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# Final Exam Notes 80218 - PHYS 1220 - 001

Clemson

GPA 3.2

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This 9 page Study Guide was uploaded by Jacob Edwards on Saturday April 23, 2016. The Study Guide belongs to 80218 - PHYS 1220 - 001 at Clemson University taught by Lih-sin The in Winter 2016. Since its upload, it has received 96 views. For similar materials see Physics with Calculus I in Physics 2 at Clemson University.

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Date Created: 04/23/16

Final Exam Study Guide Rules for dimensional analysis 1. Dimensional analysis when the dimensions of a quantity instead of the value are used to solve a problem; this is often used to check a problem 2. Dimensions can be treated as algebraic symbols for example: a. V=l*w*h each is a unit of length so volume is equal to L 3 3. The quantities can only be added or subtracted if they have the same dimensions 4. Each side of the equation must have the same dimensions 5. Trigonometric functions only apply to dimensionless quantities 6. Logarithms and exponential functions only apply to dimensionless quantities This is a good way to check your answer or work to an an2wer if you don’t have a clue what to do. Know what units compose a N [kgm/s ], J [Nm], W [J/s]. Rules for significant figures 1. When multiplying or dividing the answer will have the same amount of significant figures as the number in the equation with the least amount of significant figures a. (12.5)*(2.0)=25 2. When adding or subtracting the answer has the same amount of decimal places as the number with the least amount of decimal places in the problem 3. Numbers such as e and π don’t place restrictions on results 4. Scientific notation can be used to help avoid confusion with significant figures 5. Keep extra significant figures when solving the equation and only bring to the proper amount after the equation has been solved 6. When the answer begins with a 1 it is okay to keep an extra significant figure Chances are this won’t be a huge factor in the final but it’s still a handy thing to know. Five Kinematic Equations Five Rotational Equations These are very similar to the translational ones except d=θ, v=ω, and a=α. A very useful formula to use to work between the two formulas is: v=ωr Newton’s Laws 1. “Every body continues in its state of rest or of uniform motion in a straight line, unless it is compelled to change that state by forces impressed on it.” 2. F=ma 3. F[Bon A] [AonB]Equal and opposite reaction. Two and Three Dimensional Motion Position: r=xi+yj+zk ∆ r=∆ xi+∆ yj+∆zk Displacement: ∆r Average Velocity: v= ∆ t v= dr Instantaneous Velocity: dt ∆ v dv Acceleration: a= Or a= ∆t dt v (¿¿0tcos(θ))i+(v tsin θ − gt ) j Projectile motion: 0 2 r=¿ 2 v0 Range: Rmax= g Forces Drawing a picture is especially useful when given a problem that involves multiple forces. This will add a visual that will help with understanding the math you’ll be doing. Frictional Force Static friction F =μ F smax s N o μs - The coefficient of friction (dependent on what the surface is) o FN - Normal Force Kinetic Friction F kμ Fk N o μk -coefficient of kinetic friction (unitless) Acceleration FT−F k a x m o FT- The force of tension Rolling friction F =μ F r r N o μr -coefficient of rolling friction Drag force F =−bv D o b- the experimental constant. It is negative to show it’s moving opposite velocity (v). For the drag force on blunt objects (What we mostly will work with) drag force is shown by: 1 2 F D CρAv 2 o C- The drag coefficient o ρ- The density of the medium A- The object’s perpendicular cross-sectional area o v- The speed of the objects Centripetal Force The magnitude of centripetal force is found by: 2 F =m v C r If the origin is in a polar coordinate system v2 F C−m ^ r Keplers Laws Planetary Motion r +r a= p A 2 1. r p The perihelion distance 2. r A The aphelion distance 3. This answer comes in Astronomical Units or AU 4. 1 AU=1.5E11 meters o Were only going to consider the first and the third since they’re the ones formulas apply to. o In Kepler’s third law it says that the square of a planet’s period T is proportional to the cube of its semi-major axis T[yr ] [AU ] o o Newton’s law of universal gravity shows: m1m 2 o F GG 2 r o G- The gravitational constant= 6.673E-11 o The gravitational field of any source can be found from: Unitvector pointingaway −GM o g r)= (¿particl) r2 o G-Gravitational constant o M- Mass o r- Radius Energy o Kinetic Energy K= mv 2 o 2 o This answer is in Joules o Potential Energy o PE=mgh o To find gravitational potential energy: m1m 2 o U G( )G r o Elastic potential energy: U e k x 2 o 2 o Or 1 2 o U e k y 2 o K= the spring constant o The sum of a systems Potential and Kinetic energy is its mechanical energy. This is shown by: E=K+U o o The conservation of mechanical energy is shown by: ∆ K+∆U=0 o Momentum o Momentum moves in the same direction as velocity and is given by the equation: p=mv o The initial momentum is always the same as the final momentum. In the problems that ask for recoil motion use the following formula often times in momentum problems they are paired with translational kinematic equations: m1v1=m 2 2 Rocket An open systems is a system that gains or loses mass. This is the formula to us for the rocket or a similar system: ∆ M Fthrust ) ∆t Torque and Inertia Torque can also be found by taking the cross product of two vectors. The magnitude of Torque is given by: R=ABsinφ Inertia is the torque divided by the angular acceleration. In translational motion the more massive particle has the most rotational inertia Mass distribution with respect to the rotational axis also impacts rotational inertia; the farther away the mass is from the rotational axis the more rotational inertia will exist. Rotational inertia can be found by: n 2 I=i=1m ii M is the mass of the particle r is the distance from the rotational axis On a continuous object the rotational inertia is: I= r dm ∫ The parallel axis theorem where M is mass, h is the perpendicular distance between the new axis and the axis through the center of mass and I is the rotational inertia around the center of mass CM 2 I=I CM+M h The center of mass is found by: m 2 xCM =( )x2 m 1m 2 For center of mass multiply the associated mass with the associated x coordinate. Kinetic Energy for rotating object: K = I ω 2 r 2 Conservation of energy holds true for rotational motion similar to translational motion K iK +ri+WiK +K +U f∆E rf f th K is the translational kinetic i K riis rotational kinetic Newtons second law is shown through torque and is the sum of all torque. If the system is conserved torque initial is equal to torque final Equilibrium Two conditions need to be met for an object to be at equilibrium: dp 1. Ftot dt Ftot dL 2. τtotdt τtot These are all that is needed to solve and equilibrium problem. To determine the radius for the torque always set a reference point for the initial radius. I think this is easier if an endpoint is chosen. Cross Product You may need to know this for torque. Cross Product: To do cross product the simplest way is to use matrices. If r=1i +1j and F=2i+3j the r x F= i j k 1 0 1 0 1 1 1 1 0 =i 3 0 − j2 0 +k2 3 2 3 0 The discriminate of each matrix is found. (ac-bd) So the answer to this problem is k. Fluid Mechanics Density ρ=(mass/volume) Force Pressue= Area The density of water is 1000 kg/m Most problems can be solved using the following two equations in some fashion: A1v1=A 2 2 And Bernoulli’s Equation: 1 2 1 2 P 1 ρ2 +ρg1y =P +1ρv +2gy2 2 2 ρ= density of fluid y= height Think back to conservation of energy to use this one. It’s the same thing. This formula can also be used for flowing air.

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