Physics 111 Notes
vx=v0x+axt
x=x0+v0xt+(at2)/2
vx2=v0x2+2ax(x-x0)
x=((v0+v)/2)t
Acceleration is always toward the concave of the path.
X=(v0cosa0)t
Y=(v0sina0)t-(gt2)/2
Vx=v0cosa0
Vy=v0sina0-gt
For uniform circular motion the speed is constant and the acceleration is perpendicular to
the velocity.
Arad=v2/R so,
F=mv2/R
Arad=4(pi2)R/T2
V=2(pi)R/T
VP/A-x=VP/B-x+VB/A-x
Chapter 4
Normal force is perpendicular to surface.
Friction force is parallel to surface.
Newton's 1st law is only valid in inertial frames.
Chapter 5
Mus=fsmax/nb
When Dv2=mg terminal velocity is reached
Vt=(mg/d)1/2
Vmax=(μgR)1/2
Forms of energy
-
Mechanical
o Kinetic (motion)
o Potential (position)
Work links force and energy
Work=0 when displacement is perpendicular
Work values add
Wnet=ΔKE
KE=1/2mv2
PE=mgy
KEi+PEi=KEf+PEf
Fs=-kx
PEs=1/2kx2
P=W/t=Fv
Chapter 7
Wgrav=-ΔUgrav
The total mechanical energy is the sum of KE and PE
E=K+Uel+Wother
F=-(U'pi+U'pj+U'pk)
Exam Problem Topics
Newton's Laws
Conservation of Energy
Momentum
Work
Chapter 9
ϴ=s/r
Ω=ϴ/tWA
α=ω/t
v=ωr
a=αr
a=v2/r=r2ω2/r=rω2
τ=Fd
Στ=Iα
I=Σmr2
KEr=1/2Iω2
L=angular momentum=Iω
Chapter 10
∆𝐿
Στ=TR=Iα= ∆𝑡
L=mvd
Chapter 11
Stress=elastic modulus*strain;
𝑌∆𝐿
F/A= 𝐿
𝐹
Shear stress=𝐴
∆𝑥
Shear strain= ℎ
Shear modulus=S
𝑜
𝐹 𝑆∆𝑥
=
𝐴
ℎ
Volume strees=∆𝑃
∆𝑉
Volume strain= 𝑉
Bulk modulus=-B
∆𝑃 = −
𝐵∆𝑉
𝑉
Chapter 13
𝑭=
𝑮(𝒎𝟏 𝒎𝟐 )
𝒓𝟐
G=Newton's Constant=6.673x10-11
𝒈=
Actual PE=−
𝑮𝑴
𝒓𝟐
𝑮𝑴𝑬 𝒎
𝒓
𝟏
𝒗𝒆𝒔𝒄
𝟐𝑮𝑴 𝟐
=(
)
𝒓
𝒗𝒐𝒓𝒃
𝑮𝑴 𝟏/𝟐
=(
)
𝒓
𝑻𝟐 = 𝑲𝒂𝟑
K=2.97x10-19
𝟑
𝟐𝝅𝒂𝟐
𝑻=
(𝑮𝑴)𝟏/𝟐
𝑅𝑠𝑐ℎ =
𝑓=
2𝐺𝑀
𝑐2
1
T
𝐿
𝐼
T = 2π√ = 2𝜋√
𝑔
𝑚𝑔𝐿
𝑘
𝜔 = 2𝜋𝑓 = √
𝑚
𝑥 = 𝐴𝑐𝑜𝑠𝜔𝑡
𝑣 = −𝐴𝜔𝑠𝑖𝑛𝜔𝑡
𝑎 = −𝐴𝜔2 𝑐𝑜𝑠𝜔𝑡
Energy is conserved in SHM
𝐸=
𝑣 = ±√
𝑘𝐴2
2
𝑘 2
(𝐴 − 𝑥 2 )
𝑚
𝑣𝑚𝑎𝑥 = ±𝜔𝐴
𝑘=−
𝐹𝑥
∆𝑥
𝐹𝑠 = −𝑘𝑥 = 𝑚𝑎𝑥
𝐹𝑡 = −𝑚𝑔𝑠𝑖𝑛𝜃