Rotational Energy and Torque
1D Motion:
Rocket Propulsion:
𝑥𝑓 = 𝑥𝑖 + 𝑣𝑥 𝑡 | 𝑣𝑥𝑓 = 𝑣𝑥𝑖 + 𝑎𝑡 | 𝑣 =
𝑑 𝑑𝑡
𝑥 | 𝑎=
𝑑 𝑑𝑡
𝑣=
𝑑2 𝑑𝑡 2
𝑥
𝑣𝑓 − 𝑣𝑖 = 𝑣𝑒 ln(
𝑀𝑖
𝑀𝑓
1
) [ve is ejection velocity of fuel (assume constant) and
𝑥𝑓 = 𝑥𝑖 + 𝑣𝑥𝑖 𝑡 + 𝑎𝑡 2 |𝑣𝑥𝑓 2 = 𝑣𝑥𝑖 2 + 2𝑎∆𝑥
Mi and Mf are the initial and final masses of the rocket]
Projectile Motion:
𝐹𝑇 = 𝑣𝑒
2
𝑡=
𝑣𝑖 𝑠𝑖𝑛𝜃𝑖 𝑔
𝑣𝑖 2 𝑠𝑖𝑛2𝜃𝑖
|𝑅 =
𝑔
|ℎ=
𝑣𝑖 2 𝑠𝑖𝑛2 𝜃𝑖 2𝑔
[Air time, max range and max
height, must start and finish at same height]
Forces
Δ𝑚 Δ𝑡
[Thrust on a rocket] 𝐹𝑇 = 𝑚(𝑎𝑖 + 𝑔) [a is initial net accel]
Momentum: 𝑝⃗ = 𝑚𝑣⃗ [Vector momentum which is conserved in a system and can be 𝑠 split into components] 𝐼⃗ = Δ𝑝⃗ = 𝐹𝑎𝑣 Δ𝑡 = ∫𝑠 𝑓 𝐹 𝑑𝑡 [Impulse] 𝑖
𝑑𝑝 𝐹⃗𝑛𝑒𝑡 = = 0 𝑎⃗ = 0 [Newton’s first law]
Elastic collision: momentum and kinetic energy is conserved
𝑑𝑡
𝐹⃗ = 𝑚𝑎⃗ [Newton’s second law] | 𝐹⃗1 𝑜𝑛 2 = −𝐹⃗2 𝑜𝑛 1 [Newton’s third law, remember they act on different objects!] 𝐹𝑓 = 𝜇𝐹𝑁 [Friction]
On an inclined plane 𝜇𝑠 = 𝑡𝑎𝑛𝜃
Uniform Circular Motion 𝑎𝑐 =
𝑣2
|∑ 𝐹𝑐 =
𝑟
𝑚𝑣 2 𝑟
=
4𝜋2 𝑟 𝑇2
[Centripetal Force where T
is period of rotation, remember that centripetal force is comprised of other forces, it is not a new force itself]
Inelastic: momentum is conserved but not kinetic energy Perfectly inelastic: objects stick together after collision 𝑚1 𝑣⃗1 + 𝑚1 𝑣⃗1 = (𝑚1 + 𝑚2 )𝑣⃗𝑓
Elastic Collisions (special cases): 𝑣1𝑓 =
𝑚1 −𝑚2
𝑣 𝑚1 +𝑚2 1𝑖
+
2𝑚2 𝑚1 +𝑚2
𝑣2𝑖 | 𝑣2𝑓 =
2𝑚1 𝑣 𝑚1 +𝑚2 1𝑖
+
𝑚2 −𝑚1 𝑚1 +𝑚2
𝑣2𝑖
If 𝑚1 = 𝑚2 = 𝑚 Then 𝑣1𝑓 = 𝑣2𝑖 & 𝑣2𝑓 = 𝑣1𝑖 If 𝑣2𝑖 = 0 Then 𝑣1𝑓 =
𝑚1 −𝑚2
𝑣 𝑚1 +𝑚2 1𝑖
&𝑣2𝑓 =
2𝑚1 𝑣 𝑚1 +𝑚2 1𝑖
𝑥 𝑥 m⁄s = rads⁄s [Conversion between m/s to rads/s 𝑟 where r is radius]
If 𝑣2𝑖 = 0 & 𝑚2 ≫ 𝑚1 Then 𝑣1𝑓 = −𝑣1𝑖 [Ex: Ball bounce off a wall]
𝑣 2 = 𝑔𝑙𝑠𝑖𝑛𝜃𝑡𝑎𝑛𝜃 [Speed around a conical pendulum of string length l]
If 𝑣2𝑖 = −𝑣1𝑖 & 𝑚2 ≫ 𝑚1 Then 𝑣1𝑓 = −3𝑣1𝑖 & 𝑣2𝑓 = 𝑣2𝑖 [Ex Ball bounces off a moving truck]
𝑣2
𝑡𝑎𝑛𝜃 =
𝑟𝑔
[Angle of banked curve]
Energy 𝑠
1
𝑣𝑚𝑎𝑥 = √𝜇𝑠 𝑔𝑟 = √𝑡𝑎𝑛𝜃𝑔𝑟 [Max speed around curve]
𝑊 = 𝐹Δ𝑑 cos 𝜃 = ∫𝑠 𝑓 𝐹𝑠 𝑑𝑠 [Work] | 𝐸𝑘 = 𝑚𝑣 2 [Kinetic energy] 2
Resistive Drag
𝑈𝑔 = 𝑚𝑔𝑦 [Gravitational potential energy]
𝑖
1
𝑅 = − 𝐷𝑝𝐴𝑣 2 [Resistive drag where D
1
2
is unit-less drag coefficient, p is density, A is cross-sectional area and v is speed]
2
(includes all potential energies not only gravitational)]
𝑉𝑇 = √
2𝑚𝑔 𝐷𝑝𝐴
[Terminal velocity of object
𝑣
𝑅 = 𝑚𝑔( )2 [Another formula for resistive drag] 𝑣𝑇
Center of mass of a system of particles
𝑟⃗𝑐𝑚 =
∑𝑖 𝑥𝑖 𝑚𝑖 ∑𝑖 𝑚𝑖 ∑𝑖 𝑟⃗𝑖 𝑚𝑖 ∑𝑖 𝑚𝑖
𝑥1 𝑚 1 + 𝑥2 𝑚 2
=(
𝑚1 +𝑚2
) [Center of mass of a system of particles]
[Generalization of above to any vector e.g. force, velocity]
𝐿
𝑥𝑐𝑚 = [Center of mass of a rod where L is the length of the rod] 2 2
𝑥𝑐𝑚 = 𝑎 [Center of mass of a right triangle where a is the adjacent side] 3
𝑧𝑐𝑚 = ℎ [Center of mass of a cone where h is the vertical height] 4
𝑑 𝑑𝑥
𝑈𝑠 [Hooke’s law: force of spring at position x, k is spring 1
constant [k] N/m] | 𝑈𝑠 = 𝑘𝑥 2 [Elastic energy of a spring] 2
Work done by a conservative force on a particle moving between any two points is independent of the path taken. This force is associated with a potential energy (ex: gravity, spring force) Non-conservative forces change the mechanical energy (ex: friction) 𝑃=
[Remember center of mass does not divide equally sized masses (baseball bat ex)]
3
2
𝐹𝑠 = −𝑘𝑥 = −
of mass m, drag coefficient D, density p and cross-sectional area A]
𝑥𝑐𝑚 =
1
𝑚𝑣𝑓 2 + 𝑚𝑔𝑦𝑓 = 𝑚𝑣𝑖 2 + 𝑚𝑔𝑦𝑖 [Conservation of mechanical energy
∆𝑊 Δ𝑡
= 𝐹⃗ ∙ 𝑣⃗ [P]=Watts [1hp = 746W]
Rotational Motion: Replace x with 𝜃, v with 𝜔 and a with 𝛼 in all 1D kinematics 𝑠 = 𝑟𝜃 | 𝑣 = 𝑟𝜔 | 𝑎𝑡 = 𝑟𝛼 | 𝑎𝑐 = 𝜔2 𝑟 𝑎𝑡𝑜𝑡𝑎𝑙 = √𝑎𝑡 2 + 𝑎𝑐 2 = 𝑟√𝛼 2 + 𝜔 4
Rotational Energy and Torque 1
𝐸𝑘𝑟 = 𝐼𝜔2 [Rotational kinetic energy] 2
𝐼 = 𝜌 ∫ 𝑟 2 𝑑𝑣 = ∫ 𝑟 2 𝑑𝑚 = Σ𝑚𝑖 𝑟 2 [p is density, last one is for point masses 𝑈𝑔 = −𝐺𝑀𝐸 𝑚 [Gravitational potential, don’t forget the negative sign!] 𝑟 in a system] Units of [I] = kg●m2 𝜏 = 𝑟𝐹𝑠𝑖𝑛𝜃 | 𝜏⃗ = 𝑟⃗ × 𝐹⃗ [𝜏] = m∙N (Not a Joule!) 𝜏 = 𝐼𝛼 =
𝑑𝐿
| 𝑊 = 𝜔𝜃 | 𝑃 = 𝜏𝜔 [P]=Watts
𝑑𝑡
Moments of Inertia for Other Rigid Objects
𝑅=
2𝐺𝑀
Schwarzschild radius (radius at which point nothing can escape)
𝑐2
Simple Harmonic Motion and Waves: Conditions of SHM: Restoring force and motion is linear in x 𝑥(𝑡) = 𝐴 cos(𝜔𝑡 + 𝜙) | 𝑣(𝑡) = −𝐴𝜔 sin(𝜔𝑡 + 𝜙) | 𝑎(𝑡) = −𝜔2 𝑥(𝑡) Make sure to adjust phi to initial conditions 𝜔= √
𝑘
| 𝑇=
𝑚
2𝜋 𝜔
= 2𝜋√
𝑚 𝑘
| 𝑓=
1 𝑇
=
𝜔
=
2𝜋
1 2𝜋
√
𝑘
𝑚
𝐿
𝑇 = 2𝜋√ [Period of a pendulum of length L] | 𝜃(𝑡) = 𝜃𝑚𝑎𝑥 cos(𝜔𝑡 + 𝜙) 𝑔
𝐼𝑝𝑖𝑣𝑜𝑡
𝑇 = 2𝜋√
𝑚𝑔𝑑
[Period of a physical pendulum where d is the distance from
the center of mass and pivot point] Conditions of mechanical waves: Source of disturbance, medium to be disturbed and physical mechanism for elements to interact Remember wave speed is a property of medium 𝑦(𝑥, 𝑡) = 𝐴 sin(𝑘𝑥 − 𝜔𝑡 + 𝜙) [or kx + ωt for wave moving left] 𝑇
𝑣 = 𝜆𝑓 | 𝑣 = √ [T is tension in string, mu is linear mass density kg/m] 𝜇
𝐼𝑃𝐴 = 𝐼𝐶𝑀 + 𝑀𝐷2 [Parallel-Axis theorem]
Angular Momentum: ⃗⃗ = 𝑟⃗ × 𝑝⃗ = 𝑟⃗ × 𝑚𝑣⃗ | 𝐿
⃗⃗ 𝑑𝐿 𝑑𝑡
A reflected wave undergoes a phase change of pi. (i.e. it inverts itself) If the wave string is loose at one end, no phase change occurs. 𝐸=
= 0 ⇒ No torques applied and angular
1
1
𝜇𝜔2 𝐴2 𝜆 | 𝑃 = 𝜇𝜔2 𝑣𝐴2
2
2
momentum is conserved, ex uniform circular motion
Sound Waves:
[You can use angular momentum in linear motion but linear and translational momentum are NOT interchangeable p ≠ L]
𝑠(𝑥, 𝑡) = 𝑠𝑚𝑎𝑥 cos(𝑘𝑥 − 𝜔𝑡) | Δ𝑝(𝑥, 𝑡) = Δ𝑝𝑚𝑎𝑥 cos(𝑘𝑥 − 𝜔𝑡 + 𝜙) 𝑘=
𝐿 = 𝑟𝑚𝑣 sin 𝜃 = 𝐼𝜔 𝜔𝑝 =
2𝜋 𝜆
| 𝜔 = 2𝜋𝑓 | Superposition: Algebraically add both wave equations
Superposition of two identical waves: 𝑦 =
𝑚𝑔𝑥
[Precession frequency where x is the x distance between the
𝐼𝑡𝑜𝑝 +𝜔𝑡𝑜𝑝
𝜙
𝜙
2
2
2𝐴 cos( ) sin(𝑘𝑥 − 𝜔𝑡 + ) where phi is the
center of mass and the point of rotation]
phase shift between the two waves
Gravitation:
Standing waves: 𝑦 = 2𝐴 sin(𝑘𝑥) cos(𝜔𝑡)
𝐹𝑔 =
𝐺𝑚1 𝑚2 (𝑟1∙2 )2
[Remember r is distance between centers]
−𝐺𝑚1 𝑚2 𝐹⃗1∙2 = ⃗1∙2 [Vector representation of gravity] 3 𝑟 |𝑟1∙2 |
𝑔=
𝐺𝑀 𝑟2
[Gravitational field strength]
where 𝑓0 =
𝑣 𝜆
𝑣
Kepler’s second law: Angular momentum is conserved in planetary motion
Beats: 𝑦 = 2𝐴 cos(2𝜋
𝐺𝑀𝑠
) 𝑟 3 Where T is period of orbit, r is the radius
𝐼= 𝑣𝑜𝑟𝑏 = √
𝑟
2𝐺𝑀𝐸
𝑣𝑒𝑠𝑐 = √
𝑟
[Speed of an orbit, r away from the center of the earth] [Escape speed from r distance from center of earth]
4𝐿
Beat freq 𝑓𝑏 = |𝑓1 − 𝑓2 | 𝑓1 −𝑓2 2
𝑡) cos(2𝜋
1
1
2
2
𝑓1 +𝑓2 2
𝑡)
𝐸𝑡𝑟𝑎𝑛𝑠 = 𝜇𝜔2 𝐴2 𝜆 | 𝐸𝑙𝑜𝑛𝑔 = 𝐿2 𝜌𝜔2 𝐴2 𝑣
of the circular orbit or half the long dimension of an elliptical orbit. 𝐺𝑀𝐸
The ith harmonic
Note that in closed air columns, there is no even harmonics (i.e no 2nd or 4th) 𝑓𝑖 = (𝑖)𝑓0 = (𝑖)
Kepler’s third law: 𝑇 2 = (
2𝐿
| 𝜆 = 2𝐿 | L is length of the string
Kepler’s first law: Planets move in ellipses with the sun at one focus
4𝜋2
𝑣
𝑓𝑖 = (𝑖 + 1)𝑓0 = (𝑖 + 1)
𝑃 𝐴
1
1
2
𝑟2
= 𝜌𝜔2 𝐴2 𝑣 | 𝐼 ∝ 𝑣±𝑣𝑜
𝑓 ′ = 𝑓(
𝑣±𝑣𝑠
| Β = 10 log10 (
𝐼
10−12
)
) -vo when moving away from source, +vo moving towards
+vs when source moves away from observer, -vs moving towards observer By: Kareem Halabi
Rocket Propulsion:
𝑥𝑓 = 𝑥𝑖 + 𝑣𝑥 𝑡 | 𝑣𝑥𝑓 = 𝑣𝑥𝑖 + 𝑎𝑡 | 𝑣 =
𝑑 𝑑𝑡
𝑥 | 𝑎=
𝑑 𝑑𝑡
𝑣=
𝑑2 𝑑𝑡 2
𝑥
𝑣𝑓 − 𝑣𝑖 = 𝑣𝑒 ln(
𝑀𝑖
𝑀𝑓
1
) [ve is ejection velocity of fuel (assume constant) and
𝑥𝑓 = 𝑥𝑖 + 𝑣𝑥𝑖 𝑡 + 𝑎𝑡 2 |𝑣𝑥𝑓 2 = 𝑣𝑥𝑖 2 + 2𝑎∆𝑥
Mi and Mf are the initial and final masses of the rocket]
Projectile Motion:
𝐹𝑇 = 𝑣𝑒
2
𝑡=
𝑣𝑖 𝑠𝑖𝑛𝜃𝑖 𝑔
𝑣𝑖 2 𝑠𝑖𝑛2𝜃𝑖
|𝑅 =
𝑔
|ℎ=
𝑣𝑖 2 𝑠𝑖𝑛2 𝜃𝑖 2𝑔
[Air time, max range and max
height, must start and finish at same height]
Forces
Δ𝑚 Δ𝑡
[Thrust on a rocket] 𝐹𝑇 = 𝑚(𝑎𝑖 + 𝑔) [a is initial net accel]
Momentum: 𝑝⃗ = 𝑚𝑣⃗ [Vector momentum which is conserved in a system and can be 𝑠 split into components] 𝐼⃗ = Δ𝑝⃗ = 𝐹𝑎𝑣 Δ𝑡 = ∫𝑠 𝑓 𝐹 𝑑𝑡 [Impulse] 𝑖
𝑑𝑝 𝐹⃗𝑛𝑒𝑡 = = 0 𝑎⃗ = 0 [Newton’s first law]
Elastic collision: momentum and kinetic energy is conserved
𝑑𝑡
𝐹⃗ = 𝑚𝑎⃗ [Newton’s second law] | 𝐹⃗1 𝑜𝑛 2 = −𝐹⃗2 𝑜𝑛 1 [Newton’s third law, remember they act on different objects!] 𝐹𝑓 = 𝜇𝐹𝑁 [Friction]
On an inclined plane 𝜇𝑠 = 𝑡𝑎𝑛𝜃
Uniform Circular Motion 𝑎𝑐 =
𝑣2
|∑ 𝐹𝑐 =
𝑟
𝑚𝑣 2 𝑟
=
4𝜋2 𝑟 𝑇2
[Centripetal Force where T
is period of rotation, remember that centripetal force is comprised of other forces, it is not a new force itself]
Inelastic: momentum is conserved but not kinetic energy Perfectly inelastic: objects stick together after collision 𝑚1 𝑣⃗1 + 𝑚1 𝑣⃗1 = (𝑚1 + 𝑚2 )𝑣⃗𝑓
Elastic Collisions (special cases): 𝑣1𝑓 =
𝑚1 −𝑚2
𝑣 𝑚1 +𝑚2 1𝑖
+
2𝑚2 𝑚1 +𝑚2
𝑣2𝑖 | 𝑣2𝑓 =
2𝑚1 𝑣 𝑚1 +𝑚2 1𝑖
+
𝑚2 −𝑚1 𝑚1 +𝑚2
𝑣2𝑖
If 𝑚1 = 𝑚2 = 𝑚 Then 𝑣1𝑓 = 𝑣2𝑖 & 𝑣2𝑓 = 𝑣1𝑖 If 𝑣2𝑖 = 0 Then 𝑣1𝑓 =
𝑚1 −𝑚2
𝑣 𝑚1 +𝑚2 1𝑖
&𝑣2𝑓 =
2𝑚1 𝑣 𝑚1 +𝑚2 1𝑖
𝑥 𝑥 m⁄s = rads⁄s [Conversion between m/s to rads/s 𝑟 where r is radius]
If 𝑣2𝑖 = 0 & 𝑚2 ≫ 𝑚1 Then 𝑣1𝑓 = −𝑣1𝑖 [Ex: Ball bounce off a wall]
𝑣 2 = 𝑔𝑙𝑠𝑖𝑛𝜃𝑡𝑎𝑛𝜃 [Speed around a conical pendulum of string length l]
If 𝑣2𝑖 = −𝑣1𝑖 & 𝑚2 ≫ 𝑚1 Then 𝑣1𝑓 = −3𝑣1𝑖 & 𝑣2𝑓 = 𝑣2𝑖 [Ex Ball bounces off a moving truck]
𝑣2
𝑡𝑎𝑛𝜃 =
𝑟𝑔
[Angle of banked curve]
Energy 𝑠
1
𝑣𝑚𝑎𝑥 = √𝜇𝑠 𝑔𝑟 = √𝑡𝑎𝑛𝜃𝑔𝑟 [Max speed around curve]
𝑊 = 𝐹Δ𝑑 cos 𝜃 = ∫𝑠 𝑓 𝐹𝑠 𝑑𝑠 [Work] | 𝐸𝑘 = 𝑚𝑣 2 [Kinetic energy] 2
Resistive Drag
𝑈𝑔 = 𝑚𝑔𝑦 [Gravitational potential energy]
𝑖
1
𝑅 = − 𝐷𝑝𝐴𝑣 2 [Resistive drag where D
1
2
is unit-less drag coefficient, p is density, A is cross-sectional area and v is speed]
2
(includes all potential energies not only gravitational)]
𝑉𝑇 = √
2𝑚𝑔 𝐷𝑝𝐴
[Terminal velocity of object
𝑣
𝑅 = 𝑚𝑔( )2 [Another formula for resistive drag] 𝑣𝑇
Center of mass of a system of particles
𝑟⃗𝑐𝑚 =
∑𝑖 𝑥𝑖 𝑚𝑖 ∑𝑖 𝑚𝑖 ∑𝑖 𝑟⃗𝑖 𝑚𝑖 ∑𝑖 𝑚𝑖
𝑥1 𝑚 1 + 𝑥2 𝑚 2
=(
𝑚1 +𝑚2
) [Center of mass of a system of particles]
[Generalization of above to any vector e.g. force, velocity]
𝐿
𝑥𝑐𝑚 = [Center of mass of a rod where L is the length of the rod] 2 2
𝑥𝑐𝑚 = 𝑎 [Center of mass of a right triangle where a is the adjacent side] 3
𝑧𝑐𝑚 = ℎ [Center of mass of a cone where h is the vertical height] 4
𝑑 𝑑𝑥
𝑈𝑠 [Hooke’s law: force of spring at position x, k is spring 1
constant [k] N/m] | 𝑈𝑠 = 𝑘𝑥 2 [Elastic energy of a spring] 2
Work done by a conservative force on a particle moving between any two points is independent of the path taken. This force is associated with a potential energy (ex: gravity, spring force) Non-conservative forces change the mechanical energy (ex: friction) 𝑃=
[Remember center of mass does not divide equally sized masses (baseball bat ex)]
3
2
𝐹𝑠 = −𝑘𝑥 = −
of mass m, drag coefficient D, density p and cross-sectional area A]
𝑥𝑐𝑚 =
1
𝑚𝑣𝑓 2 + 𝑚𝑔𝑦𝑓 = 𝑚𝑣𝑖 2 + 𝑚𝑔𝑦𝑖 [Conservation of mechanical energy
∆𝑊 Δ𝑡
= 𝐹⃗ ∙ 𝑣⃗ [P]=Watts [1hp = 746W]
Rotational Motion: Replace x with 𝜃, v with 𝜔 and a with 𝛼 in all 1D kinematics 𝑠 = 𝑟𝜃 | 𝑣 = 𝑟𝜔 | 𝑎𝑡 = 𝑟𝛼 | 𝑎𝑐 = 𝜔2 𝑟 𝑎𝑡𝑜𝑡𝑎𝑙 = √𝑎𝑡 2 + 𝑎𝑐 2 = 𝑟√𝛼 2 + 𝜔 4
Rotational Energy and Torque 1
𝐸𝑘𝑟 = 𝐼𝜔2 [Rotational kinetic energy] 2
𝐼 = 𝜌 ∫ 𝑟 2 𝑑𝑣 = ∫ 𝑟 2 𝑑𝑚 = Σ𝑚𝑖 𝑟 2 [p is density, last one is for point masses 𝑈𝑔 = −𝐺𝑀𝐸 𝑚 [Gravitational potential, don’t forget the negative sign!] 𝑟 in a system] Units of [I] = kg●m2 𝜏 = 𝑟𝐹𝑠𝑖𝑛𝜃 | 𝜏⃗ = 𝑟⃗ × 𝐹⃗ [𝜏] = m∙N (Not a Joule!) 𝜏 = 𝐼𝛼 =
𝑑𝐿
| 𝑊 = 𝜔𝜃 | 𝑃 = 𝜏𝜔 [P]=Watts
𝑑𝑡
Moments of Inertia for Other Rigid Objects
𝑅=
2𝐺𝑀
Schwarzschild radius (radius at which point nothing can escape)
𝑐2
Simple Harmonic Motion and Waves: Conditions of SHM: Restoring force and motion is linear in x 𝑥(𝑡) = 𝐴 cos(𝜔𝑡 + 𝜙) | 𝑣(𝑡) = −𝐴𝜔 sin(𝜔𝑡 + 𝜙) | 𝑎(𝑡) = −𝜔2 𝑥(𝑡) Make sure to adjust phi to initial conditions 𝜔= √
𝑘
| 𝑇=
𝑚
2𝜋 𝜔
= 2𝜋√
𝑚 𝑘
| 𝑓=
1 𝑇
=
𝜔
=
2𝜋
1 2𝜋
√
𝑘
𝑚
𝐿
𝑇 = 2𝜋√ [Period of a pendulum of length L] | 𝜃(𝑡) = 𝜃𝑚𝑎𝑥 cos(𝜔𝑡 + 𝜙) 𝑔
𝐼𝑝𝑖𝑣𝑜𝑡
𝑇 = 2𝜋√
𝑚𝑔𝑑
[Period of a physical pendulum where d is the distance from
the center of mass and pivot point] Conditions of mechanical waves: Source of disturbance, medium to be disturbed and physical mechanism for elements to interact Remember wave speed is a property of medium 𝑦(𝑥, 𝑡) = 𝐴 sin(𝑘𝑥 − 𝜔𝑡 + 𝜙) [or kx + ωt for wave moving left] 𝑇
𝑣 = 𝜆𝑓 | 𝑣 = √ [T is tension in string, mu is linear mass density kg/m] 𝜇
𝐼𝑃𝐴 = 𝐼𝐶𝑀 + 𝑀𝐷2 [Parallel-Axis theorem]
Angular Momentum: ⃗⃗ = 𝑟⃗ × 𝑝⃗ = 𝑟⃗ × 𝑚𝑣⃗ | 𝐿
⃗⃗ 𝑑𝐿 𝑑𝑡
A reflected wave undergoes a phase change of pi. (i.e. it inverts itself) If the wave string is loose at one end, no phase change occurs. 𝐸=
= 0 ⇒ No torques applied and angular
1
1
𝜇𝜔2 𝐴2 𝜆 | 𝑃 = 𝜇𝜔2 𝑣𝐴2
2
2
momentum is conserved, ex uniform circular motion
Sound Waves:
[You can use angular momentum in linear motion but linear and translational momentum are NOT interchangeable p ≠ L]
𝑠(𝑥, 𝑡) = 𝑠𝑚𝑎𝑥 cos(𝑘𝑥 − 𝜔𝑡) | Δ𝑝(𝑥, 𝑡) = Δ𝑝𝑚𝑎𝑥 cos(𝑘𝑥 − 𝜔𝑡 + 𝜙) 𝑘=
𝐿 = 𝑟𝑚𝑣 sin 𝜃 = 𝐼𝜔 𝜔𝑝 =
2𝜋 𝜆
| 𝜔 = 2𝜋𝑓 | Superposition: Algebraically add both wave equations
Superposition of two identical waves: 𝑦 =
𝑚𝑔𝑥
[Precession frequency where x is the x distance between the
𝐼𝑡𝑜𝑝 +𝜔𝑡𝑜𝑝
𝜙
𝜙
2
2
2𝐴 cos( ) sin(𝑘𝑥 − 𝜔𝑡 + ) where phi is the
center of mass and the point of rotation]
phase shift between the two waves
Gravitation:
Standing waves: 𝑦 = 2𝐴 sin(𝑘𝑥) cos(𝜔𝑡)
𝐹𝑔 =
𝐺𝑚1 𝑚2 (𝑟1∙2 )2
[Remember r is distance between centers]
−𝐺𝑚1 𝑚2 𝐹⃗1∙2 = ⃗1∙2 [Vector representation of gravity] 3 𝑟 |𝑟1∙2 |
𝑔=
𝐺𝑀 𝑟2
[Gravitational field strength]
where 𝑓0 =
𝑣 𝜆
𝑣
Kepler’s second law: Angular momentum is conserved in planetary motion
Beats: 𝑦 = 2𝐴 cos(2𝜋
𝐺𝑀𝑠
) 𝑟 3 Where T is period of orbit, r is the radius
𝐼= 𝑣𝑜𝑟𝑏 = √
𝑟
2𝐺𝑀𝐸
𝑣𝑒𝑠𝑐 = √
𝑟
[Speed of an orbit, r away from the center of the earth] [Escape speed from r distance from center of earth]
4𝐿
Beat freq 𝑓𝑏 = |𝑓1 − 𝑓2 | 𝑓1 −𝑓2 2
𝑡) cos(2𝜋
1
1
2
2
𝑓1 +𝑓2 2
𝑡)
𝐸𝑡𝑟𝑎𝑛𝑠 = 𝜇𝜔2 𝐴2 𝜆 | 𝐸𝑙𝑜𝑛𝑔 = 𝐿2 𝜌𝜔2 𝐴2 𝑣
of the circular orbit or half the long dimension of an elliptical orbit. 𝐺𝑀𝐸
The ith harmonic
Note that in closed air columns, there is no even harmonics (i.e no 2nd or 4th) 𝑓𝑖 = (𝑖)𝑓0 = (𝑖)
Kepler’s third law: 𝑇 2 = (
2𝐿
| 𝜆 = 2𝐿 | L is length of the string
Kepler’s first law: Planets move in ellipses with the sun at one focus
4𝜋2
𝑣
𝑓𝑖 = (𝑖 + 1)𝑓0 = (𝑖 + 1)
𝑃 𝐴
1
1
2
𝑟2
= 𝜌𝜔2 𝐴2 𝑣 | 𝐼 ∝ 𝑣±𝑣𝑜
𝑓 ′ = 𝑓(
𝑣±𝑣𝑠
| Β = 10 log10 (
𝐼
10−12
)
) -vo when moving away from source, +vo moving towards
+vs when source moves away from observer, -vs moving towards observer By: Kareem Halabi
