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Formula Sheet
Charles Duan Aaron Lee
1 Kinematics
Velocity-distance relation under constant accelera- tion. Given an initial velocity v 0 and a distance d:
v^2 = v 02 + 2ad
Projectile motion distance:
d =
v^2 sin 2θ g
Center of mass:
rcm =
M
i
rimi =
M
rdm
2 Relativity
2.1 Kinematics
Let A be the “fixed” observer and B an observer mov- ing with velocity v relative to A.
Loss of simultaneity. For two clocks a distance L apart in A’s frame that are reading the same time in that frame, the “rear” clock from B’s point of view will be faster by a factor of:
∆t =
Lv c^2
Beta and gamma factors:
β =
v c
, γ =
1 − β^2
, γ > 1
Time dilation and length contraction:
tB = γtA; LB =
LA
γ
Velocity addition. Say A observes a motion of veloc- ity vA, then the velocity with respect to B is:
vB =
vA + v 1 + vvA/c^2
Lorentz transformations. Given that A has a coordi- nate system of (x, y, z, t), the coordinate system for B is (x′, y, z, t′) where:
∆x = γ(∆x′^ + v∆t′), ∆t = γ
∆t′^ +
v c^2
∆x′
Time-space invariant. Given, for two events:
∆s^2 ≡ c^2 ∆t^2 − ∆x^2
The value ∆s^2 is the same in any frame of reference.
2.2 Dynamics
We are given an observer and some system of mass m moving at a speed v. Momentum and energy:
p = γmv, E = γmc^2
Energy-momentum relations:
m^2 c^4 = E^2 − p^2 c^2 ,
p E
v c^2 Energy/momentum for photons:
E = pc
Lorentz transformations for energy. Given a frame of reference moving with speed v, that measures for a system E′^ and p′, we find in the nonmoving frame:
E = γ(E′^ + vp′), p = γ
p′^ +
v c^2
E′
Relativistic force:
F = γ^3 ma =
dp dt
dE dx
3 Rotational Motion
Rolling without slipping:
α = a r , ω =
v r
Moment of inertia:
I =
miri^2 =
r^2 dm
Parallel axis theorem. Given an axis of rotation par- allel to an axis through the center of mass:
Ip = Icm + M d^2
Perpendicular axis theorem. Given a planar object, with the z axis normal to it:
Iz = Ix + Iy
Definition of torque:
~τ = r × F; τ = rF sin θ
Torque and angular acceleration:
∑ τ = Iα
Definition of angular momentum:
L = r × p =
(r × v)dm; L = rp sin θ
Torque and angular momentum. Given a center of rotation that is fixed either in an inertial frame or on the center of mass:
~τ = dL dt
Angular momentum and velocity. For a system that is only rotating about a single axis:
L = Iω
Angular impulse. In a system where a force is applied at a constant distance r from the point of rotation:
∆L = r∆p
Translation and rotation. Given an object with an- gular momentum L′^ about its own center of mass, the angular momentum about any other center is:
L = M Rcm × vcm + L′
4 Harmonic Motion
Given a differential equation of the form y′′^ = −ω^2 y, the solution will be:
y = A cos(ωt + φ)
The constant ω is the angular frequency. The period T and frequency ν are:
T =
2 π ω
, ν =
T
ω 2 π
For a spring, ω =
k/m; for a pendulum, ω =
g/l. For a physical pendulum with moment of inertia at the pivot a distance d from the cm, ω =
mgd/I.
Damped motion. Consider a damping force F = −bv and a harmonic force F = kx. Then define:
ω 02 = k m
, γ = b 2 m
; ω′^2 = ω 02 − γ^2 , Ω^2 = γ^2 − ω 02
There are three possible cases:
Under γ < ω 0 x(t) = Ae−γt^ cos(ω′t + φ) Over γ > ω 0 x(t) = Ae−(γ+Ω)t^ + Be−(γ−Ω)t Critical γ = ω 0 x(t) = e−γt(A + Bt) Driven oscillation. In addition to the damping −bv and harmonic kx forces, consider a driving force Fd(t) = F cos ωdt:
x(t) =
F
mR
cos(ωdt − φ)
with the following constants:
R^2 =
ω 02 − ωd^2
2 γωd ω 02 − ωd^2
5 Universal Gravitation
Newton’s Law of gravitation:
F = −
Gm 1 m 2 r^2 where G = 6. 6726 × 10 −^11
N · m^2 kg^2 The units of G can also be m^2 kg−^1 s−^2. Gravitational potential:
U (r) = −
Gm 1 m 2 r Kepler’s Laws. The planets move in elliptical orbits, they sweep out equal areas over equal times, and for an orbit with semimajor axis a and period T :
T 2 =
4 π^2 a^3 GMsun
6 Fictitious Forces
In an accelerated reference frame R with rotation ω~, the force on an object is the sum of the real forces on it and the following “fictitious forces”:
Translational : −m
d^2 R dt^2 Centrifugal : −m~ω × (~ω × r) Coriolis : − 2 m~ω × v
Azimuthal : −m
d~ω dt
× r
