Physics 221 SI Exam 2 Review

15) Conservation of Mechanical Energy

a) Work: W net=ΔK=F⃑net . Δ r⃑=12mv2−1/2mv0

2 work positive if increase KE negative if a

decreasei) 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+U total∨Ef=Ei c) Potential Energy

a. Equations for these types of problemsGravitionional Potential Energy=U=mgh∨¿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=12k x2

ii. Typically described by a changed in elongation or compression of a springIn General for a force in the x direction : U=−∫ Fdx and therefore −dUdx

=F so in a graph of potential energy the Force can be found by the

derivative of the curve

16) Non-conservative Worka) 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=W non−conservative or W non−conservative=FFriction∆x Force

If only conservative forces involved ∆ E=∆ KE+∆U=0 aka energy is conserved

17) Energy Diagramsa) As displayed in the problem from earlier find if the particle is in equilibrium by determining if

Force is 0 by -dUdx

=F=0

b) Then is it stable or unstable equilibrium (concave up or concave down) - d2Ud x2 =+¿−¿ 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 collisionsa) Linear Momentum p=mv , momentum conserved if pi=p f∨∑ pi=∑ pf if the system

contains multiple objectsb) A change in momentum is described as Impulse created by a net external force

J=∆ p=∫Fext dtc) 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 vbma+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−v i2=−(v f 1−v f 2)

20. Center of Mass

a. the equation for the center of mass of a system is rcm=∑imiri

∑imiri

b. which can help us coincidentally find the acceleration and velocity of center of mass by

taking the derivative with respect to time obtaining vcm=∑imi v i

∑imi v i

and acm=∑imiai

∑imiai

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=∑ipi=∑

imi v i=mtotal vcm

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=∑i

n

mi ri2]or [I=I cm+M d2] 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

K rot=12I ω2. 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 T net=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 toi. L=r × p

ii. L=Iω

iii. T=dLdt

iv. When net torque is zero angular momentum is conserved and Li=Lf

25. Staticsa. Basic equations to know in Static Situations

i. ∑M=0∨∑ 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. ∑ F x=0 iii. ∑ F y=0

26. Stress Strain & Elasticity

a. Pressure is described as a force per unit area the SI unit is Pa=Pascal∨ Nm2

i. P=F⊥

A , Patm=1.01 .∗105 Pa

ii. Pressure at a depth P|¿|=Patm+ρgh¿

iii. Pgauge=ρghat a depth

b. Stress has the same units as pressure and is often calculated in the same way

i. 3 types of stress: shear, tension, compression always a σ= FA

c. Strain basically tells us a ratio of change in length to the original

i. ε= ΔLL0

= change∈lengthoriginallength linear strain

d. Young’s Modulus E=σε= StressStrain 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. Fluidsa. Equations

Book problems