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Physics 115 2014 Final Review Calculus is about “rates of change”. A TIME RATE is anything divided by time. CHANGE is expressed by using the Greek letter, Delta, D. For example: Average SPEED is simply the “RATE at which DISTANCE changes”. The MEANING? d (kt 3 ) 2 3kt dt For example, if t = 2 seconds, using x(t) = kt3=(1)(2)3= 8 meters. The derivative, however, tell us how our DISPLACEMENT (x) changes as a function of TIME (t). The rate at which Displacement changes is also called VELOCITY. Thus if we use our derivative we can find out how fast the object is traveling at t = 2 second. Since dx/dt = 3kt2=3(1)(2)2= 12 m/s Derivative of a power function Unit Vector Notation iˆ - unit vector 1 in the x direction ˆj - unit vector 1 in the y direction kˆ - unit vector 1 in the z direction The proper terminology is to use the “hat” instead of the arrow. So we have i-hat, j-hat, and k-hat which are used to describe any type of motion in 3D space. How would you write vectors J and K in unit vector notation? J 2iˆ 4 ˆj K 2iˆ 5 ˆj Example A boat moves with a velocity of 15 m/s, N in a river which flows with a velocity of 8.0 m/s, west. Calculate the boat's resultant velocity with respect to due north. Rv 82 152 17 m / s 8.0 m/s, W 15 m/s, N Rv 8 Tan 0.5333 15 Tan1 (0.5333) 28.1 17 m / s @ 28.1 WofN The Final Answer : 17 m / s 118.1 8iˆ m / s 15 ˆj m / s Dot Products in Physics Consider this situation: A force F is applied to a moving object as it transverses over a frictionless surface for a displacement, d. As F is applied to the object it will increase the object's speed! But which part of F really causes the object to increase in speed? It is |F|Cos θ ! Because it is parallel to the displacement d In fact if you apply the dot product, you get (|F|Cos θ)d, which happens to be defined as "WORK" (check your equation sheet!) A B A B cos W F x F x cos Work is a type of energy and energy DOES NOT have a direction, that is why WORK is a scalar or in this case a SCALAR PRODUCT (AKA DOT PRODUCT). Example Suppose a person moves in a straight line from the lockers( at a position x = 1.0 m) toward the physics lab(at a position x = 9.0 m). “To the right” is taken as positive, as shown below The answer is positive so the person must have been traveling horizontally in the positive direction. Example Suppose the person turns around! The answer is negative so the person must have been traveling horizontally in the negative direction What is the DISPLACEMENT for the entire trip? Dx x final xinitial 1.0 1.0 0m What is the total DISTANCE for the entire trip? 8 8 16 m Instantaneous Velocity Instantaneous velocity is a measure of an object’s displacement per unit time at a particular point in time. Example: A body’s position is defined as: 4ˆ ˆ d (7t i j ) dx t v dt dt v lim Dt 0 Dx Dt dx v dt x(t ) 7t 3iˆ 4ˆ j , v(t ) ? t 3 4 ˆ ˆ v(t ) [21t i 2 j ] m / s t 2 Instantaneous Acceleration Instantaneous velocity is a measure of an object’s velocity per unit time at a particular point in time. If the velocity of an object is defined as: v(t ) [21t 2iˆ 4 ˆ j ] m / s, a(t ) ? 2 t 4 ˆ ˆ d (21t i 2 j ) dv t a dt dt 2 a lim Dt 0 dv a dt 8 [ 42tiˆ 3 ˆj ] m / s / s t Dv Dt What do the “signs”( + or -) mean? Quantity Positive Negative Displacement Your position has changed toward the positive. Your position has changed toward the negative. Velocity You are traveling in the +x or +y direction. You are traveling in the –x or –y direction. Acceleration If moving in the positive direction, you are speeding up. If moving in the negative direction, you are slowing down. If moving in the positive direction, you are slowing down. . If moving in the negative direction, you are speeding up. The 3 Kinematic equations There are 3 major kinematic v v at o equations than can be used to describe the 2 1 motion in DETAIL. All are x xo vo t 2 at used when the 2 2 acceleration is CONSTANT. v v 2a( x x ) o o Kinematics for the VERTICAL Direction All 3 kinematics can be used to analyze one dimensional motion in either the X direction OR the y direction (assuming UP is the +-direction.) v = vo + at ® vy = voy - gt 2 2 1 1 x = xo + vox t + at ® y = yo + voy t - gt 2 2 2 2 2 2 v = vox + 2a(x - xo ) ® vy = voy - 2g(y - yo ) Examples A stone is dropped at rest from the top of a cliff. It is observed to hit the ground 5.78 s later. How high is the cliff? What do I know? v = 0 m/s oy g = 9.8 m/s2 yo=0 m t = 5.78 s What do I want? y=? Which variable is NOT given and NOT asked for? Final Velocity! y = yo + voyt - 1 gt 2 2 y (0)(5.78) 4.9(5.78) 2 y -163.7 m H =163.7m Examples A pitcher throws a fastball with a velocity of 43.5 m/s. It is determined that during the windup and delivery the ball covers a displacement of 2.5 meters. This is from the point behind the body when the ball is at rest to the point of release. Calculate the acceleration during his throwing motion. What do I know? vo= 0 m/s x = 2.5 m v = 43.5 m/s What do I want? a=? Which variable is NOT given and NOT asked for? TIME v v 2a( x xo ) 2 2 o 43.5 0 2a(2.5 0) 2 a 2 378.5 m/s/s Examples How long does it take a car at rest to cross a 35.0 m intersection after the light turns green, if the acceleration of the car is a constant 2.00 m/s/s? What do I know? vo= 0 m/s x = 35 m a = 2.00 m/s/s What do I want? Which variable is NOT given and NOT asked for? Final Velocity t=? x xo vox t 1 at 2 2 35 0 (0) 1 (2)t 2 2 t 5.92 s Examples A car accelerates from 12.5 m/s to 25 m/s in 6.0 seconds. What was the acceleration? What do I know? vo= 12.5 m/s v = 25 m/s t = 6s What do I want? Which variable is NOT given and NOT asked for? DISPLACEMENT a=? v vo at 25 12.5 a(6) a 2.08 m/s/s Summary There are 3 types of MOTION graphs • Displacement(position) vs. Time • Velocity vs. Time • Acceleration vs. Time There are 2 math tools used to get more information from the graphs • Slope • Area Summary v (m/s) x (m) a (m/s/s) area = x t (s) t (s) area = v t (s) Comparing and Sketching graphs One of the more difficult applications of graphs in physics is when given a certain type of graph and asked to draw a different type of graph List 2 adjectives to describe the SLOPE or VELOCITY 1. The slope is CONSTANT 2. The slope is POSITIVE x (m) t (s) How could you translate what the SLOPE is doing on the graph ABOVE to the Y axis on the graph to the right? v (m/s) t (s) Example v (m/s) x (m) t (s) t (s) 1st line • The slope is constant • The slope is “-” 2nd line • The slope is “0” 3rd line • The slope is “+” • The slope is constant Example – Graph Matching What is the SLOPE(a) doing? a (m/s/s) The slope is increasing v (m/s) t (s) a (m/s/s) t (s) t (s) a (m/s/s) t (s) A pictorial representation of forces complete with labels. FN Ff T T W1,Fearth1 or m1g m2g •Weight(mg) – Always drawn from the center, straight down •Force Normal(FN) – A surface force always drawn perpendicular to a surface. •Tension(T or FT) – force in ropes and always drawn AWAY from object. •Friction(Ff)- Kinetic friction is always opposing the motion. Ff FN mg Since the Fnet = 0, a system moving at a constant speed or at rest MUST be at EQUILIBRIUM. TIPS for solving problems • Draw a FBD • Resolve anything into COMPONENTS • Write equations of equilibrium • Solve for unknowns 10-kg box is being pulled across the table to the right at a constant speed with a force of 50N at an angle of 30 degrees above the horizontal. a) Calculate the Force of Friction Fax Fa cos 50cos30 43.3N a) Calculate the Normal Force Ff Fax 43.3N FN Ff Fa Fay 30 Fax mg FN m g! FN Fay m g FN m g Fay (10)(9.8) 50 sin 30 FN 73N Springs – Hooke’s Law One of the simplest type of simple harmonic motion is called Hooke's Law. This is primarily in reference to SPRINGS. Fs µDx k = Constant of Proportionality k = Spring Constant(Unit:N/m) Fs = kDx (and it opposes the stretch/compression) The negative sign only tells us that “F” is what is called a RESTORING FORCE, in that it works in the OPPOSITE direction of the displacement. Hooke’s Law from a Graphical Point of View Fs = kx Suppose we had the following data: x(m) Force(N) 0 0 0.1 12 0.2 24 0.3 36 0.4 48 Fs x k = Slope of a F vs. x graph k= Force vs. Displacement y = 120x + 1E-14 R2 = 1 80 70 0.5 60 0.6 72 Force(Newtons) 60 50 k =120 N/m 40 30 20 10 0 0 0.1 0.2 0.3 0.4 Displacement(Meters) 0.5 0.6 0.7 Example A load of 50 N attached to a spring hanging vertically stretches the spring 5.0 cm. The spring is now placed horizontally on a table and stretched 11.0 cm. What force is required to stretch the spring this amount? Fs kx 50 k (0.05) k 1000 N/m Fs kx Fs (1000)(0.11) Fs 110 N The acceleration of an object is directly proportional to the NET FORCE and inversely proportional to the mass. a FNET a 1 a m FNET FNET m a m FNET F Tips: •Draw an FBD •Resolve vectors into components •Write equations of motion by adding and subtracting vectors to find the NET FORCE. Always write larger force – smaller force. •Solve for any unknowns A 10-kg box is being pulled across the table to the right by a rope with an applied force of 50N. Calculate the acceleration of the box if a 12 N frictional force acts upon it. FN Ff mg Fa In which direction, is this object accelerating? The X direction! So N.S.L. is worked out using the forces in the “x” direction only FNet m a Fa F f m a 50 12 10a a 3 .8 m / s 2 A mass, m1 = 3.00kg, is resting on a frictionless horizontal table is connected to a cable that passes over a pulley and then is fastened to a hanging mass, m2 = 11.0 kg as shown below. Find the acceleration of each mass and the tension in the cable. FN FNet m a m2 g T m2 a T T m1g T m1a m2 g m1a m2 a m2 g m2 a m1a m2 g a(m2 m1 ) m2g a m2 g (11)(9.8) 7.7 m / s 2 m1 m2 14 FNet m a m2 g T m2 a T m1a T (3)(7.7) 23.1 N Where does the calculus fit in? dv d 2x F ma m m dt dt There could be situations where you are given a displacement function or velocity function. The derivative will need to be taken once or twice in order to get the acceleration. Here is an example. You are standing on a bathroom scale in an elevator in a tall building. Your mass is 72-kg. The elevator starts from rest and travels upward with a speed that varies with time according to: v(t ) 3t 0.20t 2 When t = 4.0s , what is the reading on the bathroom scale (a.k.a. Force Normal)? Fnet m a dv d (3t 0.20t ) a 3 0.40t FN m g m a FN m a m g dt dt FN (72)(9.8) (72)(4.6) a(4) 3 0.40(4) 4.6 m/s/s 1036.8 N 2 TWO types of Friction • Static – Friction that keeps an object at rest and prevents it from moving (no sliding) • Kinetic – Friction that acts during motion (two surfaces sliding) Force of Friction • The Force of Friction is directly related to the Normal Force. Ff µ FN m = constant of proportionality m = coefficient of friction The coefficient of Fsf £ m s FN Fkf = mk FN friction is a unitless constant that is specific to the material type and usually less than one. Example A 1500 N crate is being pushed across a level floor at a constant speed by a force F of 600 N at an angle of 20° below the horizontal as shown in the figure. Fa a) What is the coefficient of kinetic friction between the crate and the floor? F f k FN FN Fay F f Fax Fa cos 600(cos20) 563.82N FN Fay m g Fa sin 1500 20 FN 600(sin 20) 1500 1705.21N Fax 563.82 k 1705.21 k 0.331 Ff mg Example If the 600 N force is instead pulling the block FN at an angle of 20° above the horizontal as shown in the figure, what will be the acceleration of the crate. Assume that the coefficient of friction is the same as found in (a) FNet m a 20 Fax Ff Fax F f m a Fa cos FN m a Fa cos (m g Fa sin ) m a 600cos 20 0.331(1500 600sin 20) 153.1a 563.8 428.57 153.1a a 0.883 m / s 2 Fa mg Fay Inclines Ff FN m g cos mg m g sin Tips •Rotate Axis •Break weight into components •Write equations of motion or equilibrium •Solve Example Masses m1 = 4.00 kg and m2 = 9.00 kg are connected by a light string that passes over a frictionless pulley. As shown in the diagram, m1 is held at rest on the floor and m2 rests on a fixed incline of angle 40 degrees. The masses are released from rest, and m2 slides 1.00 m down the incline in 4 seconds. Determine (a) The acceleration of each mass (b) The coefficient of kinetic friction and (c) the tension in the string. T FN Ff m2gcos40 m2g m1 40 m2gsin40 m1g T m1 g m1a T m1a m1 g m2 g sin ( Ff T ) m2a 40 T FNET m a Example FNET m a T m1 g m1a T m1a m1 g m2 g sin ( Ff T ) m2a x vox t 1 at 2 2 1 0 1 a ( 4) 2 2 a 0.125 m / s 2 T 4(.125) 4(9.8) 39.7 N m2 g sin F f T m2 a m2 g sin F f (m1a m1 g ) m2 a m2 g sin k FN m1a m1 g m2 a m2 g sin k m2 g cos m1a m1 g m2 a m2 g sin m1a m1 g m2 a k m2 g cos k m2 g sin m1a m1 g m2 a m2 g cos k 56.7 0.5 39.2 1.125 0.235 67.57 Horizontally Launched Projectiles To analyze a projectile in 2 dimensions we need 2 equations. One for the “x” direction and one for the “y” direction. And for this we use kinematic #2. 2 1 x vox t at 2 x voxt Remember, the velocity is CONSTANT horizontally, so that means the acceleration is ZERO! 1 2 y = - gt 2 Remember that since the projectile is launched horizontally, the INITIAL VERTICAL VELOCITY is equal to ZERO. Horizontally Launched Projectiles Example: A plane traveling with a horizontal velocity of 100 m/s is 500 m above the ground. At some point the pilot decides to drop some supplies to designated target below. (a) How long is the drop in the air? (b) How far away from point where it was launched will it land? What do I know? What I want to know? vox=100 m/s t=? x=? y = 500 m voy= 0 m/s g = 9.8 m/s/s y 1 gt 2 500 1 (9.8)t 2 2 2 102.04 t 2 t 10.1 seconds x voxt (100)(10.1) 1010 m Vertically Launched Projectiles There are several things you must consider when doing these types of projectiles besides using components. If it begins and ends at ground level, the “y” displacement is ZERO: y = 0 Vertically Launched Projectiles You will still use kinematic #2, but YOU MUST use COMPONENTS in the equation. vo vox voy x voxt 1 2 y = voy - gt 2 vox vo cos voy vo sin Example A place kicker kicks a football with a velocity of 20.0 m/s and at an angle of 53 degrees. (a) How long is the ball in the air? (b) How far away does it land? (c) How high does it travel? vox vo cos vox 20 cos 53 12.04 m / s 53 voy vo sin voy 20sin 53 15.97 m / s Example A place kicker kicks a football with a velocity of 20.0 m/s and at an angle of 53 degrees. (a) How long is the ball in the air? What I know vox=12.04 m/s voy=15.97 m/s y=0 g = 9.8 m/s/s y voy t 1 gt 2 0 (15.97)t 4.9t 2 2 2 15.97t 4.9t 15.97 4.9t t 3.26 s What I want to know t=? x=? ymax=? Example A place kicker kicks a football with a velocity of 20.0 m/s and at an angle of 53 degrees. (b) How far away does it land? What I know vox=12.04 m/s voy=15.97 m/s y=0 g = -9.8 m/s/s x voxt (12.04)(3.26) What I want to know t = 3.26 s x=? ymax=? 39.24 m Example A place kicker kicks a football with a velocity of 20.0 m/s and at an angle of 53 degrees. What I know vox=12.04 m/s voy=15.97 m/s y=0 g = 9.8 m/s/s What I want to know t = 3.26 s x = 39.24 m ymax=? (c) How high does it travel? CUT YOUR TIME IN HALF! y voy t 1 gt 2 2 y (15.97)(1.63) 4.9(1.63) 2 y 13.01 m Circular Motion and New’s 2nd Law 2 Recall that according to Newton’s Second Law, the acceleration is directly proportional to the Force. If this is true: v FNET m a ac r 2 mv FNET Fc r Fc Centripetal Force Since the acceleration and the force are directly related, the force must ALSO point towards the center. This is called CENTRIPETAL FORCE. NOTE: The centripetal force is a NET FORCE. It could be represented by one or more forces. So NEVER draw it in an F.B.D. Examples Top view What is the minimum coefficient of static friction necessary to allow a penny to rotate along a 33 1/3 rpm record (diameter= 0.300 m), when the penny is placed at the outer edge of the record? F f Fc FN mg Side view Ff m v2 FN r m v2 m g r v2 rg rev 1 min 33.3 * 0.555rev sec min 60 sec 1sec 1.80 sec T rev 0.555 rev 2r 2 (0.15) vc 0.524 m / s T 1.80 v2 (0.524) 2 0.187 rg (0.15)(9.8) Examples The maximum tension that a 0.50 m string can tolerate is 14 N. A 0.25-kg ball attached to this string is being whirled in a vertical circle. What is the maximum speed the ball can have (a) the top of the circle, (b)at the bottom of the circle? m v2 FNET Fc m ac r m v2 T mg r (T m g) m v2 r r (T m g) 0.5(14 (0.25)(9.8)) v m 0.25 v 5.74 m / s T mg Examples At the bottom? T mg m v2 FNET Fc m ac r m v2 T mg r (T m g) m v2 r r (T m g) 0.5(14 (0.25)(9.8)) v m 0.25 v 4.81 m / s N.L.o.G – Putting it all together m1m2 r2 G constantof proportion ality G UniversalGravitational Constant Fg 27 G 6.67x10 Fg G Nm 2 m1m2 r2 Fg m g Use this when you are on theearth Fg G m1m2 Use this when you are LEAVING the earth r2 kg 2 Kepler’s 3rd Law – The Law of Periods "The square of the period of any planet is proportional to the cube of the semi major axis of its orbit." Gravitational forces are centripetal, thus we can set them equal to each other! Since we are moving in a circle we can substitute the appropriate velocity formula! The expression in the RED circle derived by setting the centripetal force equal to the gravitational force is called ORBITAL SPEED. Using algebra, you can see that everything in the parenthesis is CONSTANT. Thus the proportionality holds true! Example A 2-kg sliding puck whose initial velocity magnitude is v1 = 10 m/s strikes a wall at a 30 degree angle and bounces off. If it leaves the wall with a velocity magnitude of v2 = 10 m/s, and if the collision takes a total of 0.02 seconds to complete, what was the average force applied to the puck by the wall? There is something you need to consider: Momentum is a VECTOR!!! Let’s look at this problem using a X-Y axis for reference Example cont’ If we did the same thing for the Y direction we would discover that the Force Net is equal to ZERO! The temptation is to treat momentum as a SCALAR...DO NOT DO THIS! SIGNS COUNT! Momentum is conserved! The Law of Conservation of Momentum: “In the absence of an unbalanced external force, the total momentum before the collision is equal to the total momentum after the collision.” po (truck ) m vo (500)(5) 2500kg * m / s po ( car ) (400)(2) 800kg * m / s po (total ) 3300kg * m / s ptruck 500* 3 1500kg * m / s pcar 400* 4.5 1800kg * m / s ptotal 3300kg * m / s Several Types of collisions Sometimes objects stick together or blow apart. In this case, momentum is ALWAYS conserved. p before p after m1v01 m2 v02 m1v1 m2 v2 When 2 objects collide and DON’T stick m1v01 m2 v02 mtotal vtotal When 2 objects collide and stick together mtotal vo (total ) m1v1 m2 v2 When 1 object breaks into 2 objects Elastic Collision = Kinetic Energy is Conserved Inelastic Collision = Kinetic Energy is NOT Conserved Work The VERTICAL component of the force DOES NOT cause the block to move the right. The energy imparted to the box is evident by its motion to the right. Therefore ONLY the HORIZONTAL COMPONENT of the force actually does WORK. When the FORCE and DISPLACEMENT are in the SAME DIRECTION you get a POSITIVE WORK VALUE. The ANGLE between the force and displacement is ZERO degrees. What happens when you put this in for the COSINE? When the FORCE and DISPLACEMENT are in the OPPOSITE direction, yet still on the same axis, you get a NEGATIVE WORK VALUE. This negative doesn't mean the direction!!!! IT simply means that the force and displacement oppose each other. The ANGLE between the force and displacement in this case is 180 degrees. What happens when you put this in for the COSINE? When the FORCE and DISPLACEMENT are PERPENDICULAR, you get NO WORK!!! The ANGLE between the force and displacement in this case is 90 degrees. What happens when you put this in for the COSINE? Example W F r W F r cos A box of mass m = 2.0 kg is moving over a frictional floor ( uk = 0.3) has a force whose magnitude is F = 25 N applied to it at an angle of 30 degrees, as shown to the left. The box is observed to move 16 meters in the horizontal direction before falling off the table. W F r W F r cos W 25 16 cos30 W 346.4 Nm W 346.4 J a) How much work does F do before taking the plunge? Example cont’ What if we had done this in UNIT VECTOR notation? F 21.65iˆ 12.5 ˆj W ( Fx rx ) ( Fy ry ) W (21.65 16) (12.5 0) W 346.4 Nm W 346.4 J much work does the FORCE Example cont’ How NORMAL do and Why? Fn W F r W F r cos There is NO WORK since “F” and “r” are perpendicular. W FN 16 cos90 Ff W 0J How much does the internal energy of the system increase? 34.08 J Elastic Potential Energy The graph of F vs.x for a spring that is IDEAL in nature will always produce a line with a positive linear slope. Thus the area under the line will always be represented as a triangle. NOTE: Keep in mind that this can be applied to WORK or can be conserved with any other type of energy. Elastic potential energy W F ( x)dx (kx)dx x x x 0 x 0 W (kx)dx k xdx x2 x W k | | x 0 W U spring 1 kx2 2 2 Elastic “potential” energy is a fitting term as springs STORE energy when there are elongated or compressed. Energy is CONSERVED! W DK D U K K o (U U o ) K K o U U o Ko U o K U Energybefore Energyafter Example A 2.0 m pendulum is released from rest when the support string is at an angle of 25 degrees with the vertical. What is the speed of the bob at the bottom of the string? Lcos L h EB = EA UO = K mgho = 1/2mv2 gho = 1/2v2 1.83 = v2 1.35 m/s = v h = L – Lcos h = 2-2cos h = 0.187 m Ballistic Pendulum • Event includes a collision during which momentum conservation is the best model and post-collision motion during which energy conservation is the best model. • Examples: last three problems on HW, Workshop 14,