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1 Physics 7B - AB Lecture 10 June 5 Overview Practice Final Problems Good news everyone! In four days you will be sitting physics exam. Oooh yes.....

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2 Quiz 4 Re-evaluation Request Due TODAY Quiz 5 & 6 Due June 9 at the time of Final Quiz 6 Rubrics on the website Review session starts TODAY. Schedule on the course website

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3 7B Final June 9 Mon 1- 3pm Review session starts TODAY.

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4 Final location Haring Hall, Rm. 1227 A through D Haring Hall, Rm. 2205 E through Q Surge 3 Bldg., Rm. 1309 R through Z Separated by family name:

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5 Final checklist Pens and pencils Calculator We will not have spare calculators, make sure you bring yours Photo ID (Student or Government ID) Without it you can’t sit the final, and you will fail the course. The pages will be separated -- write your name on every single page when you first get your final Formulas will be provided with the final

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6 Momentum (PF6, Q5, CDQ5, CDQ7) Forces: how to change momentum (PF6&7, Q4) Angular momentum (Q5,CDQ7) Torque: how to change ang mom (PF8,Q5,CDQ6) Simple Harmonic Motion: A specific type of net force Fluid (PF1&2, Q1, CDQ1) Circuits (PF4, Q2, CDQ2) Physics breaks into two separate “parts” in 7B: Then we have learned four techniques Vectors (PF5, Q3, Q4) Vectors (PF5, Q3, Q4) Components, use of trigonometry (PF5, Q5) Components, use of trigonometry (PF5, Q5) Force diagrams (PF5&7, Q6), Force diagrams (PF5&7, Q6), Extended force diagrams (PF8, Q6,CDQ6) Extended force diagrams (PF8, Q6,CDQ6) Linear/Angular Momentum Charts (Q4, CDQ5) Linear/Angular Momentum Charts (Q4, CDQ5) Energy density model No net change in energy density/around a circuit Exponential decay (PF3, Q3,CDQ4) Exponential decay (PF3, Q3,CDQ4) Osmosis (Q3) Osmosis (Q3) Forces and its relation to change in motion

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7 Today’s lecture Overview of the material, using practice final problems as examples These review notes are supposed to go over the course, but as you have seen everything before, pieces of the course can be “mixed up” -- this is good practice for the final. These notes cover a lot of the class, but not all. For example, practice final does not have any problem on diffusion. See CD Quiz 3 for an example. Diffusion occurs when there is concentration gradient of a specie of particles. Wait a while… Permeable, semipermeable membrane What flows can be particels or water depending on the membrane property

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8 Forces, Force diagram, Vectors, Components Practice Final 4 Students (m =100kg) in hammocks = 45 = 25 Hammmock + Students + Strings = single object What are the contact/non contact forces exerted on the system?

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9 Identifying forces There are contact forces and non contact forces The only non contact force we worry in 7B is gravitational pull of the Earth exerted on all objects, i.e. F Earth on ball Contact force can be exerted by anything that is in contact with your object, i.e. F string on ball F Earth on ball F y, string on ball = 100N F x, string on ball = m|a| = 10kg(1.5m/s) = 15N F string on ball

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10 Forces, Force diagram, Vectors, Components Practice Final 4 Students (m =100kg) in hammocks = 45 = 25 Hammmock + Students + Strings = single object What are the contact/non contact forces exerted on the system?

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11 Forces, Force diagram, Vectors, Components Practice Final 4 Students (m =100kg) in hammocks F Earth on hammock = 1000N F Post on hammock

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12 Forces, Force diagram, Vectors, Components Practice Final 4 Students (m =100kg) in hammocks F Earth on hammock = 1000N F Post on hammock

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13 Forces, Force diagram, Vectors, Components Practice Final 4 Students (m =100kg) in hammocks = 45 = 25 A static problem, i.e., torques as well as forces are all balanced (another way of saying this is, net torque is zero & net force is zero) They have to be balanced compnents wise.

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14 Torque, Extended force diagram Practice Final 7 Deltoid muscle is attached to the man’s arm and pulls on his arm in this direction Attached at 15cm from the shoulder joint Center of Mass of the arm at 30cm from the shoulder joint 40N = F Earth on arm Another static problem This one is harder.

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15 Torque, Extended force diagram Practice Final 7 Deltoid muscle is attached to the man’s arm and pulls on his arm in this direction Attached at 15cm from the shoulder joint Center of Mass of the arm at 30cm from the shoulder joint 40N = F Earth on arm A static problem, i.e., torques as well as forces are all balanced (another way of saying this is, net torque is zero & net force is zero) Another static problem

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16 Torque, Extended force diagram Practice Final 7 Deltoid muscle is attached to the man’s arm and pulls on his arm in this direction Attached at 15cm from the shoulder joint Center of Mass of the arm at 30cm from the shoulder joint 40N = F Earth on arm (a)What’s tangential component of the force of Deltoid muscle on the arm? (b)Draw extended force diagram

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17 Torque, Extended force diagram Practice Final 7 40N = F Earth on arm 80N = F tangential F Muscle on arm Note: An arrow is a vector, a dotted arrow is a component of a vector

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18 Torque, Extended force diagram Practice Final 7 40N = F Earth on arm 80N = F tangential F Muscle on arm F shoulder joint on arm Note: An arrow is a vector, a dotted arrow is a component of a vector

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19 Torque, Extended force diagram Practice Final 7 40N = F Earth on arm 80N = F tangential F Muscle on arm F shoulder joint on arm Now both torque and forces are balanced! Note: An arrow is a vector, a dotted arrow is a component of a vector

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20 Torque, Extended force diagram Practice Final 7 40N = F Earth on arm Muscle goes limp and the arm starts to swing (Now torque is not balanced! ) What is its after 0.1s?

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21 Torque, Extended force diagram Practice Final 7 40N = F Earth on arm Muscle goes limp and the arm starts to swing (Now torque is not balanced! ) What is its after 0.1s? | | ∆ t = |∆L| = (0.3m)(40N)(0.1sec) = 1.2Nms Lf = I = 1.2Nms So then if we knew I arm, we can figure out !

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22 We know how to find each force Every force in the problem also contributes a torque. (This torque may turn out to be zero) The magnitude of the torque is on obj = (F tangential ) r where r is the distance between where the force is applied and the pivot point F Earth on pentagon Pivot Recipes for torque

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23 F Earth on pentagon Pivot r Magnitude of torque is r F Earth on obj, perp Direction is into the screen (RHR) Do this for each force, then add all the torques up to find the net torque. (Some forces are easy: applied either at or through the pivot)

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24 Forces, Force diagram, Acceleratin/Velocity Practice Final 7 Cushion Compact shaped person jumping off a window of a burning building He falls freely for 1.5 sec, then the cushion exerts a constant force to bring the person to rest in 0.3sec Hint : Assume |v Person | = 14.7m/s right before he hits the cushion Position y (m)Velocity v (m/s)Acceleration a (m/s 2 )

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25 Forces, Force diagram, Acceleration/Velocity Practice Final 7 Cushion Compact shaped person jumping off a window of a burning building (1) Force diagram during the free fall F Earth on person 2) Force diagram while the cushion is bringing the person to rest F Earth on person F Cushion on person

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26 Forces, Force diagram, Acceleration/Velocity Practice Final 7 Cushion Compact shaped person jumping off a window of a burning building Acceleration a (m/s 2 ) (1) Force diagram during the free fall F Earth on person 2) Force diagram while the cushion is bringing the person to rest F Earth on person F Cushion on person Check whether the net force, i.e. F on person, is consistent with a person Remember F on person = m a person !

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27 Forces, Momentum Practice Final 6 m Cadillac = 2000kg 20m/s to Right m Toyota = 1000kg 20m/s to Left Initial Final m Cadillac+Toyota = 3000kg Traveling either to Right or to Left, or remain stationary Stuck together We don’t know the direction/speed of travel after the collision. What we do know is: p C+T initial = p C+T final

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28 Forces, Momentum Practice Final 6 m Cadillac = 2000kg 20m/s to Right m Toyota = 1000kg 20m/s to Left Initial Final m Cadillac+Toyota = 3000kg Stuck together |p C | = m C | v C | = 40000kgm/s | p T | = m T | v T | = 20000kgm/s pCpC pTpT | p C+T initial | = 20000kgm/s p C+T initial Total

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29 Forces, Momentum Practice Final 6 m Cadillac = 2000kg 20m/s to Right m Toyota = 1000kg 20m/s to Left Initial Final m Cadillac+Toyota = 3000kg Stuck together | p C+T initial | = 20000kgm/s p C+T initial Total p C+T final Travelling to Right at 6.66m/s 40000kgm/s 20000kgm/s pCpC pTpT

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30 Forces, Momentum Practice Final 6 m Cadillac = 2000kg 20m/s to Right m Toyota = 1000kg 20m/s to Left Initial Final m Cadillac+Toyota = 3000kg Stuck together Then find ∆p of each car from: ∆p = m ∆v = m (v final –v initial ) p C+T final Travelling to Right at 6.66m/s Pay attention to the direction of vectors when adding/subtracting 40000kgm/s 20000kgm/s pCpC pTpT

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31 Forces, Momentum Practice Final 6 m Cadillac = 2000kg 20m/s to Right m Toyota = 1000kg 20m/s to Left Initial Final m Cadillac+Toyota = 3000kg Stuck together Then find ∆p of each car from: ∆p = m ∆v = m (v final – v initial ) p C+T final Travelling to Right at 6.66m/s Pay attention to the direction of vectors when adding/subtracting ∆p Toyota comes out to be equal to ∆p Cadillac even with its smaller mass because Toyota changes its direction after the collision 40000kgm/s 20000kgm/s pCpC pTpT

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32 Forces, Momentum Practice Final 6 m Cadillac = 2000kg 20m/s to Right m Toyota = 1000kg 20m/s to Left Initial Final m Cadillac+Toyota = 3000kg Stuck together 40000kgm/s 20000kgm/s pCpC pTpT Then find ∆v of each car: p C+T final Travelling to Right at 6.66m/s Pay attention to the direction of vectors when adding/subtracting ∆v Toyota comes out to be greater than ∆v Cadillac because Toyota changes its direction after the collision

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33 Forces, Momentum Practice Final 6 m Cadillac = 2000kg 20m/s to Right m Toyota = 1000kg 20m/s to Left Initial Final m Cadillac+Toyota = 3000kg Stuck together 40000kgm/s 20000kgm/s What about F ave experienced by each car during the collision? p C+T final Travelling to Right at 6.66m/s One approach (Newton’s 1st law): F ave ∆t = ∆p Alternative approach (Newton’s 3rd law): F C on T = – F T on C !!! pCpC pTpT SAME !

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34 Fluids/Circuits Basic rules for looking at fluids/circuits : 1. Energy (density) conservation OR (the same as) 2. Current entering = current leaving 3. Pressures where two fluids systems touch are equal Voltages that are connected by wire (no circuit element in between) are equal ∆V = – IR ∆P + (1/2) ∆(v 2 ) + g∆h = E pump /volume – IR Junction rule A 1 v 1 = A 2 v 2

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35 Fluids, Circuit Practice Final 1 P 1 = 200kPa v 1 = 10m/s w = 1000kg/m 3 v 1 vs v 2 ?? A 2 = 0.5A 1

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36 Fluids, Circuit Practice Final 1 P 1 = 200kPa v 1 = 10m/s w = 1000kg/m 3 A 1 v 1 = A 2 v 2 v 2 = ( A 1 /A 2 ) v 1 = 2 v 1 = 20m/s ! A 2 = 0.5A 1

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37 Fluids, Circuit Practice Final 1 P 1 = 200kPa v 1 = 10m/s w = 1000kg/m 3 Keep raising the end 2, At what h, P 2 is equal to zero? v 2 = 20m/s ∆P + (1/2) ∆(v 2 ) + g∆h = 0

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38 Fluids, Circuit Practice Final 1 P 1 = 200kPa v 1 = 10m/s w = 1000kg/m 3 v 2 = 20m/s ∆P + (1/2) ∆(v 2 ) + g∆h = 0 Substitute P 2 = 0, to find h 2 (say h 1 = 0) h 2 = 5m.

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39 Fluids, Circuit Practice Final 2 Why does a hose have a nozzle at the end?

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40 Fluids, Circuit, Forces Practice Final 2 Why does a hose have a nozzle at the end? A 1 v 1 = A 2 v 2 !! In order to keep the flow rate constant throughout the fluid circuit, fluid velocity will increase at a narrowed nozzle. With greater velocity of the fluid coming out of the nozzle, the water will reach farther, allowing the firefighter to fight the fire far from the fire. v 2 > v 1 v2v2 v1v1

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41 Fluids, Circuit, Forces Practice Final 2 What is the direction of the net force on the small amount of water at three different locations? v 2 > v 1 v2v2 v1v1 Example of a problem that combines concepts from different models.

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42 Fluids, Circuit, Forces Practice Final 2 These are the direction of the net force on the small amount of water at three different locations. Think about how fluid velocity v is changing at each location, as net force is in the same direction as ∆ v. 0

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43 Fluids/Circuits Oops… One more technique to remember : Know how to find equivalent resistance, this is an essential technique for analyzing circuits 32 20 12 24

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44 R eq of the whole circuit = 40 Circuit Practice Final 3 32 20 12 24 V B ?? ∆V 3 = 3V

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45 R eq of the whole circuit = 40 Circuit Practice Final 3 32 20 12 24 V B ?? ∆V 3 = 3V Find I 3 = 0.25A Find ∆V 2 = 5V Find ∆V 3 = 8V Find I 1 = 0.25A Find I = 0.5A Find ∆V 4 = 12V Find ∆V B 20V Finally!

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46 Close the switch… The circuit becomes simpler Circuit Practice Final 3 32 20 12 24 20V What happens to: ∆V 1 I 2 ∆V 3 ∆V 4 ?? Two things that don’t change : V B, resistor values R eq decreases, and so I eq increases

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47 Exponential decay Practice Final 3 Circuit B has a battery having voltage V and the capacitor is uncharged. Circuit A has no battery and the capacitor is charged to a voltage V. Circuit A and B are similar but slightly different RC circuits

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48 Exponential decay Practice Final 3 Capacitor: A capacitor stores electrical energy by accumulating charge on two conducting plates. It can also release the stored energy very quickly. What was Capacitor C again??

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49 Exponential decay Anytime we see change in something is directly proportional to the amount of that something, we have exponential decay (growth) Ex. Microorganisms in a culture dish, a virus of sufficient infectivity, human population, nuclear chain reaction, charge/discharging capacitor The same statement mathematically: Solution : ± Stuff (t) (Amount of stuff at t = 0 ) e ±kt In each time interval of 1/k (time constant), amount of stuff reduces to 1/e of each previous value/ or grows by a factor of e

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50 Exponential decay Practice Final 3 Cgarged to V Uncharged C V c = Ve ±(1/RC)t

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51 Good luck! Rotation this way Direction of

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