Download SI Exam 2 Review

Physics 221 SI Exam 2 Review
15) Conservation of Mechanical Energy
1
a) Work: Wnet = ΔK = ⃑Fnet . Δr = mv 2 − 1/2mv02 work positive if increase KE negative
2
if a decrease
i) Another way to look at work is considering the direction of the Force and change in
displacement, if same direction + if opposite b) Basic concept total mechanical energy ETotal = K total + Utotal or Ef = Ei
c) Potential Energy
a. Equations for these types of problems
Gravitionional Potential Energy = U = mgh or = mgy
i. Described by a change in height. Potential energy and does not depend on the
path that the object takes, only whether it has a positive or negative change in
height
Elastic(Spring)Potential Energy = U =
1 2
kx
2
ii. Typically described by a changed in elongation or compression of a spring
In General for a force in the x direction : U = − ∫ F dx and therefore
dU
− dx = F so in a graph of potential energy the Force can be found by the
derivative of the curve
16) Non-conservative Work
a) Work done by an external force that removes energy from the system for our purposes this is
friction
Described as A loss or change in total energy through a process times the distance is it
applied over
∆𝐸 = ∆𝐾𝐸 + ∆𝑈 = 𝑊𝑛𝑜𝑛−𝑐𝑜𝑛𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑣𝑒 or 𝑊𝑛𝑜𝑛−𝑐𝑜𝑛𝑠𝑒𝑟𝑣𝑎𝑡𝑖𝑣𝑒 = 𝐹𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛 ∆𝑥 Force
If only conservative forces involved ∆𝐸 = ∆𝐾𝐸 + ∆𝑈 = 0 aka energy is conserved
17) Energy Diagrams
a) As displayed in the problem from earlier find if the particle is in equilibrium by determining if
𝑑𝑈
Force is 0 by - 𝑑𝑥 = 𝐹 = 0
𝑑2 𝑈
b) Then is it stable or unstable equilibrium (concave up or concave down) - 𝑑𝑥 2 = +𝑜𝑟 − if positive
(concave up) it is stable, if negative (concave down) its unstable
c) When looking at an Energy diagram it can sometimes be analyzed as if it were a ball rolling up
and down hills. Where the initial position helps to determine total mechanical energy of the ball
and it can only travel through the hills if it has enough energy to overcome the peak of a hill.
18) & Lecture 19 Linear Momentum and collisions
a) Linear Momentum 𝑝 = 𝑚𝑣 , momentum conserved if 𝑝𝑖 = 𝑝𝑓 𝑜𝑟 ∑ 𝑝𝑖 = ∑ 𝑝𝑓 if the system
contains multiple objects
b) A change in momentum is described as Impulse created by a net external force
𝐽 = ∆𝑝 = ∫ 𝐹𝑒𝑥𝑡 𝑑𝑡
c) Collisions
i) Perfectly Elastic collision- Kinetic energy is conserved
ii) Inelastic Collision-Kinetic Energy decreases (when objects deform in collision)
iii) “Perfectly Inelastic Collisions- Kinetic Energy decreases but can be identified by combining
Kinetic NRG and momentum equations (Objects stick together after collision)
(1) 𝑣 =
𝑚𝑎 𝑣𝑎 +𝑚𝑏 𝑣𝑏
𝑚𝑎 +𝑚𝑏
final velocity is the same for both objects after collision
iv) Explosions- Kinetic energy increases, Bodies break into parts, explosion mechanism
provides more Kinetic energy
v) Super-elastic collision- KE increases, some internal energy is transformed into KE due to
collision (Nuclear Fission, Spring release)
d) For elastic collisions another useful equation to allow to solve problems is the relative velocity
equation
𝑣𝑖1 − 𝑣𝑖2 = −(𝑣𝑓1 − 𝑣𝑓2 )
20. Center of Mass
∑ 𝑚𝑟
a. the equation for the center of mass of a system is 𝑟𝑐𝑚 = ∑𝑖 𝑚𝑖𝑟𝑖
𝑖
𝑖 𝑖
b. which can help us coincidentally find the acceleration and velocity of center of mass by
∑ 𝑚𝑣
∑ 𝑚𝑎
taking the derivative with respect to time obtaining 𝑣𝑐𝑚 = ∑𝑖 𝑚𝑖𝑣𝑖and 𝑎𝑐𝑚 = ∑𝑖 𝑚𝑖𝑎𝑖
𝑖
𝑖 𝑖
𝑖
𝑖 𝑖
c. Then multiplying the velocity equation by the sum of the masses can help us describe
how the momentum of the center of mass can be found
𝑝𝑡𝑜𝑡𝑎𝑙 = ∑ 𝑝𝑖 = ∑ 𝑚𝑖 𝑣𝑖 = 𝑚𝑡𝑜𝑡𝑎𝑙 𝑣𝑐𝑚
𝑖
𝑖
b. It should also be noted that for “tipping” problems, when the center of mass of the system
crosses over the main supporting point or fulcrum, that is the point at which an object
will tip
Center of mass example run through….
Rotation of Rigid Bodies
21. & Lecture 22 Moment of Inertia & Rigid Body Motion
a. Moment of inertia can be described as[ 𝐼 = ∑𝑛𝑖 𝑚𝑖 𝑟𝑖2]or [𝐼 = 𝐼𝑐𝑚 + 𝑀𝑑 2 ] if a
body is rotating about the same axis, a certain distance parallel to the axis of
rotation, given some standard moments about the center of mass of objects. You
will need to use those given moments of inertia on your equations page often.
b. Another addition to conservation of energy is rotational kinetic energy 𝐾𝑟𝑜𝑡 =
1
2
𝐼𝜔2 . Now objects moving down an incline with rotation must store energy in
rotation and linear kinetic energy in addition to gravitational potential energy.
Therefor the equation for Kinetic energy must include a rotational component if
applicable(Most common problem type I’ve seen)
c. Torque often an important concept to look at when analyzing problems is 𝑇 = 𝑟 ×
𝐹 or 𝑇𝑛𝑒𝑡 = 𝐼𝛼 in order to start rotation a net torque must be present
d. A net torque can provide work by 𝑊 = ∫ 𝑇𝑑𝜃 or a Torque multiplied by a Δθ arc
it is applied through
e. This section also brings back equations for circular motion from Exam 1.
Understand those on the back of the first page of your equations packet.
24. Angular Momentum
a. Equations related to
i. 𝐿 = 𝑟 × 𝑝
ii. 𝐿 = 𝐼𝜔
iii. 𝑇 =
𝑑𝐿
𝑑𝑡
iv. When net torque is zero angular momentum is conserved and 𝐿𝑖 = 𝐿𝑓
25. Statics
a. Basic equations to know in Static Situations
i. ∑ 𝑀 = 0 𝑜𝑟 ∑ 𝜏 = 0 Moments or torques can be taken about multiple
points in a problem for simplification. Remember if a force falls on the
point of rotation it provides no moment about that point (hence
eliminating that often unknown force). This can help to solve problems
where more than 3 variables are unknown. Choose your rotation points to
make solving simple for yourself.
ii. ∑ 𝐹𝑥 = 0
iii. ∑ 𝐹𝑦 = 0
26. Stress Strain & Elasticity
𝑁
a. Pressure is described as a force per unit area the SI unit is 𝑃𝑎 = 𝑃𝑎𝑠𝑐𝑎𝑙 𝑜𝑟 𝑚2
i. 𝑃 =
𝐹⊥
𝐴
, 𝑃𝑎𝑡𝑚 = 1.01.∗ 105 𝑃𝑎
ii. Pressure at a depth 𝑃𝑎𝑏𝑠 = 𝑃𝑎𝑡𝑚 + 𝜌𝑔ℎ
iii. 𝑃𝑔𝑎𝑢𝑔𝑒 = 𝜌𝑔ℎ at a depth
b. Stress has the same units as pressure and is often calculated in the same way
𝐹
i. 3 types of stress: shear, tension, compression always a 𝜎 = 𝐴
c. Strain basically tells us a ratio of change in length to the original
i. 𝜀 =
Δ𝐿
𝐿0
=
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑙𝑒𝑛𝑔𝑡ℎ
𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ
d. Young’s Modulus 𝐸 =
𝜎
𝜀
linear strain
𝑆𝑡𝑟𝑒𝑠𝑠
= 𝑆𝑡𝑟𝑎𝑖𝑛 is a property of materials and can be used to
solve for stress and strain (will be given to you in this case if it is not on your
equations page)
27. Gravitation
a. Equations pretty much covered on equations page they don’t really need to be
relisted studying the power point slides is the best way to gain a better
understanding
b. Problems
28. Fluids
a. Equations
Book problems