Wiley Plus. Final Assignment (5) Is Due Today: Before 11 pm!

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1 Wiley Plus Final Assignment (5) Is Due Today: Before 11 pm!

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3 Final Exam Review December 9, 009 3

4 What about vector subtraction? Suppose you are given the vector relation A B C RULE: The resultant vector head Always points toward the head of the vector that is being subtracted from (in this case ). A B A C This can be turned into vector Addition, by writing: A C B B A C December 9, 009 4

5 v Speed, Velocity and Acceleration In Two Dimensions Position vectors r, r 0 at t, t 0 Displacement r r r 0 r Average velocity t t o r v Instantaneous Velocity lim 0 t 0 t v v o Average acceleration t t o v Instantaneous Acceleration lim t 0 t There is an acceleration whenever there is a change of speed or direction December 9, 009 5

6 An airplane, initially flying at 150 m/s due east, makes a gradual turn at constant speed towards the south. After 15 seconds, the plane is pointing due south. What is its average acceleration? December 9, 009 6

7 December 9, t a v v x x x t a t v x x x x t v v x x x x x x o x v v x x a Equations of Kinematics in Two Dimensions 1) ) 3) 4) t a v v y y y t a t v y y y y t v v y y y y y y o y v v y y a Same as before, only with subscripts for each direction of motion x y

8 Dec. 8, 006 Final Exam #5 A cannonball is aimed 30 degrees above the horizontal and is fired with an initial speed of 15 m/s at ground level. How far away from the cannon will the cannonball hit the ground? December 9, 009 8

9 Newton s Laws of Motion 1. Velocity is constant if a zero net force acts. Acceleration is proportional to the net force and inversely proportional to mass: a 0 if F Force and acceleration are in the same direction 0 F a so F ma m 3. Action and reaction forces are equal in magnitude and opposite in direction

10 Newton s First Law (law of inertia) Object of mass m Acceleration a 0 if F 0 Velocity v constant The velocity is constant if a zero net force acts on the mass. That is, if a number of forces act on the mass and their vector sum is zero: F net F1 F... 0 So the acceleration is zero and the mass remains at rest or has constant velocity.

11 Newton s Law of Gravitation Newton s law of universal gravitation can be applied to extended (finite size) objects such as planets. The prove requires the use of calculus (not done here). The general equation: Gm1m F g r works as long as the mass of the object is distributed around the center with spherical symmetry. In this case r is the distance between the centers of the spheres (not the surfaces).

12 Why is g=9.80 m/s? m w R Earth M E R kg m M E What is the gravitational force (weight, w) of a mass m on the earth s surface? GmM w R E ( Nm / kg 4 m ( kg ) ) 6 ( m) GM w m mg g R E At the earth s surface Above earth s surface, weight decreases with distance r from the centre of the earth as 1/r.

13 Problem 4.6 (6 th edition): The weight of an object is the same on planets A and B. The mass of planet A is 60% that of planet B. Find the ratio of the radii of the planets. m w M r

14 Static friction occurs when there is no sliding or skidding: object at rest Static Friction car moving without skidding or spinning wheels where the tire meets the road, the tire is momentarily at rest, so the friction is static Kinetic Friction There is sliding or skidding. usually less than the static friction force it is hard to get an object sliding, and easier to keep it sliding.

15 Static Friction F s N normal The static friction force f s opposes applied forces and increases as the applied force F is increased. f s rises only to a maximum value f s, x sfn, y and then the block begins to slide. coefficient force of static friction

16 Problem 4.38 (edition 6): A cup of coffee sits on a table in a plane ( s = 0.30 ). The plane accelerates. What is the maximum acceleration before the cup starts to slide? F N mg F N c a s f mg a p

17 Tension Force Tension the force within a rope or cable that is used to pull an object. A force T applied to the end of the rope is transferred to other end of the rope where it exerts the same force on the block. The block exerts an equal and opposite force on the rope (Newton s 3rd law).

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19 Problem 4.6: Can the person who is pulling the rope ever make the rope perfectly horizontal? To support the weight: mg T sin

20 Dec. 17, 005 Final Exam # A spring scale is loosely fastened to the ceiling of a railway car. When a 1.0-kg block is hung from the scale, it reads 1 Newton and is oriented as shown in the figure. What is the approximate acceleration of the car, as measured by an observer at rest on the ground, outside the car? December 9, 009 0

21 Dynamics of Uniform Circular Motion Period of circular motion: T = r/v Centripetal acceleration: a c = v /r Centripetal force: F c = ma c = mv /r For motion in a horizontal circle, equilibrium in the vertical direction, vertical forces cancel use Newton s second law to relate net horizontal force to the centripetal acceleration

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23 v1 F N 1 mg m r v m r F N v3 F N 3 mg m r v4 m r F N 4 Vertical Circular Motion 1) force toward center weight larger than mg ) force toward center 3) force toward center rider falls off if 4) force toward center v3 mg m r The rider falls off if v 3 F N 3 0 rg

24 Problem 5.40: A motorcycle is traveling up one side of a hill and down the other side. The crest is a circular arc with a radius of 45 m. Determine the maximum speed that the motorcycle can have while moving over the crest without losing contact with the road. The net downward force on the bike at the crest of the hill allows the motorbike to remain in contact with the ground. Then F N > 0. That is: net downward force = centripetal force, mv /r.

25 Conservation of Mechanical Energy In the absence of applied forces and friction: The work done by a net zero applied force is zero So, 0 KE PE And KE + PE = E = total energy is constant Total Energy is Conserved in a Closed System Other kinds of potential energy: elastic (stretched spring) electrostatic (charge moving in an electric field)

26 Work done in lifting an object Alternative view: define a different form of energy- Gravitational potential energy: Define: Mechanical energy = kinetic energy + potential energy Mechanical energy: Then: PE mgy E 1 mv mgy Work done by applied force, F, is (change in KE) + (change in PE) So, W Fh KE PE

27 Problem 6.40 A particle, starting from point A in the drawing, is projected down the curved runway. Upon leaving the runway at point B, the particle is traveling straight upward and reaches a height of 4.00 m above the floor before falling back down. Ignoring friction and air resistance, find the speed of the particle at point A.

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29 P 6.8 An extreme skier, starting from rest, coasts down a mountain that makes an angle of 5 with the horizontal. The coefficient of kinetic friction between her skis and the snow is 0.. She coasts for a distance of 15.0 m before coming to the edge of a cliff. Without slowing down, she skis off the cliff and lands downhill at a point whose vertical distance is 3 m below the edge. How fast is she going just before she lands? December 9, 009 9

30 December 9,

31 Oct. 5, 006 Midterm #8 A 50 kg girl runs up a flight of stairs in a time of 1.5 sec. The stairs are 7.0 meters long and make an angle of 7 degrees above the horizontal. What is the average power that she produced? December 9,

32 Conservation of Momentum Two isolated masses collide. The initial total momentum is: p With p 1 p p1 m1 v p m v 01 0 While the masses are in contact, they exert equal and opposite forces on each other (Newton s third law). F 1 F 1 So the impulse acting on m 1 is equal in magnitude and opposite in direction to the impulse acting on m

33 Therefore, (change in momentum = impulse) After the collision: So, the total momentum after the collision is: That is, total momentum is conserved: p 1 p p p p p v p p p v p f f1 m m p p p p p p p p p p p p p p

34 f f v m mv mv Momentum : Conserved So, Then use conservation of energy Combine the two equations and after some algebra: m mv mv v f f f v f m mv v m m m m v v m m m m v v f f v v v then m m if f f,,

35 P 7.34 The drawing shows a collision between two pucks on an air-hockey table. Puck A has a mass of kg and is moving along the x axis with a velocity of m/s. It makes a collision with puck B, which has a mass of kg and is initially at rest. The collision is not head-on. After the collision, the two pucks fly apart with the angles shown in the drawing. Find the speed of (a) puck A and (b) puck B. December 9,

36 December 9,

37 Correspondence between Linear and Angular Motion Angular Linear o t o ot 1 t x v a v v x x o o at v t o 1 at 1 o o t x x o 1 v v o t o o x x v v a o o In addition we now also have: l r and v r

38 Example: The wheels of a bicycle have an angular velocity of 0 rad/s. The brakes are applied, bringing the bicycle to a uniform stop. During braking, the angular displacement of the wheels is 15.9 revolutions. a) How much time does it take to stop? b) What is the angular acceleration of the wheels?

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41 P 9.68 A platform is rotating at an angular speed of. rad/s. A block is resting on this platform at a distance of 0.30 m from the axis. The coefficient of static friction between the block and the platform is Without any external torque acting on the system, the block is moved toward the axis. Ignore the moment of inertia of the platform and determine the smallest distance from the axis at which the block can be relocated and still remain in place as the platform rotates. December 9,

42 December 9, 009 4

43 A How-to Approach To Torque Problems 1. Identify your point of rotation (P) (usually either given or obvious) Not so obvious for free falling bodies center of gravity. Identify the point at which a force is applied and establish the distance (d) to the point of rotation 3. Identify a line (the lever arm) from the point of rotation to the line of action of the force, that is perpendicular to the line of action 4. Identify the angle (θ) between the line (d) and the force direction 5. Calculate the length of the lever arm from (d) and (θ): l = d sin (θ) 6. Calculate the torque: T = F l = F d sin (θ)

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50 P 10. An object attached to a horizontal spring is oscillating back and forth along a frictionless surface. The maximum speed of the object is 1.48 m/s, and its maximum acceleration is 7.13 m/s. How much time elapses betwen an instant when the object's speed is at a maximum and the next instant when its acceleration is at a maximum? December 9,

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56 Problem The main water line enters a house on the first floor. The line has a gauge pressure of Pa. (a) A faucet on the second floor, 6.50 m above the first floor, is turned off. What is the gauge pressure at this faucet? (b) How high could a faucet be before no water would flow from it, even if the faucet were open?

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58 Problem (a) The mass and the radius of the sun are kg and m. What is its density? (b) If a solid object is made from a material that has the same density as the sun, would it sink or float in water? Why? (c) Would a solid object sink or float in water if it were made from a material whose density was the same as that of the planet Saturn (mass = kg, radius = m)? Provide a reason for your answer.

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62 P A uniform rectangular plate is hanging vertically downward from a hinge that passes along its left edge. By blowing air at 11.0 m/s over the top of the plate only, it is possible to keep the plate in a horizontal position, as illustrated in part a of the drawing. To what value should the air speed be reduced so that the plate is kept at a 30.0 angle with respect to the vertical, as in part b of the drawing? December 9, 009 6

63 December 9,

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65 Thermal Expansion Linear expansion the increase in length, width or thickness when an object is heated. L L0T = coefficient of linear expansion Typical values for metals per C o.

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70 That s all! Make sure you: 1) Understand the concepts presented in this review and know how to apply them. ) Work some problems that are similar to those worked in this review 3) Understand that it is still your responsibility to review and understand all of the material presented in this course! Good Luck

PHYSICS 221, FALL 2011 EXAM #2 SOLUTIONS WEDNESDAY, NOVEMBER 2, 2011

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w = mg F g = τ = Frsinθ a c v f + at = v i F = ma F f p = mv W = Fd v f 2 = v i 2 + 2ad F t = Δ(mv) Impulse = F t KE = 1 2 mv 2 PE = mgh PE = 1 2 kx 2

w = mg F g = τ = Frsinθ a c v f + at = v i F = ma F f p = mv W = Fd v f 2 = v i 2 + 2ad F t = Δ(mv) Impulse = F t KE = 1 2 mv 2 PE = mgh PE = 1 2 kx 2 Wissahickon Physics Midterm x = x i + vt + 1 2 at 2 v f 2 = v i 2 + 2ad Name v f = v i + at Gm 1 m 2 w = mg F g = r 2 g = 9.8m/s 2 G = 6.67 10 11 (N m 2 /k g 2 ) F = ma F f = µn τ = Frsinθ a c = v 2 p

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