AP Physics C - Problem Drill 18: Gravitation and Circular Motion
|
|
- Shannon Knight
- 5 years ago
- Views:
Transcription
1 AP Physics C - Problem Drill 18: Gravitation and Circular Motion Question No. 1 of 10 Instructions: (1) Read the problem and answer choices carefully () Work the problems on paper as 1. Two objects some distance apart gravitationally attract each other. If the distance between them is suddenly tripled, how will that affect the gravitational attraction? Question 01 (A) 3 times as great (B) 1/3 as great (C) 1/9 as great (D) no change will occur (E) 9 times as great Since the distance is on the bottom of the fraction in the universal gravitation law, increasing the distance would lessen the force between the objects. There is a d squared in the bottom of the fraction in the universal gravitation law. Consider that affect on the final answer. C. Correct! Since the universal gravitation law is an inverse square law, tripling the distance will make the force 1/9 of its previous value. The universal gravitation law incorporates a distance term. This change in distance will affect the force somehow. Examine the universal gravitational law. Notice that there is a d squared in the bottom of the formula. Examine the universal gravitational law. m m F= G d 1 g The masses and gravitational constant remain unchanged. The distance triples. Making the distance larger will definitely lessen the force. However, the distance is squared. This emphasized that change even more. Squaring a tripled distance will reduce the force by a factor of 9. The force will be 1/9 of its previous value. The correct answer is (C).
2 Question No. of 10 Instructions: (1) Read the problem and answer choices carefully () Work the problems on paper as. Calculate the gravitational force of attraction between the earth and moon, when the moon is at distance of 3.8 x 10 5 km, the mass of the moon is kg, the mass of the earth is kg. Question 0 (A) 8.00 x N (B).1 x 10 8 N (C).1 x 10 6 N (D).1 x N (E) 8.00 x 10 8 N Gravitational force follows an inverse square law. Check that your use of exponents. C. Correct! Use the Universal Law of Gravitation and keep track of the exponents. Check that your use of exponents. Gravitational force follows an inverse square law. Check your use of exponent. Known: Distance moon to earth, d = 3.8 x 10 5 km Mass of earth, m e = kg Mass of moon, mm = kg Unknown: Force, F g =? N Define: m /kg Use the universal law of gravitation: mm e m G g F = where Universal Gravitational Constant, G = 6.67 x N d Output: kg kg kg F = N m /kg = N m /kg g 5 11 ( km) km 6 F =.1 10 N g Substantiate: Units are correct, sig figs are correct, magnitude is reasonable. The correct answer is (C).
3 Question No. 3 of 10 Instruction: (1) Read the problem statement and answer choices carefully () Work the problems on paper as 3. Harry is attracted to Sally (in a gravitational sense!). Assume that Harry and Sally have a mass of 100kg each. If they stand 4 m apart, what is the gravitational force attracting them? Question 03 (A) 4. x N (B) 4. x 10-8 N (C) 65 N (D) 1.7 x 10-7 N (E) 4. x 10 3 N Don t forget that two masses are need for the gravitation formula. In this case, both masses are 100 kg. B. Correct! Use Newton s law of universal gravitation. Be sure to input the correct constant G. Also be sure to square the distance. Don t forget the universal gravitational constant, G, in the equation. Don t forget that the distance is squared in the denominator of the gravitation formula. Check that you are using the correct exponents in the calculation. Known: Mass of Harry and Sally, m1 =m =100 kg Distance apart, d = 4 m Unknown: F g =? N Define : Use Newton s law of universal gravitation: m /kg mm 1 F G g d = where Universal Gravitational Constant, G = 6.67 x N Output: Substitute the known values and carefully calculate: -11 N m ( )( 100kg)(100kg) kg -8 F = = N g (4m) Substantiate: Units are correct, sig figs are correct, magnitude is reasonable. Notice how the units cancel to leave Newton as the final label. Also notice how the final answer is very small. This force would be negligible in almost all cases. This is due to the relatively small masses involved. The correct answer is (B).
4 Question No. 4 of 10 Instructions: (1) Read the problem and answer choices carefully () Work the problems on paper as 4. Where is the difference in weight the greatest? Question 04 (A) You would weigh the same at all points inside and on the earth. (B) Greatest difference between weight at North Pole and South Pole. (C) Greatest difference is between weight at the equator and at the poles. (D) Greatest difference is between weight at center of the earth, and weight on its surface. (E) Greatest difference is between weight at the surface and weight half way to the center of the earth. Remember weight depends on gravitational force. Weight is essentially the same at both poles. There is a slight difference in weight at the poles and equator due to flattening of the earth at the poles and the spin of the earth, but it is small. D. Correct! At the center of the earth you would be weightless. Weight does decrease the closer you get to the earth s center but consider where the weight is least. There is a slight difference in weight at the poles and equator due to flattening of the earth at the poles and the spin of the earth but it is small. Weight is essentially the same at either pole. Weight does decrease the closer you get to the earth s center, at the center (assuming that the earth s density is uniform) an object would be weightless, since the mass of the surrounding material would be equal the gravitational forces would cancel each other out giving a net gravitational force of zero, which means no weight. The correct answer is (D).
5 Question No. 5 of 10 Instructions: (1) Read the problem and answer choices carefully () Work the problems on paper as 5. The picture shows a ball on a string being swung around clockwise which arrow shows the direction at which the ball would fly off at if the string broke? Question 05 (A) A, anticlockwise around the circle. (B) B, perpendicular to the direction it was travelling when the string broke. (C) C at a tangent to the circle, in the direction it was travelling when the string broke. (D) D continuing around the circle in the original direction of travel. (E) E inwards along the radius of the circle of travel. Consider which way the ball is travelling just before the string breaks. E D A C B Consider which way the ball is travelling just before the string breaks, what forces will be acting after the string breaks. C. Correct! The ball will continue to move in the direction that it was travelling at the instant the string broke, that is at a tangent to the circle, the tangent forms an angle of 90 to the radius of the circle. There is no longer a centripetal force acting to keep the ball moving in a circle. This is the direction of the centripetal force that is present when the string is intact; when the string breaks the force is no longer present. When the string breaks there is no longer a centripetal force acting to keep the ball moving in a circle. The ball will continue to move in the direction that it was travelling at the instant the string broke, that is at a tangent to the circle, remember the tangent forms an angle of 90 to the radius of the circle. The correct answer is (C).
6 Question No. 6 of 10 Instructions: (1) Read the problem and answer choices carefully () Work the problems on paper as 6. Which of these is a correct statement about centrifugal force? Question 06 (A) Centrifugal force acts to keep an object moving in a circle. (B) Centrifugal force is a center seeking force. (C) Centrifugal force is a real force that results from an object moving in a circle. (D) Centrifugal force is the apparent force that seems to be acting to pull an object away from the center when it moves in a circle. (E) When you swing a ball on a string in a circle you apply a centrifugal force to the string, which is transferred to the ball. This is not a centrifugal force. This is Centripetal force not Centrifugal force. Centrifugal force is an apparent force. D. Correct! If such a force existed the ball from the previous questions would fly outward from the circle not at a tangent. When you swing a ball on a string in a circle you apply an inward force on the string, which is transferred to the ball, the ball exerts an equal but opposite force back on the string. BUT the force you apply is not a centrifugal force. Consider the previous question where a ball was swung around in a circle, when you do this it feels like a force is pulling on your hand, it could be interpreted that there is a force acting on the ball to pull it out from the circle. This is what we refer to as a Centrifugal force (a center fleeing force) In fact you are applying an inward force to the string, which is applied to the ball, and we know from Newton s third law the ball applies an equal but opposite force to the string that you feel on your hand. If the Centrifugal force existed we would expect that if string broke the ball would go flying out wards in the direction the force, but we know that it actually goes at a tangent to the circle. The correct answer is (D).
7 Question No. 7 of 10 Instructions: (1) Read the problem and answer choices carefully () Work the problems on paper as 7. In a cyclotron, an electromagnet exerts a force of 7.5 x N on a beam of protons. Each proton has a mass of 1.67x10-7 kg. The electromagnet causes the protons to travel in a circular path of radius 1. m. What is the velocity of the proton beam? Question 07 (A).3 x 10 7 m/s (B) 3. x 10-0 m/s (C) 5.5 x 10 7 m/s (D) 5.4 x m/s (E) 4.3 x 10-8 m/s A. Correct! Use the formula: F c = mv /r, rearrange to give v=sqrt(f c r/m). Be sure of your substitutions. You accidentally inverted the radius and mass. This answer seems very slow for protons in a cyclotron. Use the centripetal force formula. You re looking for the velocity. Algebraically rearrange to solve for this velocity. This would be faster than the speed of light. Don t forget to take the square root since you are looking for v, not v squared. Use the centripetal force formula. You re looking for the velocity. Algebraically rearrange to solve for this velocity. Known: Force on proton, F c = 7.5 x N Mass proton, m p = 1.67x10-7 kg Radius of circular path = 1. m Unknown: Velocity v =? m/s Define: Use the formula for centripetal force and rearrange to find velocity: mv F= c r v Remember that N equals a kg m/s. This gives m/s as your final answer units. = F r c m Output: Substitute values: -13 ( kg m/s )(1.m) 7 v = =.3 10 m/s kg Substantiate: Units are correct, sig figs are correct, magnitude is reasonable. The correct answer is (A).
8 Question No. 8 of 10 Instructions: (1) Read the problem and answer choices carefully () Work the problems on paper as 8. Imagine that a car is traveling in a circle on a track that is banked at a 5 angle. If the radius of the circle is 50 m, how fast could the car travel without drifting? Ignore friction and air resistance. Question 08 (A) 15 m/s (B) 3.8 x 10-5 m/s (C) 110 m/s (D) 8 m/s (E) None of the above A. Correct! The centripetal force is equal to mass times the acceleration of gravity times the tangent of the angle. Equate this to mass times velocity squared divided by radius. The masses cancel out. Substitute all the given information and solve for velocity. The g used in this problem is the acceleration from gravity, not the universal gravitational constant. This answer choice is extremely slow for any car race. The tangent of the angle must be used, not just the angle itself. Remember that the centripetal force in this case is a component of the normal force. Don t forget to take the square root since you are looking for v, not v squared. You might think that the mass is necessary for the problem. However, the mass cancels out when the two expressions for centripetal force are equated. Known: Radius of curvature, r = 50 m Banking angle, θ = 5 F c Unknown: Velocity, v=? m/s Define: The centripetal force is the force horizontal to the road, to find it we must use trig and combine with the formula for centripetal force and velocity. opp F F c c tan θ = = = adj W mg Normal force 5 5 Weight velocity Rearrange the force equation to find mv F = mg tan θ = c r v = r g tan θ v = r g tan θ v = 50 m 9.8 m/s tan 5 Output: v = 490 m /s v = 15 m/s Substantiate: Units are correct, sig figs are correct, magnitude is reasonable. The correct answer is (A).
9 Question No. 9 of 10 Instructions: (1) Read the problem and answer choices carefully () Work the problems on paper as 9. Two satellites are orbiting the earth, satellite 1 orbits is at 300 km above the Earth, satellite orbits is at 600 km above the Earth, what can be said about their relative velocities? The earth s radius is 6.4 x 10 6 m. Question 09 (A) There is insufficient information. (B) The velocity of satellite 1 will be twice that of satellite. (C) The velocity of satellite 1 will half that of satellite. (D) The velocity of satellite 1 will be 4 times that of satellite. (E) None of the above You can calculate the relative velocities; use the fact that the centrifugal force is the same as the gravitational pull. Then rearrange to find an equation for velocity. Remember that the satellites are orbiting about the center of the earth; draw a picture to help find the radius of orbits. Remember that the satellites are orbiting about the center of the earth, draw a picture to help find the radius of orbits, and then use the fact that the centrifugal force is the same as the gravitational pull. Remember that the satellites are orbiting about the center of the earth, draw a picture to help find the radius of orbits, and then use the fact that the centrifugal force is the same as the gravitational pull. Then rearrange to find an equation for velocity. E. Correct! Remember that the satellites are orbiting about the center of the earth, draw a picture to help find the radius of orbits, and then use the fact that the centrifugal force is the same as the gravitational pull. Then rearrange to find an equation for velocity. You can calculate the relative velocities; use the fact that the centrifugal force is the same as the gravitational pull. Then rearrange to find an equation for velocity. F G = F c Consider any satellite of mass m, and an earth mass of M m M m v G = r r M v = G r The velocity of the satellites does not depend on its mass. Remember that the satellites are orbiting about the center of the earth so the radius of orbits must include the radius of the earth, do not forget to convert from km to m for satellites. r t = r e + r o r 1 = 6.4 x 10 6 m x 10 6 m =6.7 x 10 6 m r = 6.4 x 10 6 m x 10 6 m =7.0 x 10 6 m If we compare velocity of satellite 1 to velocity of satellite v 1 M M r 6 7.0x 10 m v r r r x 10 m = G G = = = 1.0 The correct answer is (E).
10 Question No. 10 of 10 Instructions: (1) Read the problem and answer choices carefully () Work the problems on paper as needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed. 10. If the Hubble space telescope orbits 600 km above the earth, how fast is it orbiting above the earth? The radius of the earth is 6.4 x 10 6 m. The mass of the earth is 6 x 10 4 kg. Question 10 (A).4 x 10 3 m/s (B).9 x 10 9 m/s (C).6 x 10 4 m/s (D) 7.6 x 10 3 m/s (E) Cannot calculate without mass of telescope. Equate the centripetal force to the gravitational force. The mass of the satellite cancels out. This number is way too large, greater than the speed of light. Be sure to use the universal gravitational constant for G. This value isn t the acceleration from gravity 9.8 m/s/s that was used so often before. Don t just use the satellite altitude for the radius. The middle of the satellite orbit is the center of the earth. Add the satellite altitude to the radius of the earth for a more accurate radius. D. Correct! Equate the centripetal force to the gravitational force. The mass of the satellite cancels out. Rearrange for velocity, and then substitute the known values. For the radius, be sure to add the altitude of the satellite to the radius of the earth. This will give the distance of the satellite from the center of the earth. You might think that the mass of the telescope is necessary, but it isn t. Like in so many equations, the mass cancels out. Equate the centripetal force to the gravitational force. Known: Orbit radius, r o = 600 km Earth s radius, r e = 6.4 x 10 6 m Earth s mass M = 6 x 10 4 kg Unknown: Velocity of the satellite, v =? m/s Define: First, change the telescopes altitude into meters. Next add this distance to the radius of the earth. This will give the total distance from the center of the earth to the telescope. r t = r e + r o The centripetal force on the satellite equals the gravitational pull on the telescope. F G = F c m M m v G = r r M v = G r Remember: G = 6.67 x N m /kg Output: r o = 600 km = 6 x 10 5 m r t = 6.4 x 10 6 m + 6 x 10 5 m = 7 x 10 6 m x 10 kg v = 6.67 x 10 N m /kg 6 7 x 10 m v = 7.6 x 10 3 m/s Substantiate: Units are correct, sig figs are correct, magnitude is reasonable. The correct answer is (D). 4
Circular Motion and Gravitation Notes 1 Centripetal Acceleration and Force
Circular Motion and Gravitation Notes 1 Centripetal Acceleration and Force This unit we will investigate the special case of kinematics and dynamics of objects in uniform circular motion. First let s consider
More informationChapter 5 Lecture Notes
Formulas: a C = v 2 /r a = a C + a T F = Gm 1 m 2 /r 2 Chapter 5 Lecture Notes Physics 2414 - Strauss Constants: G = 6.67 10-11 N-m 2 /kg 2. Main Ideas: 1. Uniform circular motion 2. Nonuniform circular
More informationCircular Motion and Gravitation Notes 1 Centripetal Acceleration and Force
Circular Motion and Gravitation Notes 1 Centripetal Acceleration and Force This unit we will investigate the special case of kinematics and dynamics of objects in uniform circular motion. First let s consider
More information6. Find the centripetal acceleration of the car in m/s 2 a b c d e. 32.0
PHYSICS 5 TEST 2 REVIEW 1. A car slows down as it travels from point A to B as it approaches an S curve shown to the right. It then travels at constant speed through the turn from point B to C. Select
More informationCircular Motion & Gravitation MC Question Database
(Questions #4,5,6,27,37,38,42 and 58 each have TWO correct answers.) 1) A record player has four coins at different distances from the center of rotation. Coin A is 1 cm away, Coin B is 2 cm away. Coin
More information5. A car moves with a constant speed in a clockwise direction around a circular path of radius r, as represented in the diagram above.
1. The magnitude of the gravitational force between two objects is 20. Newtons. If the mass of each object were doubled, the magnitude of the gravitational force between the objects would be A) 5.0 N B)
More informationMultiple Choice (A) (B) (C) (D)
Multiple Choice 1. A ball is fastened to a string and is swung in a vertical circle. When the ball is at the highest point of the circle its velocity and acceleration directions are: (A) (B) (C) (D) 2.
More informationAP Physics 1 Lesson 9 Homework Outcomes. Name
AP Physics 1 Lesson 9 Homework Outcomes Name Date 1. Define uniform circular motion. 2. Determine the tangential velocity of an object moving with uniform circular motion. 3. Determine the centripetal
More informationCIRCULAR MOTION AND GRAVITATION
CIRCULAR MOTION AND GRAVITATION An object moves in a straight line if the net force on it acts in the direction of motion, or is zero. If the net force acts at an angle to the direction of motion at any
More informationPSI AP Physics B Circular Motion
PSI AP Physics B Circular Motion Multiple Choice 1. A ball is fastened to a string and is swung in a vertical circle. When the ball is at the highest point of the circle its velocity and acceleration directions
More informationProficient. a. The gravitational field caused by a. The student is able to approximate a numerical value of the
Unit 6. Circular Motion and Gravitation Name: I have not failed. I've just found 10,000 ways that won't work.-- Thomas Edison Big Idea 1: Objects and systems have properties such as mass and charge. Systems
More informationUniform Circular Motion. Uniform Circular Motion
Uniform Circular Motion Uniform Circular Motion Uniform Circular Motion An object that moves at uniform speed in a circle of constant radius is said to be in uniform circular motion. Question: Why is uniform
More informationChapter 5 Review : Circular Motion; Gravitation
Chapter 5 Review : Circular Motion; Gravitation Conceptual Questions 1) Is it possible for an object moving with a constant speed to accelerate? Explain. A) No, if the speed is constant then the acceleration
More informationUpon collision, the clay and steel block stick together and move to the right with a speed of
1. A 2.0-kilogram ball traveling north at 4.0 meters per second collides head on with a 1.0-kilogram ball traveling south at 8.0 meters per second. What is the magnitude of the total momentum of the two
More informationLecture Presentation. Chapter 6 Preview Looking Ahead. Chapter 6 Circular Motion, Orbits, and Gravity
Chapter 6 Preview Looking Ahead Lecture Presentation Chapter 6 Circular Motion, Orbits, and Gravity Text: p. 160 Slide 6-2 Chapter 6 Preview Looking Back: Centripetal Acceleration In Section 3.8, you learned
More informationBlueberry Muffin Nov. 29/30, 2016 Period: Names:
Blueberry Muffin Nov. 9/30, 016 Period: Names: Congratulations! 1. To solve the problems, use your etextbook, physical textbooks, physics websites, your Sketchbooks.. Show your thinking through calculations,
More informationBlueberry Muffin Nov. 29/30, 2016 Period: Names:
Blueberry Muffin Nov. 29/30, 2016 Period: Names: Congratulations! 1. To solve the problems, use your etextbook, physical textbooks, physics websites, your Sketchbooks. 2. Show your thinking through calculations,
More informationChapter 5 Circular Motion; Gravitation
Chapter 5 Circular Motion; Gravitation Units of Chapter 5 Kinematics of Uniform Circular Motion Dynamics of Uniform Circular Motion Highway Curves, Banked and Unbanked Nonuniform Circular Motion Centrifugation
More informationChapter 8. Centripetal Force and The Law of Gravity
Chapter 8 Centripetal Force and The Law of Gravity Centripetal Acceleration An object traveling in a circle, even though it moves with a constant speed, will have an acceleration The centripetal acceleration
More informationUnit 5 Circular Motion and Gravitation
Unit 5 Circular Motion and Gravitation In the game of tetherball, the struck ball whirls around a pole. In what direction does the net force on the ball point? 1) Tetherball 1) toward the top of the pole
More informationPHYSICS. Chapter 8 Lecture FOR SCIENTISTS AND ENGINEERS A STRATEGIC APPROACH 4/E RANDALL D. KNIGHT Pearson Education, Inc.
PHYSICS FOR SCIENTISTS AND ENGINEERS A STRATEGIC APPROACH 4/E Chapter 8 Lecture RANDALL D. KNIGHT Chapter 8. Dynamics II: Motion in a Plane IN THIS CHAPTER, you will learn to solve problems about motion
More informationIn this lecture we will discuss three topics: conservation of energy, friction, and uniform circular motion.
1 PHYS:100 LECTURE 9 MECHANICS (8) In this lecture we will discuss three topics: conservation of energy, friction, and uniform circular motion. 9 1. Conservation of Energy. Energy is one of the most fundamental
More informationAssignment - Periodic Motion. Reading: Giancoli, Chapter 5 Holt, Chapter 7. Objectives/HW:
Assignment - Periodic Motion Reading: Giancoli, Chapter 5 Holt, Chapter 7 Objectives/HW: The student will be able to: 1 Define and calculate period and frequency. 2 Apply the concepts of position, distance,
More informationCentripetal force keeps an Rotation and Revolution
Centripetal force keeps an object in circular motion. Which moves faster on a merry-go-round, a horse near the outside rail or one near the inside rail? While a hamster rotates its cage about an axis,
More informationCircular Motion (Chapter 5)
Circular Motion (Chapter 5) So far we have focused on linear motion or motion under gravity (free-fall). Question: What happens when a ball is twirled around on a string at constant speed? Ans: Its velocity
More informationAP Physics 1 Lesson 10.a Law of Universal Gravitation Homework Outcomes
AP Physics 1 Lesson 10.a Law of Universal Gravitation Homework Outcomes 1. Use Law of Universal Gravitation to solve problems involving different masses. 2. Determine changes in gravitational and kinetic
More informationPRACTICE TEST for Midterm Exam
South Pasadena AP Physics PRACTICE TEST for Midterm Exam FORMULAS Name Period Date / / d = vt d = v o t + ½ at 2 d = v o + v 2 t v = v o + at v 2 = v 2 o + 2ad v = v x 2 + v y 2 = tan 1 v y v v x = v cos
More informationTYPICAL NUMERIC QUESTIONS FOR PHYSICS I REGULAR QUESTIONS TAKEN FROM CUTNELL AND JOHNSON CIRCULAR MOTION CONTENT STANDARD IB
TYPICAL NUMERIC QUESTIONS FOR PHYSICS I REGULAR QUESTIONS TAKEN FROM CUTNELL AND JOHNSON CIRCULAR MOTION CONTENT STANDARD IB 1. A car traveling at 20 m/s rounds a curve so that its centripetal acceleration
More informationPhysics 111: Mechanics Lecture 9
Physics 111: Mechanics Lecture 9 Bin Chen NJIT Physics Department Circular Motion q 3.4 Motion in a Circle q 5.4 Dynamics of Circular Motion If it weren t for the spinning, all the galaxies would collapse
More informationAlgebra Based Physics Uniform Circular Motion
1 Algebra Based Physics Uniform Circular Motion 2016 07 20 www.njctl.org 2 Uniform Circular Motion (UCM) Click on the topic to go to that section Period, Frequency and Rotational Velocity Kinematics of
More informationCircular Motion PreTest
Circular Motion PreTest Date: 06/03/2008 Version #: 0 Name: 1. In a series of test runs, a car travels around the same circular track at different velocities. Which graph best shows the relationship between
More informationChapter 2. Forces & Newton s Laws
Chapter 2 Forces & Newton s Laws 1st thing you need to know Everything from chapter 1 Speed formula Acceleration formula All their units There is only 1 main formula, but some equations will utilize previous
More informationName (please print): UW ID# score last first
Name (please print): UW ID# score last first Question I. (20 pts) Projectile motion A ball of mass 0.3 kg is thrown at an angle of 30 o above the horizontal. Ignore air resistance. It hits the ground 100
More informationChapter 7. Preview. Objectives Tangential Speed Centripetal Acceleration Centripetal Force Describing a Rotating System. Section 1 Circular Motion
Section 1 Circular Motion Preview Objectives Tangential Speed Centripetal Acceleration Centripetal Force Describing a Rotating System Section 1 Circular Motion Objectives Solve problems involving centripetal
More informationChapter 5 Circular Motion; Gravitation
Chapter 5 Circular Motion; Gravitation Units of Chapter 5 Kinematics of Uniform Circular Motion Dynamics of Uniform Circular Motion Highway Curves, Banked and Unbanked Newton s Law of Universal Gravitation
More informationWhat path do the longest sparks take after they leave the wand? Today we ll be doing one more new concept before the test on Wednesday.
What path do the longest sparks take after they leave the wand? Today we ll be doing one more new concept before the test on Wednesday. Centripetal Acceleration and Newtonian Gravitation Reminders: 15
More informationUniform Circular Motion AP
Uniform Circular Motion AP Uniform circular motion is motion in a circle at the same speed Speed is constant, velocity direction changes the speed of an object moving in a circle is given by v circumference
More informationChapter 8: Newton s Laws Applied to Circular Motion
Chapter 8: Newton s Laws Applied to Circular Motion Centrifugal Force is Fictitious? F actual = Centripetal Force F fictitious = Centrifugal Force Center FLEEing Centrifugal Force is Fictitious? Center
More informationGravity. Gravity and Newton. What really happened? The history of Gravity 3/9/15. Sir Isaac Newton theorized the Law of Gravitation in 1687
3/9/15 Gravity and Newton Gravity What really happened? Probably the more correct version of the story is that Newton, upon observing an apple fall from a tree, began to think along the following lines:
More informationUniform Circular Motion
Circular Motion Uniform Circular Motion Uniform Circular Motion Traveling with a constant speed in a circular path Even though the speed is constant, the acceleration is non-zero The acceleration responsible
More informationComments about HW #1 Sunset observations: Pick a convenient spot (your dorm?) Try to get 1 data point per week Keep a lab notebook with date, time,
Comments about HW #1 Sunset observations: Pick a convenient spot (your dorm?) Try to get 1 data point per week Keep a lab notebook with date, time, weather, comments Mark down bad weather attempts Today:
More informationChapter 5. A rock is twirled on a string at a constant speed. The direction of its acceleration at point P is A) B) P C) D)
A 1500 kg car travels at a constant speed of 22 m/s around a circular track which has a radius of 80 m. Which statement is true concerning this car? A) The velocity of the car is changing. B) The car is
More informationPreparing for Six Flags Physics Concepts
Preparing for Six Flags Physics Concepts uniform means constant, unchanging At a uniform speed, the distance traveled is given by Distance = speed x time At uniform velocity, the displacement is given
More informationUCM-Circular Motion. Base your answers to questions 1 and 2 on the information and diagram below.
Base your answers to questions 1 and 2 on the information and diagram The diagram shows the top view of a 65-kilogram student at point A on an amusement park ride. The ride spins the student in a horizontal
More informationChapter 6 Circular Motion, Orbits and Gravity
Chapter 6 Circular Motion, Orbits and Gravity Topics: The kinematics of uniform circular motion The dynamics of uniform circular motion Circular orbits of satellites Newton s law of gravity Sample question:
More informationChapter 7: Circular Motion
Chapter 7: Circular Motion Spin about an axis located within the body Example: Spin about an axis located outside the body. Example: Example: Explain why it feels like you are pulled to the right side
More informationAP Physics 1 Chapter 7 Circular Motion and Gravitation
AP Physics 1 Chapter 7 Circular Motion and Gravitation Chapter 7: Circular Motion and Angular Measure Gravitation Angular Speed and Velocity Uniform Circular Motion and Centripetal Acceleration Angular
More informationCentripetal Force Review. 1. The graph given shows the weight of three objects on planet X as a function of their mass.
Name: ate: 1. The graph given shows the weight of three objects on planet X as a function of their mass. 3. If the circular track were to suddenly become frictionless at the instant shown in the diagram,
More informationAn object moving in a circle with radius at speed is said to be undergoing.
Circular Motion Study Guide North Allegheny High School Mr. Neff An object moving in a circle with radius at speed is said to be undergoing. In this case, the object is because it is constantly changing
More informationChapter Six News! DO NOT FORGET We ARE doing Chapter 4 Sections 4 & 5
Chapter Six News! DO NOT FORGET We ARE doing Chapter 4 Sections 4 & 5 CH 4: Uniform Circular Motion The velocity vector is tangent to the path The change in velocity vector is due to the change in direction.
More informationASTRONAUT PUSHES SPACECRAFT
ASTRONAUT PUSHES SPACECRAFT F = 40 N m a = 80 kg m s = 15000 kg a s = F/m s = 40N/15000 kg = 0.0027 m/s 2 a a = -F/m a = -40N/80kg = -0.5 m/s 2 If t push = 0.5 s, then v s = a s t push =.0014 m/s, and
More informationHonors Assignment - Circular and Periodic Motion
Honors Assignment - Circular and Periodic Motion Reading: Chapter 5, and 11 1 through 11 5 Objectives/HW: Assignment #1 M: # 1 6 Assignment #2 M: # 7 15 Assignment #3 Text: Chap 5 # 6, 12 M: # 17 22 Assignment
More informationIn the y direction, the forces are balanced, which means our force equation is simply F A = F C.
Unit 3: Dynamics and Gravitation DYNAMICS Dynamics combine the concept of forces with our understanding of motion (kinematics) to relate forces to acceleration in objects. Newton s Second Law states that
More information3 UCM & Gravity Student Physics Regents Date
Student Physics Regents Date 1. Which diagram best represents the gravitational forces, Fg, between a satellite, S, and Earth? A) B) 4. Gravitational force exists between point objects and separated by
More informationName St. Mary's HS AP Physics Circular Motion HW
Name St. Mary's HS AP Physics Circular Motion HW Base your answers to questions 1 and 2 on the following situation. An object weighing 10 N swings at the end of a rope that is 0.72 m long as a simple pendulum.
More informationNo Brain Too Small PHYSICS
MECHANICS: CIRCULAR MOTION QUESTIONS CIRCULAR MOTION (2016;1) Alice is in a car on a ride at a theme park. The car travels along a circular track that is banked, as shown in the diagram. On the diagram,
More informationChapter 6: Systems in Motion
Chapter 6: Systems in Motion The celestial order and the beauty of the universe compel me to admit that there is some excellent and eternal Being, who deserves the respect and homage of men Cicero (106
More informationUNIVERSITY OF SASKATCHEWAN Department of Physics and Engineering Physics
UNIVERSITY OF SASKATCHEWAN Department of Physics and Engineering Physics Physics 111.6 MIDTERM TEST #2 November 15, 2001 Time: 90 minutes NAME: STUDENT NO.: (Last) Please Print (Given) LECTURE SECTION
More informationIn the last lecture the concept of kinetic energy was introduced. Kinetic energy (KE) is the energy that an object has by virtue of its motion
1 PHYS:100 LETUE 9 MEHANIS (8) I. onservation of Energy In the last lecture the concept of kinetic energy was introduced. Kinetic energy (KE) is the energy that an object has by virtue of its motion KINETI
More informationQuestion 01. A. Incorrect! This is not Newton s second law.
College Physics - Problem Drill 06: Newton s Laws of Motion Question No. 1 of 10 1. Which of the options best describes the statement: Every object continues in a state of rest or uniform motion in a straight
More informationBasic Physics Content
Basic Physics Content The purpose of these 38 questions is to let me know what your initial grasp is of the material that lies ahead. This is not graded, so don t stress out. Enjoy thinking about and answering
More informationCircular Motion and Gravitation Practice Test Provincial Questions
Circular Motion and Gravitation Practice Test Provincial Questions 1. A 1 200 kg car is traveling at 25 m s on a horizontal surface in a circular path of radius 85 m. What is the net force acting on this
More information66 Chapter 6: FORCE AND MOTION II
Chapter 6: FORCE AND MOTION II 1 A brick slides on a horizontal surface Which of the following will increase the magnitude of the frictional force on it? A Putting a second brick on top B Decreasing the
More informationThink of a car turning a corner, or the fun carnival ride, or a satellite orbiting the earth.
Uniform Circular Motion Objects moving in curved(arc, semi circular, circular) path at constant speeds. Think of a car turning a corner, or the fun carnival ride, or a satellite orbiting the earth. When
More informationII. Universal Gravitation - Newton 4th Law
Periodic Motion I. Circular Motion - kinematics & centripetal acceleration - dynamics & centripetal force - centrifugal force II. Universal Gravitation - Newton s 4 th Law - force fields & orbits III.
More informationPreview. Circular Motion and Gravitation Section 1. Section 1 Circular Motion. Section 2 Newton s Law of Universal Gravitation
Circular Motion and Gravitation Section 1 Preview Section 1 Circular Motion Section 2 Newton s Law of Universal Gravitation Section 3 Motion in Space Section 4 Torque and Simple Machines Circular Motion
More informationHW Chapter 5 Q 7,8,18,21 P 4,6,8. Chapter 5. The Law of Universal Gravitation Gravity
HW Chapter 5 Q 7,8,18,21 P 4,6,8 Chapter 5 The Law of Universal Gravitation Gravity Newton s Law of Universal Gravitation Every particle in the Universe attracts every other particle with a force that
More informationCircular Motion and Gravitation. Centripetal Acceleration
Circular Motion and Gravitation Centripetal Acceleration Recall linear acceleration 3. Going around a curve, at constant speed 1. Speeding up vi vi Δv a ac ac vi ac 2. Slowing down v velocity and acceleration
More informationContents. Objectives Circular Motion Velocity and Acceleration Examples Accelerating Frames Polar Coordinates Recap. Contents
Physics 121 for Majors Today s Class You will see how motion in a circle is mathematically similar to motion in a straight line. You will learn that there is a centripetal acceleration (and force) and
More informationPHYSICS. Chapter 8 Lecture FOR SCIENTISTS AND ENGINEERS A STRATEGIC APPROACH 4/E RANDALL D. KNIGHT Pearson Education, Inc.
PHYSICS FOR SCIENTISTS AND ENGINEERS A STRATEGIC APPROACH 4/E Chapter 8 Lecture RANDALL D. KNIGHT Chapter 8. Dynamics II: Motion in a Plane IN THIS CHAPTER, you will learn to solve problems about motion
More informationPreview. Circular Motion and Gravitation Section 1. Section 1 Circular Motion. Section 2 Newton s Law of Universal Gravitation
Circular Motion and Gravitation Section 1 Preview Section 1 Circular Motion Section 2 Newton s Law of Universal Gravitation Section 3 Motion in Space Section 4 Torque and Simple Machines Circular Motion
More informationPhysics 1100: Uniform Circular Motion & Gravity
Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Physics 1100: Uniform Circular Motion & Gravity 1. In the diagram below, an object travels over a hill, down a valley, and around a loop the loop at constant
More information2016 PHYSICS FINAL REVIEW PACKET
2016 PHYSICS FINAL REVIEW PACKET EXAM BREAKDOWN CHAPTER TOPIC # OF QUESTIONS 6 CONSERVATION OF ENERGY 22 7 MOMENTUM/COLLISIONS 17 5 CIRCULAR MOTION GRAVITY/SATELLITE MOTION 30 11 WAVES 24 - ELECTROMAGNETISM/MISC./LABS
More informationCircular Motion.
1 Circular Motion www.njctl.org 2 Topics of Uniform Circular Motion (UCM) Kinematics of UCM Click on the topic to go to that section Period, Frequency, and Rotational Velocity Dynamics of UCM Vertical
More informationP - f = m a x. Now, if the box is already moving, for the frictional force, we use
Chapter 5 Class Notes This week, we return to forces, and consider forces pointing in different directions. Previously, in Chapter 3, the forces were parallel, but in this chapter the forces can be pointing
More informationCircular Orbits. Slide Pearson Education, Inc.
Circular Orbits The figure shows a perfectly smooth, spherical, airless planet with one tower of height h. A projectile is launched parallel to the ground with speed v 0. If v 0 is very small, as in trajectory
More informationUniform Circular Motion
Slide 1 / 112 Uniform Circular Motion 2009 by Goodman & Zavorotniy Slide 2 / 112 Topics of Uniform Circular Motion (UCM) Kinematics of UCM Click on the topic to go to that section Period, Frequency, and
More informationPractice Test for Midterm Exam
A.P. Physics Practice Test for Midterm Exam Kinematics 1. Which of the following statements are about uniformly accelerated motion? Select two answers. a) If an object s acceleration is constant then it
More informationExperiment #7 Centripetal Force Pre-lab Questions Hints
Experiment #7 Centripetal Force Pre-lab Questions Hints The following are some hints for this pre-lab, since a few of these questions can be a little difficult. Note that these are not necessarily the
More informationLinear vs. Rotational Motion
Linear vs. Rotational Motion Every term in a linear equation has a similar term in the analogous rotational equation. Displacements: s = r θ v t ω Speeds: v t = ω r Accelerations: a t = α r Every point
More informationExperiencing Acceleration: The backward force you feel when your car accelerates is caused by your body's inertia. Chapter 3.3
Experiencing Acceleration: The backward force you feel when your car accelerates is caused by your body's inertia. Chapter 3.3 Feeling of apparent weight: Caused your body's reaction to the push that the
More informationPhysics Midterm Review Sheet
Practice Problems Physics Midterm Review Sheet 2012 2013 Aswers 1 Speed is: a a measure of how fast something is moving b the distance covered per unit time c always measured in units of distance divided
More informationhttps://njctl.org/courses/science/ap-physics-c-mechanics/attachments/summerassignment-3/
AP Physics C Summer Assignment 2017 1. Complete the problem set that is online, entitled, AP C Physics C Summer Assignment 2017. I also gave you a copy of the problem set. You may work in groups as a matter
More informationThe Force of Gravity exists between any two masses! Always attractive do you feel the attraction? Slide 6-35
The Force of Gravity exists between any two masses! Always attractive do you feel the attraction? Slide 6-35 Summary Newton s law of gravity describes the gravitational force between A. the earth and the
More information(d) State the effect on the magnitude of the centripetal force in the following cases:
YEAR 12 PHYSICS: UNIFORM CIRCULAR MOTION ASSIGNMENT NAME: QUESTION 1 (a) A car of mass 1200 kg rounds a bend of radius 50m at a speed of 20ms -1. What centripetal acceleration does it experience? (b) Calculate
More informationCircular Motion & Gravitation FR Practice Problems
1) A mass m is attached to a length L of string and hung straight strainght down from a pivot. Small vibrations at the pivot set the mass into circular motion, with the string making an angle θ with the
More informationCircular Motion Dynamics Concept Questions
Circular Motion Dynamics Concept Questions Problem 1: A puck of mass m is moving in a circle at constant speed on a frictionless table as shown above. The puck is connected by a string to a suspended bob,
More informationChapter 8. Dynamics II: Motion in a Plane
Chapter 8. Dynamics II: Motion in a Plane Chapter Goal: To learn how to solve problems about motion in a plane. Slide 8-2 Chapter 8 Preview Slide 8-3 Chapter 8 Preview Slide 8-4 Chapter 8 Preview Slide
More informationB) v `2. C) `2v. D) 2v. E) 4v. A) 2p 25. B) p C) 2p. D) 4p. E) 4p 2 25
1. 3. A ball attached to a string is whirled around a horizontal circle of radius r with a tangential velocity v. If the radius is changed to 2r and the magnitude of the centripetal force is doubled the
More informationPhysics Semester 2 Final Exam Review Answers
Physics Semester 2 Final Exam Review Answers A student attaches a string to a 3 kg block resting on a frictionless surface, and then pulls steadily (with a constant force) on the block as shown below.
More informationEQUATIONS OF MOTION: NORMAL AND TANGENTIAL COORDINATES (Section 13.5)
EQUATIONS OF MOTION: NORMAL AND TANGENTIAL COORDINATES (Section 13.5) Today s Objectives: Students will be able to apply the equation of motion using normal and tangential coordinates. APPLICATIONS Race
More informationCircular/Gravity ~ Learning Guide Name:
Circular/Gravity ~ Learning Guide Name: Instructions: Using a pencil, answer the following questions. The Pre-Reading is marked, based on effort, completeness, and neatness (not accuracy). The rest of
More informationEQUATIONS OF MOTION: NORMAL AND TANGENTIAL COORDINATES
EQUATIONS OF MOTION: NORMAL AND TANGENTIAL COORDINATES Today s Objectives: Students will be able to: 1. Apply the equation of motion using normal and tangential coordinates. In-Class Activities: Check
More informationChapter: The Laws of Motion
Table of Contents Chapter: The Laws of Motion Section 1: Newton s Second Law Section 2: Gravity Section 3: The Third Law of Motion 1 Newton s Second Law Force, Mass, and Acceleration Newton s first law
More informationPOGIL: Newton s First Law of Motion and Statics. Part 1: Net Force Model: Read the following carefully and study the diagrams that follow.
POGIL: Newton s First Law of Motion and Statics Name Purpose: To become familiar with the forces acting on an object at rest Part 1: Net Force Model: Read the following carefully and study the diagrams
More informationAnnouncements 15 Oct 2013
Announcements 15 Oct 2013 1. While you re waiting for class to start, see how many of these blanks you can fill out. Tangential Accel.: Direction: Causes speed to Causes angular speed to Therefore, causes:
More informationUnit 1: Equilibrium and Center of Mass
Unit 1: Equilibrium and Center of Mass FORCES What is a force? Forces are a result of the interaction between two objects. They push things, pull things, keep things together, pull things apart. It s really
More informationConstant Acceleration. Physics General Physics Lecture 7 Uniform Circular Motion 9/13/2016. Fall 2016 Semester Prof.
Physics 22000 General Physics Lecture 7 Uniform Circular Motion Fall 2016 Semester Prof. Matthew Jones 1 2 Constant Acceleration So far we have considered motion when the acceleration is constant in both
More informationAcceleration in Uniform Circular Motion
Acceleration in Uniform Circular Motion The object in uniform circular motion has a constant speed, but its velocity is constantly changing directions, generating a centripetal acceleration: a c v r 2
More information(a) On the dots below that represent the students, draw and label free-body diagrams showing the forces on Student A and on Student B.
2003 B1. (15 points) A rope of negligible mass passes over a pulley of negligible mass attached to the ceiling, as shown above. One end of the rope is held by Student A of mass 70 kg, who is at rest on
More information