular acceleration
The equations for constant angular acceleration are
similar to those for constant linear acceleration
replace θ for x, ω for v, and α for a,
missing
v = v0 + at
ω = ω0 + αt
θ – θ0
x – x0 = v0t + ½ at2
θ – θ0 = ω0t + ½ αt2
ω
v2 = v02 + 2a(x-x0)
ω2 = ω02 +2α (θ - θ0)
t
x – x0 = ½ (v0 + v)t
θ – θ0 = ½(ω0 − ω)t
α
x – x0 = vt − ½ at2
θ – θ0 = ωt − ½ αt2
ω0
Kinetic Energy of Rotation
Consider a rigid body rotating around a fixed axis as a
collection of particles with different linear speed, the
total kinetic energy is
K = Σ ½ mi vi2 = Σ ½ mi (ω ri )2 = ½ ( Σ mi ri 2) ω2
Define rotational inertia ( moment of inertia) to be
I=Σ
mi ri2
I = ∫ r 2dm
ri : the perpendicular distance between mi and the
rotation axis
Then K = ½Iω2
Compare to the linear case: K = ½mv2
Newton’s second law for rotation
n
v v
v I: rotational inertia
v
τnet = ∑ ri × Fi = Iα α: angular acceleration
i =1
n v
v
Compare to the linear equation: Fnet = ∑ Fi = mav
i =1
Angular momentum with respect to point O for a particle
of mass m and linear momentum p is defined as:
v r r
r r
l = r × p = m (r × v )
Magnitude: l = r p sin φ = r mv sin φ
The Kinetic Energy of Rolling
View the rolling as pure rotation around P, the kinetic energy
K = ½ IP ω2
parallel axis theorem: Ip = Icom +MR2
so
K = ½ Icomω2 + ½ MR2ω2
since vcom = ωR
K = ½ Icomω2 + ½ M(vcom)2
½ Icomω2 : due to the object’s rotation about its center of mass
½ M(vcom)2 : due to the translational motion of its center of mass
Chapter 12 Equilibrium & Elasticity
Equilibrium: P = constant and L = constant
Static equilibrium: Objects that are
not moving either in translation or
rotation. P = 0 L = 0
Requirements of equilibrium
r
r
dP
Fnet =
dt
r
r dL
τ =
dt
r
Fnet = 0 (balance of forces)
r
τ net = 0 (balance of torques)
The equation of continuity
∆V = A ∆x = A v ∆t
∆V = A1v1∆t = A2v2∆t
A1v1 = A2v2 (Continuity equation)
or:
Av = constant
Bernoulli’s equation
Bernoulli’s equation is based on
Conservation of Mechanical Energy
=
Conserved
Potential energy per volume
Kinetic energy per volume
Pressure
p + ½ ρv2 + ρgy = constant
p1 + ½ ρv12 + ρgy1 = p2 + ½ ρv22 + ρgy2 = constant
Bernoulli’s equation
Bernoulli’s equation
based on Conservation of Mechanical Energy
p + ½ ρv2 + ρgy = constant
For fluid at rest, v = 0
p + ρgy = constant
p2 = p1 + ρg(y1-y2)
For y = constant,
p + ½ ρv2 = constant
if v increases, then p decreases
Thermodynamics -- 1st Law
Recall the definition of work:
W = ∫ p dV ⇒ dW = pdV
1st Law of Thermodynamics
E = Q − W => ∆E = Q − W