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MID TERM REVIEW PART IV NOTES FOR MID-TERM EXAM Please review these formulas in preparation for the mid-term exam. The table that follows summarizes the formulas and gives an example of each formula. Motion v =d/t = vaverage d = vt t = d/v vaverage = (vi + vf)/2 a = v/t = (vf-vi)/t vf = vi +at vf2 = vi2 + 2ad d = vit + ½at2 Projectile Motion tup = tdown vtop = 0 = vV – gtup tup = vV/g ttot = 2tup = 2vV/g vAV to top = (vV + vtop)/2 = vV/2 dH = vHttot = 2vVvH/g dV = vav*tup = ½vVtup = ½vV2/g Vertical Distance as a Function of Horizontal Distance vVtop = 0m/s Vertical Distance Dv 20.0 dV time to top of arc = ttop = vV/g total time from launch to landing = ttot = 2tup = 2vV/g vV average vertical velocity from launch to top = vAV = (vV + vVtop)/2 = (1/2)vV dV = vAV*ttop = (1/2)vV*vV/g = 1/2(vV)2/g distance traveled horizontally = dH = vH*ttot = 2vHvV/g - dH 12 .0 18 .0 24 .0 30 .0 36 .0 42 .0 48 .0 54 .0 60 .0 66 .0 72 .0 78 .0 84 .0 90 .0 96 .0 10 2. 0 10 8. 0 11 4. 0 12 0. 0 - 6. 0 vH Horizontal Distance Dh Forces Fnet = 0 W = mg Fgrav = Gm1m2/r2 Newton’s 1st Law Weight Law of Gravitation [R13/R23] = [T12/T22] Kepler’s 3rd Law Fslid = slid*Fnormal F = -kx Fnet = ma F1 on 2 = – F2 on 1 g = GMearth/rearth2 Fstatic = static*Fnormal Sliding Friction Fcentripetal = mv2/r Hooke’s Law (Spring Restoring Force) Newton’s 2nd Law Newton’s 3rd Law Acceleration due to gravity on Earth Static Friction Centripetal Force Formula Distance/Velocity/Acceleration v = d/t The symbol “” means “change in” the quantity the immediately follows. Sometimes this is simply written as: Definitions and Units Application Examples v = velocity d = distance t = time This formula is used when only the total distance traveled is known and the total time it took to move that distance. This formula gives only an average velocity. 1) Car moves 500 meters in 100 seconds. What is the average speed or velocity? v = 500m/100s or v = 5 meters/sec v = d/t Metric system units: v: meters/sec (m/s) d: meters (m) t: seconds (s or sec) and it is understood that d refers to the distance traveled and t is the time from an initial time t = 0 Velocity (or speed) is the change in distance (d) divided by the elapsed time (t). vaverage = (vi + vf)/2 vaverage = average velocity vf = final velocity vi = initial velocity a = v/t = (vf-vi)/t Metric system units: v: meters/sec (m/s) a = acceleration vf = final velocity vi = initial velocity t = time Metric system units: a: meters/sec2 (m/s2) v: meters/sec (m/s) t: seconds (sec, s) vf2 = vi2 + 2ad Acceleration is the change in speed or velocity (v) divided by the time (t) it took to change the speed or velocity. a = acceleration vf = final velocity vi = initial velocity t = time d = distance Metric system units: a: meters/sec2 (m/s2) v: meters/sec (m/s) t: seconds (sec, s) d: meters (m) Knowing any two terms yields the 3rd: v = d/t d = vt t = d/v This formula is used when an object is accelerated or decelerated and only the initial and final velocities are known. 2) Car has an average speed of 50m/s and sustains this for 60 seconds, how far did it travel? d = (50m/s)*(60s) = 3000m 3) Car travels 500 meters with an average velocity of 50m/s. How long did this take? t = 500m/50m/s = 10s Car starts at 10m/s and speeds up to 30m/s. The average speed is: vaverage = (10m/s + 30m/s)/2 = 20m/s This formula is used when both initial and final velocities are known and the elapsed time is known. 1) Initial speed of a car is 20m/s and the final speed is 40m/s and it takes 10s to do this. Find a; a = (40m/s – 20m/s)/10s = 2m/s2 Knowing any two terms yields the 3rd: 2) The initial speed of a car is 20m/s and it has an acceleration of 5m/s2 for 5s. What is the final speed? vf = 20m/s + (5m/s2)*5s = 45m/s a = v/t v = (vf-vi) = at or vf = vi + at t = v/a This formula is used as a shortcut to find the final speed or velocity when the initial speed is known as well as the acceleration and the time. Knowing any three of the terms allows us to find the 4th: vf2 = vi2 + 2ad vi2 = vf2 – 2ad d = (vi2 – vf2)/2a a = (vi2 – vf2)/2d 3) A car changes its speed by 30m/s when it has an acceleration of 6m/s2. How long did this take? t = (30m/s)/(6m/s2) = 5s 1) A runner’s initial speed is 5m/s and he accelerates at 5m/s2 over a distance of 7.5 meters. What is his final speed? vf2 = (5m/s)2 + 2*(5m/s2)(7.5m) = 100m2/s2 or vf = 10m/s 2) Same problem except the final speed was 10m/s. What was the initial speed? vi2 = (10m/s)2 – 2*(5m/s2)(7.5m) = 25m2/s2 or vi = 5m/s 3) A car’s initial speed was 3m/s and its final speed was 5m/s and it traveled a distance of 8m. What was the acceleration? a = [(5m/s)2 – (3m/s)2]/(2*8m) a = (25m2/s2 – 9m2/s2)/16m = 1m/s2 4) Same problem except a is given as 1m/s. What was the distance? d = [(5m/s)2 – (3m/s)2]/(2*1m/s2) d = (25m2/s2 – 9m2/s2)/2m = 8m Formula d = vit + ½at2 Definitions and Units a = acceleration vi = initial velocity t = time d = distance Metric system units: a: meters/sec2 (m/s2) v: meters/sec (m/s) t: seconds (sec, s) d: meters (m) Application This formula is used to find the distance when the time, acceleration and initial speed are known. Knowing any three of the terms allows us to find the 4th: ½at2 d = vit + vi = d/t – ½at a = 2(d – vit)/t2 t = is the solution to a messy quadratic equation and rarely used. Here is the solution: Examples 1) A rock is dropped downwards with an initial speed of 5m/s and the acceleration due to gravity is 10m/s2. How fare did it fall in 3 seconds? d = (5m/s)*(3s) + ½(10m/s2)*(3s)2 d = 15m + 45m = 60m 2) Same problem except d = 60m is given. What was the initial velocity? vi = (60m/3s) – ½(10m/s2)(3s) vi = 20m/s – 15m/s = 5m/s t = [-vi +/- (vi2 + 2ad)1/2]/a 3) Same problem except vi = 5m/s and d = 60m. What was the acceleration if the time was 3 seconds? a = 2[60m – (5m/s)*3s]/(3s)2 = (2*45m)/(9s2) = 10m/s2 4) Same problem. What was the time if the distance as 60m, initial speed was 5m/s and the acceleration was g = 10m/s2? t = [-5m/s +/- (25m2/s2 +2*10m/s2*60m)1/2]/10m/s2 t = [-5m/s +/- 35m/s]/10m/s2 = (30m/s)/10m/s2 = 3 seconds Formula Projectile Motion tup = tdown Definitions and Units Application Examples tup = time from launch to the maximum height This formula is used in projectile motion. Questions can be phrased as: (a) what is the time to maximum height? (b) what is the total time if the time to the top is t seconds? (c) if total time from launch to landing is t seconds, what is the time to the top? 1) time to top for a projectile is 5 seconds; what is the time from the top to landing? and ttot = 2tup Metric system units: t: seconds (sec, s) tup = tdown tup = 5 seconds, so tdown = 5 seconds 2) time to top for a projectile is 10 seconds; what is total flight time? tup = tdown so ttot = 2ttop ttot = 2*10seconds = 20 seconds 3) total time of flight is 30 seconds; what is the time to the top of the flight path? vtop = 0 = vV – gtup and tup = vV/g ttot = 2vv/g vtop = vertical velocity at maximum height vv = initial vertical velocity g = acceleration due to gravity tup – time to maximum height Metric system units: vtop: meters/sec (m/s, m/sec) vv: meters/sec (m/s, m/sec) g: meters/sec2 (m/s2, m/sec2) tup: seconds (sec, s) vAV to top = (vV + vtop)/2 = vV/2 dH = vHttot = 2vVvH/g vtop = vertical velocity at maximum height vv = initial vertical velocity tup – time to maximum height Metric system units: vtop: meters/sec (m/s, m/sec) vv: meters/sec (m/s, m/sec) vAV: meters/sec (m/s, m/sec) vH = horizontal velocity vv = initial vertical velocity dH – horizontal distance g = acceleration due to gravity Metric system units: vH: meters/sec (m/s, m/sec) vv: meters/sec (m/s, m/sec) dH: meters (m) g: meters/sec2 (m/sec2, m/s2) This formula is used to find the time to the top. At the maximum height, the projectile’s velocity is precisely 0m/sec. Knowing that we can then calculate the time to the top given the initial speed, vv. The second formula, derived from the first, states that the time to the top is the initial vertical velocity divided by the acceleration due to gravity. The third formula is derived from the time up/total time relations; it states that total time of flight is twice time up, using the initial velocity and g for the calculation. This formula allows for a simpler calculation for distance to the top of the projectile path. The average velocity is, in reality, a constant velocity. That means we can use the average velocity in a simpler distance/time – velocity formula. This formula requires the intermediate step of finding the total time. Once that is known, the horizontal distance is found from the product of a constant horizontal velocity and the total time of flight. tup = (1/2)*ttot tup = (1/)*30 seconds = 15 seconds 1) If the initial vertical velocity is 30m/s and g = 10m/s2, what is the time to the top? tup = vv/g = (30m/s)/(10m/s2) = 3sec 2) If the time to the top is 5 seconds and g = 10m/s2, what was the initial vertical velocity? vv = g*tup = (10m/s2)*(5sec) = 50 meters 3) If the initial vertical velocity is 20m/s and g = 10m/s2, what is the total time of flight? ttot = 2vv/g = 2*(20m/s)/(10m/s2) ttot = 4 seconds 1) The initial vertical velocity of a projectile is 50m/s. What is the average velocity to the top of the projectile’s path? vAV = (vV + vtop)/2 = vV/2 vAV = (50m/s)/2 = 25m/s 1) What is dH if vH = 40m/s and the time to the top of the path is 10 seconds? dH = vH*ttot = vH*2*tup dH = (40m/s)*2*(10sec) = 800m Formula Projectile Motion dV = vav*tup dV = ½vVtup dV = ½vV2/g Definitions and Units Application Examples vAV = average velocity vv = initial vertical velocity dv = horizontal distance g = acceleration due to gravity tup = time to top This formula is shown in three variations, all equivalent. Each formula is found by substituting in the formulas for vAV and tup. 1) What is the vertical distance from launch to highest point if the average velocity in the vertical direction is 10m/s and the time to the top is 10 seconds? dV = vav*tup = (10m/s)*(10s) = 100m Metric system units: vAV: meters/sec (m/s, m/sec) vv: meters/sec (m/s, m/sec) dv: meters (m) g: meters/sec2 (m/sec2, m/s2) tup: seconds (sec, s) 2) If the initial vertical velocity is 20m/s and the time to the top is 10 seconds, what is the maximum vertical height? dV = ½vVtup = ½ (20m/)*10s = 100m 3) If the initial vertical velocity is 20m/s and g = 10m/s2, what is the maximum height reached? Special case when object lands at height different from launch height d = vvttot + ½ gttot2 vv = initial vertical velocity d = distance above or below original launch height. Always take original launch height = 0. g = acceleration due to gravity ttot = total time of flight Use the quadratic formula to fund the total time of flight. To do this one should execute the following steps: First, re-write the distance formula as follows: ½ gttot2 + vvttot - d = 0 Metric system units: vv: meters/sec (m/s, m/sec) d: meters (m) g: meters/sec2 (m/sec2, m/s2) ttot: seconds (sec, s) Second, divide through by the coefficient of t2 which is ½g ttot2 + 2(vv//g)ttot - 2d/g = 0 dV = ½vV2/g = ½ (20m/s)2/10m/s2 = 100m 1) An object is launched with an initial vertical speed of 20m/sec and an initial horizontal speed of 20 m/sec. It lands on a valley floor which is 100 meters below the launch height. What is the total time of flight? ttot = {-(vv/g)[(vv/g)2+(2d/g)]1/2)} ttot = (20/10) {(20/10)2 + (2*100/10)}1/2 ttot = 2 {4 + 20}1/2 Third, in the quadratic formula (x =( -b [b2 – 4ac]1/2)/2a) we now have a = 1, b = 2vv/g and c = -2d/g. Plug in and get this solution for ttot: ttot ={-(2vv/g)[(2vv/g)2 -4(2d/g)]1/2)}/2 Fourth, carry out the division by 2 in the result: ttot = {-(vv/g)[(vv/g)2+(2d/g)]1/2)} ttot = 2 4.90 = 6.90 seconds The positive root gives the total time. The negative root gives the time when the object would have been launched from the landing height (time before actual launch) with the same speed as the object has when it hits the ground. If the object landed at the same height as the launch height, then d = 0 and our formula reduces to; ttot = {-(vv/g)[(vv/g)2]1/2)} Note that the minus sign in the lead term vanishes since g and vv point opposite to each other, i.e., g is negative. The lead term is the time to the top. The term in the square root adjusts it by adding the extra time to get to the landing height. When the landing height = launch height (d = 0) then we get the two familiar solutions of ttot = 0 sec and ttot = 2ttop = 2vv/g. ttot = 2vv/g = 2ttop which is 4 seconds. Formula Vectors F FV FH FH = Fcos() FV = Fsin() The key triangles you should remember which may possibly appear on the mid-term and will certainly be found on the PSAT, SAT and maybe the HSPAs: 3-4-5 5x 4x All vectors in physics can be broken down into two perpendicular components. A vector requires both a magnitude and a direction to specify it. A magnitude is a number; a direction is something like an angle, or North-South-East-West. Metric system units: Both the resultant vector and the component vectors have the same units. 3x 5-12-13 13x 12x 5x 30o-60o-90o 2x Definitions and Units F is called the resultant force vector and FH and FV are called the components of the force vector F. The force vector is pointed at an angle with respect to the horizontal. 60o 1x 30o F: force vectors all have Newtons (N) as units: v: velocity vectors all have m/s (meters/sec, meters/s) as units a: acceleration vectors all have meters/sec2 (m/s2, meters/s2) as units d: distance vectors all have meters (m) as units x3 45o-45o-90o x2 45o 1x 45o Application The vectors we have encountered this far in this course are: 1) Force 2) Acceleration 3) Velocity 4) Distance Of particular note: the magnitude of the velocity vector is called the speed. We use components of vectors to simplify the analysis of forces and motion. We break down vectors into the x and y or horizontal and vertical components. Once a vector has been broken down into its components, we can analyze the motion or the forces independently in each direction. For each of the four special s in column 1 as well as for the trigonometry relations for vectors, you can verify that the relations hold by using the Pythagorean Theorem. For each I wrote out the Pythagorean Theorem and verified that the ratio of the sides in each triangle is correct: Vectors are often written in bold type or with an arrow above them in texts: Trig relation: F or F F2 = (Fsin(q))2 + (Fcos(q))2 F2 = F2*(sin2(q) + cos2(q)) F2 = F2 *(1) = F2 The rules for drawing vectors in diagrams: Special s: 1x In all vector s we use the Pythagorean Theorem: F2 = FH2 + FV2 where F is the resultant vector and FV and FH are the vertical and horizontal coordinates of the resultant vector. a) each vector is represented as an arrow: (5x)2 = (3x)2 + (4x)2 25x2 = 9x2 + 16x2 = 25x2 25x2 = 25x2 b) the length of the arrow is proportional to the magnitude of the vector; c) the arrow points in the direction the vector operates; d) all vectors representing the same kind of quantity (force or velocity, etc)in the same diagram must use the same scale. e) resultant vectors and their components are drawn “head to tail” as in the vector diagrams in column 1. (13x)2 = (5x)2 + (12x)2 169x2 = 25x2 + 144x2 = 169x2 169x2 = 169x2 (2x)2 = (1x)2 + x3 )2 4x2 = x2 + 3x2 = 4x2 4x2 = 4x2 (x2 )2 = (1x)2 + (1x)2 = 2x2 2x2 = 2x2 Examples 1) the horizontal component of a force is 30N and the vertical component is 40N, what is the resultant force vector? For those who have had some geometry, these two components are the legs of a 3-4-5 triangle, magnified by a common factor of 10 (3*10 = 30, 4*10 = 40). Hence, the resultant vector is 50N (5*10). The angle it operates at is found from the inverse sine function on a calculator: sin-1(40N/50N) = = 53.13o. (The other triangles you might need to know are shown in column 1.) 2) a force of 100N is applied at an angle of 30o above the horizontal. What are the horizontal and vertical components? FH = Fcos(30o) = 100N*0.866 = 86.6N FV = Fsin(30o) = 100N*(0.5) = 50N 3) additional applications include the inclined plane problem which will be done below in the force section. Notes for Mid-term Exam Formula Newton’s First Law or the Law of Inertia Definitions and Units “An object will stay at rest or move in a straight line with a constant speed unless acted upon by an unbalanced force” Application This law is the basis of classical mechanics. By definition, if an object either does any of the following: a) moves when it had previously been at rest; b) changes direction; or c) speeds up or slows down; then a force acts on it. Newton’s Second Law: F = the net force m = the mass of the object a = acceleration of the object If a net force is acting on an object, then the acceleration the object experiences is proportional to the force and inversely proportional to the mass of the object. F = ma Metric system units: F: Newtons m: kilograms a: m/s2 (meters/sec2, meters/s2) Examples 1) a car stops suddenly and a passenger without a seatbelt strikes the windshield; 2) The net force of an object at rest on a table is 0 Newtons. 3) An object on a string and swung in a circle at a constant speed still has a force acting on it since the object’s direction changes every instant. 4) The net force acting on a car moving in s straight line at a constant speed is 0 Newtons. 1) A brick of mass = 5kg is accelerated at 5m/s2. What is the force acting on the brick? F = ma = (5kg)*(5m/s2) = 25N The three formulas are; 2) A mass of 10kg experiences a force of 150N. What is the acceleration? F = ma a = F/m m = F/a a = F/m = 150N/10kg = 15m/s2 3) A force of 200N on an object results in its acceleration of 20m/s2. What is the mass of the object? m = F/a = 200N/20m/s2 = 10kg W = mg W = weight of an object g = acceleration due to gravity = 9.8m/s2 and on tests and quizzes, we use 10m/s2 m = mass of an object Metric system units: W: Newtons m: kilograms g: m/s2 (meters/sec2, meters/s2 This is a specific example of Newton’s 2nd Law. Any mass, m, has a weight equal to mg whether the object is moving or not. 1) A brick has a mass = 5kg.What is its weight? W = ma = (5kg)*(10m/s2) = 50N The three formulas are; 2) A mass of 10kg has a weight of 150N. What is the acceleration due to gravity? W = mg g = F/m m = W/g g = W/m = 150N/10kg = 15m/s2 NOTE: the values we use for g are only valid on the surface of the earth. At other locations – on the moon, another planet, the Sun the value for g changes. It can be higher or lower. (See the section on Gravitation.) 3) An object has a weight 200N What is the mass of the object? m = W/g = 200N/10m/s2 =210kg Formula Newton’s Third Law: F1 on 2 = – F2 on 1 Definitions and Units Both forces use Newtons as the units in the metric system. The force of object 1 on object 2 is equal in magnitude but opposite in direction to the force of object 2 on object 1. The Normal Force, Fnormal, is the force a table or surface exerts back on an object in contact with the table or surface. The object can be at rest or moving. There can be other forces acting. Fnormal is equal to the net downward force acting on the object and this net downward force is the force the table or surface experiences. It pushes back on the object with a force equal to the net downward force. The minus sign in Newton’s Third law indicates the forces are opposite in direction. A Free Body Diagram is a diagram of all the forces acting on an object. Free Body Diagrams Fnormal = 10N Ffric = 5N W = mg = 10N Fpull = 5N Each force arrow is drawn in the proper direction and the length of each arrow is proportional to the magnitude of the force. Each force must be labeled with the proper name and the magnitude, if known, should be shown as well. Application This law is often misunderstood. To understand it properly one must draw “free body diagrams” which are simple pictures showing all forces acting on a single body or object. (See next topic.) The key is to understand that one needs to consider all the forces acting on a single object. To understand an object’s motion we need only know the forces acting on the object, not the forces it exerts on another object or objects. Examples 1) A mass of 10kg sits on a table at rest. What is the force of the table on the object? The objects weight is W = mg = (10kg)*(10m/s2) = 100N The object is at rest and its weight presses down on the table. The table presses back with a force equal in magnitude, 100N, but opposite in direction, i.e., up. The two forces acting on the object, then, are its weight, acting down, and the force of the table up on the object, which, in this example, equals the weight. This latter force is the Fnormal or normal force. 2) A baseball is hit by a bat with a force of 500N. What is he force of the baseball on the bat? Fbatonball = 500N = – Fballonbat Fballonbat = 500N acting opposite in direction to the Fbatonball force. Free Body Diagrams are essential for depicting forces on an object. From the diagram one then notes separately all of the horizontal forces and then all of the vertical forces. This includes finding the horizontal and vertical components of forces that act at angles. By doing that, one can then write equations to find the NET horizontal and NET vertical forces acting on an object. This takes the form of Newton’s 2nd law for the horizontal and vertical directions. A 1kg mass is at rest on a table. Then a force Fpull of 5N is applied to the right. A friction force opposing motion of 5N results. Draw the Free Body Diagram and calculate the net forces acting on the object. The picture in column 1 shows this example. Since the mass is at rest, the forces in the up and down directions must balance. So W equals the normal force of the table on the object. Vertical forces: W = mg = (1kg)*(10m/s2) = 10N, down W = Fnormal = 10N, up Net FV = W – Fnormal = 10N – 10N = 0N Horizontal forces: Fpull = 5N to the right Ffric = 5N to the left Net FH = Fpull – Ffric = 5N – 5N = 0N Formula Fstatic = static*Fnormal Fslid = slid*Fnormal Definitions and Units Friction forces act opposite in direction to the motion of an object. A frictional force by itself cannot create motion, but it can act to impede it. Friction is also needed to create traction for car tires, runners’ shoes, and almost anything that must move along a surface. Friction is always proportional to the normal force between two surfaces. Fstatic – the force of static friction Fslid = the force of sliding friction = the coefficient of friction Fnormal = the normal force between the two surfaces static > slid And always ≤ 1.0 Metric system units: Fslid, Fstatic and Fnorm: Newtons (N) : no units. is the ratio of two forces and has no units. Application The procedure to solve friction problems is as follows: 1) Draw a free Body Diagram and carefully note all forces acting vertically and horizontally. 2) Calculate the normal force. To do this, remember that the normal force is the force that makes the total net force in the vertical direction equal to zero. If, before you calculate the normal force, you find that the net vertical force is already positive, that means the object is moving away from the surface and there is then no normal force. 3) Calculate the appropriate friction force. If the object is stationary, then the appropriate force is the static friction force. If the horizontal forces are sufficient to overcome the static force, then the appropriate force is the sliding friction force. Examples 1) A 10kg mass is on a table. A force, Fdown, of 5N also pushes down on the mass from above. A force of 10N pulls to the right. The coefficient of static friction = 0.4. Is this pulling force sufficient to move the object? Find the normal force. The forces acting down on the table are the weight of the object W and the extra 5N force: Fnormal = W + Fdown = (10kg)*(10m/s2) + 5N = 105 N Fstatic = static*Fnormal = (0.4)*(105N) = 42N Fstatic > Fpull so the object does not move. 2) An object with weight = 50N is on a table. A force, Fup, of 10N is pulling up on the object. The coefficient of sliding friction = 0.3. A force, Fpull, of 20N pulls to the right. What is the net force on the object in the horizontal direction? Find the normal force. Fnormal = W – Fup = 50N – 10N = 40N and this force acts up on the object. Fslid = slid*Fnormal = (0.3)*(40N) = 12N Net FH = Fpull – Fslid = 20N – 12N = 8N Formula Gravitation Fgrav = Gm1m2/r2 g = GMearth/rearth2 Definitions and Units Application Examples Fgrav is the gravitational force of attraction between two masses m1 and m2 separated by a distance equal to r Any two objects are attracted to each other via the gravitational force. The force is one of attraction. 1) What is the gravitational force between mass 1 = 100kg and mass 2 = 500kg if the distance between them is 1000 meters? It is an inverse square law and because of that, when the distance between the two masses doubles, the force is reduced to one fourth of its previous value. Fgrav = Gm1m2/r2 = (6.67x10-11Nm2/kg2)*(100kg)* (500kg)/(1000)2 = (6.67x10-11)*(5x104)/(1*106) = 3.335x10-12 N Fgrav – gravitational force m1 = mass of first object m2 = mass of second object R = distance between the two objects G = universal constant of gravitation = 6.67x10-11 Nm2/kg2 Metric system units: F: Newtons (N) m1, m2: kilograms (kg) G: Nm2/(kg) r: meters (m) The gravitational force is a very weak force but it does operate over vast distances, galaxies and beyond. The second equation defines the acceleration due to gravity, g, at the surface of the earth in terms of G, the radius of the earth and the mass of the earth. When gravity holds an object in orbit, such a the moon around the earth, the force of gravity is also a centripetal force and can be equated to the centripetal force formula, mv2/r, to find the tangential velocity of the object in orbit. 2) If the distance between the two objects doubles, what happens to the force? Fnew = Gm1m2/(2r)2 = ¼ Gm1m2/r2 3) Let mass m1 be 50kg and let two masses, m2 = 50kg and m3 = 200kg. Let the distance between m1 and m2 be 50m and between m1 and m3 be 100m. Are the two gravitational forces acting on m1 equal? r12 m2 r13 m1 m3 Fg2on1 = Gm1m2/r122 Fg30n1 = Gm1m3/r232 We want to know if these forces are equal and this really is the question whether m2/r122 = m3/r122 since all other factors cancel out (G and m1 cancel out). So, does m2/r122 = m3/r132 or does (50kg)/(50m)2 = (200kg)/(100m)2 Divide both sides by 50kg: 1/(50m)2 = 4/(100m)2 Multiply both sides by (100m)2:\ which leaves only the number 4 on the right hand side of the equation (100m)2/(50m)2 = (100m/50m)2 = 22 = 4 So they are equal! Formula Circular motion Fcentripetal = mv2/r f = #revolutions/time elapsed T = 1/f v = r v = Circumference/Period v = 2r/T v = r = 2f v = r r Definitions and Units Application Examples Fcent is the Centripetal Force. Centripetal force is an inward-directed force, i.e., towards the center of the circle, acting on an object moving in a circle. These formulas completely describe circular motion with a constant angular or constant tangential velocity. 1) a mass of 10kg moves in a circle of radius r = 3 meters at a tangential speed of 9m/s. What are (a) the centripetal force, (b) angular velocity, (c) frequency, (d) period. Fcent = centripetal force v = tangential or linear velocity m = mass of the object moving in a circle r = radius of the circle f = frequency which is the number of revolutions an object makes per second. T = the period of the circular motion and T = the inverse of the frequency Fcent acts towards the center; v is tangent to the circle. Hence v and Fcent are always perpendicular to each other. = angular velocity and it is related to the frequency by a factor of 2. Metric system units: F: newtons (N) v: m/s (meters/s, meters/sec) m: kilograms (kg) r: meters (m) f: revs/s (revs/sec) T: seconds (sec, s) : radians/s (rads/s, rads/sec) The centripetal force is the force that causes an object too move in a circle and, therefore, it can be generated by any of the following: a) gravitational force as between the Earth and the Moon for example; b) tension, as in a string; c) normal force, as in a wall pushing against an object; d) friction, as in the friction on a road allowing a car to move in a circle. A commonly used word is centrifugal force which is not a real force. What we experience in circular motion is the sensation of being pushed to the outside, as in a car going around a curve. This is actually your inertia wanting to continue moving in a straight line but you are prevented from doing so by a seatbelt or the side of a car. (a) Fcent = mv2/r = (10kg)*(9m/s)2/(3m) = 270N (b) v = r so w = v/r = (9m/s)/(3m) = 3 radians/sec (c ) f = /2 = (3m/s)/ = 0.955 revs/sec (d) T = 1/f = 1/0.955 = 1.05 seconds 2) a mass of 2kg on a string is spun in a circle of radius 4m. What is the tangential or linear velocity if the centripetal force is 18N? Fcent = mv2/r or v2 = r*F/m v2 = (4m)*(18N)/(2kg) 36m2/s2 v = 6m/s 3) what is the radius of the circle an object of mass = 100kg is moving in if the force causing the circular motion is the force of friction is 200N and the tangential speed is 10m/s? Fcent = Ffric = mv2/r so r = mv2/Ffric r = 100*(10m/s)2/200N = 50m 4) Let m1 be a mass orbiting a second mass, m2, at a distance r. What is the tangential velocity in terms of m1, m2, r and G the gravitational constant? Fgrav = Fcent Gm1*m2/r2 = m1v2/r and cancel a m1/r on both sides Gm2/r = v2 v = (Gm2/r)1/2 The velocity of m1 is independent of the objects mass and depends on the m2, r and G. Formula Bracket analysis Fcentripetal = mv2/r Fnew = [ ]*[ ]2 * Forig [ ] F = centripetal force Each set of brackets[ ] stands in for one of the parameters in the centripetal force formula First pair of brackets stands in for the mass Second pair stands in for the velocity Third pair stands in for the radius Definitions and Units Procedure Fcent is the Centripetal Force. Problems are posed as follows: Mass increases by 50% Speed decreases 50% Radius doubles Application Examples Instead of carrying out messy calculations, it is far easier to manipulate simple fractions to find the new force as a multiple of the original force. 1) Mass increases by 50% Speed decreases 50% Radius doubles What is the new Centripetal Force as a MULTIPLE of the original Centripetal Force Step 1: mass multiple = 1.5 or 3/2 Speed multiple = 0.5 or ½ Radius multiple = 2 Step 3: Plug in the numbers: Fnew = [1.5]*[0.5]2 * Forig [2] Fnew = (3/16)*Forig = 0.1875*Forig Step 1: Determine the multiple of each of the original parameters Step 2: Plug in the multiples into the appropriate set of brackets Step 3: Carry out the simple math 2) Mass remains the same Speed triples Radius is cut in half Step 1: mass multiple = 1 Speed multiple = 3 Radius multiple = ½ Step 3: Plug in the numbers: Fnew = [1]*[3]2 * Forig [1/2] Fnew = 18*Forig Same analysis is used for gravitational force problems Fgrav is the Gravitational Force. Problems are posed as follows: Fgrav = Gm1m2/r2 Fnew = [ ]*[ ] * Forig [ ]2 F = centripetal force Each set of brackets[ ] stands in for one of the parameters in the centripetal force formula First pair of brackets stands in for the first mass Second pair stands in for the second mass Third pair stands in for the distance between the masses Mass increases by 50% Speed decreases 50% Radius doubles What is the new Gravitational Force as a MULTIPLE of the original Gravitational Force Instead of carrying out messy calculations, it is far easier to manipulate simple fractions to find the new force as a multiple of the original force. 1) Mass1 increases by 50% Mass2 decreases 50% Distance doubles Step 1: mass1 multiple = 1.5 or 3/2 Mass2 multiple = 0.5 or ½ Distance multiple = 2 Step 3: Plug in the numbers: Fnew = [1.5]*[0.5] * Forig [2]2 Fnew = (3/16)*Forig = 0.1875*Forig Step 1: Determine the multiple of each of the original parameters Step 2: Plug in the multiples into the appropriate set of brackets Step 3: Carry out the simple math 2) Mass1 remains the same Mass2 triples Distance is cut in half Step 1: mass1 multiple = 1 Mass2 multiple = 3 Distance multiple = ½ Step 3: Plug in the numbers: Fnew = [1]*[3] * Forig [1/2]2 Fnew = 12*Forig Formula Hooke’s Law F = -kx k = Hooke’s or spring constant x = distance spring stretches or compresses x cm Definitions and Units Application Examples F is the force exerted by a spring when it is compressed or stretched. The force the spring exerts is always opposite to the stretch or compression force. Springs are used in scales. When properly calibrated they can be used to measure weights. Springs are common in grocery stores as well as the more common bathroom scales. 1) What force is needed to stretch a spring a distance of 5cm if the spring constant is 6N/cm? 1) If a spring is stretched by pulling on it, then the spring is pulling back. F = kx k = F/x x = F/k 2) If the spring is compressed by a force, then the spring pushes against the force. Springs can be combined in series or in parallel. Metric system units: F: Newtons (N) x: cm or meters k: N/m or N/cm depending on type of problem Series Springs Series springs are springs connected in a line. If the spring constants are known, then we look for a single spring that can replace the two springs and have the same effect, i.e., the same net stretch distance. (1/k*) = (1/k1) + (1/k2) k* = (k1*k2)/(k1+k2) Parallel Springs Parallel springs are springs that have one weight connected to them and all are stretched the same distance. F = kx = 6*5 = 30N 2. What is the spring constant if a force of 10N stretches a spring scale a distance of 3.5cm? k = F/x = 10/3.5 = 2.857N/cm 3. What distance will a spring stretch if it has a Hooke’s constant = 12N/cm and a mass of 8.5kg is hung from the spring? x = F/k = (mg)/x = 85/12 = 7.083cm 4. What force is needed to stretch two springs “in series” a distance of 8cm if k1 = 4N/cm and k2 = 5N/cm? Ftotal = k*x k* = (4*5)/(4+5) = 20/9 Ftotal = (20/9)*(8) = 160/9 Ftotal = 17.78N 5. What is the equivalent spring constant if a force of 12N stretches a pair of series springs a total distance of 6cm? If one of the Hooke’s constants is 4N/cm, what is the second one? Ftotal = k*x k* = Ftotal/x = 12/6 = 2N/cm (1/k*) = (1/k1) +(1/k2) (1/2) = (1/4) + (1/k2) (1/4) = 1/k2 k2 = 4N/cm 6. What force is needed to stretch two springs “in parallel” a distance of 8cm if k1 = 4N/cm and k2 = 5N/cm? If the spring constants are known, then we look for a single spring that can replace the two springs and have the same effect, i.e., the k* = k1 + k2 Ftotal = k*x = (4+5)*8 = 72N 2. What is the equivalent spring constant if a force of 12.5N stretches a pair of parallel springs a distance of 2.5cm? If one of the Hooke’s constants is 3.5N/cm, what is the second one? Ftotal = k*x k* = Ftotal/x= 12.5/2.5 = 5N/cm k* = k1 + k2 k2 = k* – k1 = 5 – 3.5 = 1.5N/cm Formula Kepler’s 3rd Law [R13/R23] = [T12/T22] R1 and R2 are the radii of orbits of two planets or satellites and T1 and T2 are their respective periods of orbit. Definitions and Units Application Examples Kepler’s 3rd law was originally derived from the elliptical orbits the planets have as the move around the Sun. A circle is a special case of an ellipse. The derivation using ellipses is somewhat more difficult. Kepler’s 3rd Law allows periods of orbit of distant planets or satellites to be found simply knowing the radius of the orbit and the period of a known orbiting body. For example, the Moon orbits the Earth and its distance is known. Hence, a satellite orbiting the Earth at some known height (from the center of the Earth) wil have a period of orbit found from Kepler’s 3rd law. What is the distance from the center of the Earth for a geosynchronous satellite? Using R1 as the radius of the orbit of the satellite about the Earth and R2 as the radius of the orbit of the Moon about the Earth and T1 as the period the satellite and T2 as the period the Moon’s orbit (27.3 days) then we have: Solve for R1: Metric system units: These are ratios of like quantities and hence the ratios themselves do not have any units. The radii can be measured in any consistent set of units – meters, miles, kilometers. The periods likewise must be in consistent units such as seconds, days, months, years. As long as both radii are in the same units and both periods are in the same units, then the ratios work. Use the Moon as the reference orbit: R2 = 1 Moon distance = 1MU = 3.8x108 meters. Use the unit “MU” until the very end for calculations. T2 for the Moon is 27.3 days. T1 for the satellite is 1 day since it remains stationary over the Earth’s surface. R13/R23 = T12/T22 R1 = R2*[T1/T2]2/3 (the exponent 2/3 means take the cube root of the square of the quantity in the brackets) R1 = 1MU*[1/27.3]2/3 (T1/T2)2 = (R1/R2)3 R1 = 1MU*(0.1103) = 0.1103MUs Then T1 is: Plug in for MU: T1 = T2*((R1/R2)3/2 R1 = (1.103x10-1)*(3.8x108) Note the exponent 3/2 means take the square root of the cube of the quantity. Hence, if R1 and R2 are known, then one knows the period the orbit of the satellite about the Earth. Knowing the period and the circumference of the satellites orbit (2R1) then the tangential or orbital velocity of the satellite is v = d/t = 2R1/T1 and use the known value of R1 and the calculated value for T1. A specific case is when the satellite is in geosynchronous orbit about the Earth. That means its period is the same as the Earth’s time for one rotation, or one day. Knowing the Moon’s period (27.3 days) and the satellite’s period (1 day) one can find the radius required for geosynchronous orbit from Kepler’s 3rd law. R1 = 4.19x107meters Since the radius of the Earth is 6.4x106 meters we can calculate the height, H., of the satellite above the Earth as: H = R1 – RE = 41.9x106 – 6.4x106 = 35.5x106m = 35,500 kilometers. This number, by the way, is a very accurate estimate of the required altitude above the Earth. (This is about 22,000 miles.) These satellites are used for cable TV and satellite TV. Cable operators receive programming from satellites and re-broadcast over their coax and fiber networks it to their subscribers. DISH and DirecTV deliver signals directly from satellites to the home or business location. All of these TV satellites are in geosynchronous orbit.