Useful Equations
Introduction
I’ve just gone through and compiled the important equations from the ends of the various chapters, and the equations given on the quizzes
so far. Hope this helps :).
Chapter 1: Introduction and vectors
Trig
sohcahtoa
opposite
hypotenuse
adjacent
cos θ =
hypotenuse
opposite
tan θ =
adjacent
sin θ =
(1)
(2)
(3)
Vectors
R = A + B = (Ax + Bx )î + (Ay + By )ĵ + (Az + Bz )k̂
(4)
Chapter 2: Motion in one dimension
Definitions
Average quantities
∆x
∆t
∆v
aavg ≡
∆t
vavg ≡
(5)
(6)
Instantaneous quantities
dx
dt
dv
a≡
dt
v≡
(7)
(8)
Equations of motion
For constant velocity problems (integrating the instantaneous velocity definition)
x f = xi + vxt
(9)
And for constant acceleration problems (integrating the instantaneous acceleration definition twice, and manipulating a bit).
vx f = vxi + axt
1
x f = xi + (vxi + vx f )t
2
1
x f = xi + vxit + axt
2
v2x f = v2xi + 2ax (x f − xi )
Where eqn. 13 comes from solving eqn 12 for t using the quadratic formula and plugging the result into eqn 10.
The quadratic formula says that
√
−b ± b2 − 4ac
x=
2a
Yeilds two values of x that solve the quadratic equation
0 = ax2 + bx + c
1
(10)
(11)
(12)
(13)
(14)
(15)
Chapter 3: Motion in two dimensions
Constant acceleration
Applying our 1D equations of motion to each direction, we get the multidimensional formulas:
v f = vi + at
1
r f = ri + vit + at
2
In the special case of projectile motion with ax = 0 and ay = g these reduce to
vx f = vxi = constant
x f = xi + vxt
vy f = vyi − gt
1
y f = yi + vyi − gt 2
2
We could apply our accelerations to all four of the constant 1D acceleration equations in chapter 2, but you get the idea...
(16)
(17)
(18)
(19)
(20)
(21)
Circular motion
A particle moving in a circle of radius r with velocity v has a centerward acceleration of
ac =
v2
r
(22)
Frames of reference
If an observer O0 is moving with velocity vO0 O with respect to observer O, their measurements of the velocity of a particle located at
point P are related according to
vPO = vPO0 + vOO0
(23)
Chapter 4: The laws of motion
Newton’s laws
1. An object in motion will remain in motion unless acted upon by an outside force.
2. ∑ F = ma
3. For every force there is an equal and opposite reaction force.
Note that the two forces referenced in the 3rd law belong to two different free body diagrams (FBDs). For example, the sun and earth
attract each other gravitationally. Let F be the magnitude of the force, and î be the direction from the earth to the sun. The force on the
earth due to the sun (showing up on the earth’s FBD) is F î and the force on the sun due to the earth (showing up on the sun’s FBD) is
−F î.
Chapter 5: More applications of Newton’s laws
Friction
Let µs and µk be the static and kinetic coefficients of friction (respectively) between an object and a surface, and let FN be the normal
force on the object due to the surface. The respective forces of friction are given by
Fs f ≤ µs FN
Fk f = µk FN
(24)
(25)
Drag
Objects moving through viscous materials (air, water, etc.) experience a velocity dependent resistive force
Fdrag = −bv
Where b depends on the particular system under consideration.
2
(26)
Chapter 6: Energy and energy transfer
Work
In general
W≡
Z rf
F · dr
(27)
F · îdx
(28)
ri
Or for the one dimensional case
W≡
Z xf
xi
Where the dot product A · B is defined as
A · B ≡ |A||B| cos θ
(29)
Where θ is the angle between the vectors. So in the 1D, constant force-and-angle case
W = F · ∆r
(30)
Fs = −kx
(31)
The force from a spring is given by Hooke’s law
So the work done by a spring from xi to x f is
Ws =
Z xf
xi
1
1
(−kx)dx = kxi2 − kx2f
2
2
(32)
Kinetic energy
1
K = mv2
2
(33)
Wnet = K f − Ki = ∆K
(34)
Work-kinetic energy theorem
Power
W
∆t
dW
dE
P≡
= F·v =
dt
dt
Pavg ≡
(35)
(36)
Chapter 7: Potential energy
Conservative forces
A force is conservative if the work it does on a particle is independent of the path the particle takes between two given points. The
potential energy change is the inverse of the work done by the force
∆U = U f −Ui = −
Z xf
xi
Fx dx
(37)
Or, taking the derivative of both sides with respect to x
dU
dx
The gravitational potential of a particle under a gravitational force Fg = mg is (relative to the energy at some height hr = 0)
Fx = −
Ug = −
(38)
Z h
(−mg)dy = mgh
(39)
0
The spring potential of a particle under a spring force Fs = −kx is (relative to the unstretched energy at xr = 0)
Z x
1
(−kx)dx = kx2
2
0
The gravitational potential energy of a particle under Newton’s gravitational force FG is (relative to the energy at rr = ∞)
Z r
Z r
−Gm1 m2
1
−Gm1 m2 r
−Gm1 m2
UG = −
dr = Gm1 m2
dr =
=
2
2
r
r
r
r
∞
∞
∞
Us = −
And so on for any other conservative forces...
3
(40)
(41)
Mechanical energy
The total mechanical energy at any moment is
Emech ≡ K +U
(42)
Ei = Ki +Ui = E f = K f +U f
∆K = −∆U
(43)
(44)
Conserving energy using this formula
Which is identical to the work-kinetic energy theorem from Chapter 6 (eqn 34 with Wnet = −∆U). The only difference between these
two approaches is that nonconservative forces do not have really well defined “potential energies”.
Equilibria
In a potential energy diagram, a point of stable equilibrium is a local minimum, a point of unstable equilibrium is a local maximum, and
a region of neutral equilibrium is a region of constant potential energy.
Chapter 8: Momentum and collisions
Linear momentum
Momentum
is conserved
p ≡ mv
(45)
∑ pi = ∑ p f
(46)
Impulse
The impulse-momentum theorem
I=
Z t2
t1
∑ Fdt = ∆p
(47)
Types of collisions
• Inelastic collision: kinetic energy is not conserved.
• Perfectly inelastic collision: the particles stick together afterwards.
• Elastic collision: kinetic energy is conserved.
In all types of collisions (without external forces), momentum is conserved.
Center of mass
∑i mi ri
∑i mi
rCM =
(48)
Chapter 10: Rotational motion
Definitions
dθ
dt
dω
α≡
dt
ω≡
4
(49)
(50)
Relations to linear quantities
(All of the linear quanties are tangential)
st = rθ
vt = rω
at = rα
(51)
(52)
(53)
For constant α problems, all of the constant acceleration equations from Chapter 2 are still valid with the proper substitutions.
Analogs of other linear properties
The rotational motion equivalent of mass is the moment of inertia
I = ∑ mi ri2
(54)
i
Z
I=
ρr2 dV
(55)
And the moments of inertia for some common shapes (Table 10.2, page 300)
Hoop/ring about axis
Solid cylinder about axis
Hollow cylinder about axis
Long thin rod perp to axis through center
Long thin rod perp to axis through end
Solid sphere about diameter
Thin spherical shell about diameter
Rectangular plate perp to one side through center
MR2
1
2
2 MR
1
2
2
2 M(Ri + Ro )
1
2
12 ML
1
2
3 ML
2
2
5 MR
2
2
3 MR
1
2
2
12 M(a + b )
The kinetic energy of a rotating body is given by
1
K = Iω2
(56)
2
(Note that the kinetic energy of any given particle should be expressed as either a linear or a rotational kinetic energy, not as both at
once. You can use linear kinetic energy for one particle and rotational kinetic energy for another to find the kinetic energy at a single
point in time.)
The rotational equivalent to force is torque
τ ≡ r×F
(57)
Where the cross product between two vectors defined by
c
A × B ≡ |A||B| sin θrhr
(58)
c is a unit vector in the direction specified by the right-hand rule.
Where rhr
The analog to Newton’s second law is
dL
∑ τ = Iα = dt
The angular momentum L is given by
L ≡ r × p = Iω
(59)
(60)
Angular momentum is conserved if there are no external torques on the system. As with rotational kinetic energy, you should be careful
to avoid double counting by using both rotational and linear momentum to compute the momentum of the same particle, although here
the units do not match (another reason to keep track of your units!).
Quiz 1: Vector addition and projectile motion
v22 = v21 + 2a(x2 − x1 )
v2 = v1 + at
1
x2 = x1 + v1t + at 2
2
5
(From 13)
(From 10)
(61)
(62)
(From 12)
(63)
Quiz 2: Projectile motion, Newton’s laws, and reference frames
1
y = y0 + v0 (sin θ)t − gt 2
2
x = v0 (cos θ)t
vy = v0 (sin θ) − gt
v2x = v2x0 + 2ax ∆x
dp
∑ Fext = dt
∑ Fext = ma
R
= (v20 sin 2θ)/g
V12 = V1G + VG2
(From 21)
(64)
(From 19)
(From 20)
(65)
(66)
(From 13)
(67)
(From Newton’s 2nd law and 45)
(68)
(Newton’s 2nd law)
(69)
(Range eqn 3.16 from book page 76)
(From 23)
(70)
(71)
Where they have also used sohcahtoa to get vxi = v0 cos θ and vyi = v0 sin θ.
Quiz 3: Friction, momentum, work, kinetic energy, and springs
∆W = ∆KE
(From 34)
(72)
KE = 1/2mv
(From 33)
(73)
∆Wext = 1/2kx
F = −kx
(From 32 with x f = 0)
(From 31)
(74)
(75)
F.dx
(From 28)
(76)
A.B = AB cos θ
∆W12 = F.X
(From 29)
(From 30)
(77)
(78)
2
∆W12 =
2
Z 2
1
6