DOCSLIB.ORG

Physics 106 Group Problems Summer 2015 Circular Motion and Newton’S Law of Gravity

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Physics 106 Group Problems Summer 2015 Circular Motion and Newton’S Law of Gravity Physics 106 Group Problems Summer 2015 Circular motion and Newton's law of gravity Name: TA: 1. (5 points) One of the many planets found outside our solar system is Kepler-23C. Its radius is 3.2 times greater than Earths and it has a mass 860 times greater than Earths. What would be the gravitational acceleration on Kepler-23Cs surface in gs, where a g = 9.8 m=s2? Solution: ME g = G 2 RE M23C g23C = G 2 R23C 860ME g23C = G 2 (3:2RE) ME g23C = 84G 2 RE g23C = 84g 2. (5 points) Two masses m1 and m2 are held a distance d apart and the gravitational force between the two of them is measured. They are then moved a distance 4d apart. By what factor has the gravitational attraction between the two changed? Solution: m m F = G 1 2 NEW (4d)2 1 F = F NEW 16 SOLUTIONS 1 3. (5 points) Suppose one night the radius of the earth doubled but its mass stayed the same. What would be an approximate new value for the free-fall acceleration at the surface of the earth? Solution: ME g = G 2 RE ME gNEW = G 2 2RE 1 ME gNEW = G 2 4 RE 1 g = g NEW 4 4. (5 points) Two identical stars, at a fixed distance 2R apart, revolve in a circle about their mutual center of mass, as shown to the right. Each star has mass M and speed v. What is the relation among M,R and v? Solution: Fc = Fgrav v2 M 2 M = G R (2R)2 GM v2 = 4R SOLUTIONS 2 5. (5 points) A flywheel is initially rotating at !0 =20 rad=s and has a constant angular acceleration. After 9.0 s it has rotated through θ=450 rad. What is the angular acceleration α of the wheel? Solution: 2 ∆θ = !0t + 1=2αt ∆θ − ! t α = 2 0 t2 α = 6:7rad=s2 6. (5 points) A wheel of radius R=0.5 m rolls without sliding on a horizontal surface. Starting from rest, the wheel moves with constant angular acceleration α =6 rad=s2. What is the distance traveled by the center of the wheel from t =0 to t =3 s ? Solution: Whenever the problem mentions rolling without sliding or slipping that is a hint for us to use the relation between angular and linear acceleration a = αr and velocity v = !r. a = αr 1 ∆x = x + v t + at2 0 0 2 1 ∆x = 0 + 0 + at2 2 ∆x = 13:5m SOLUTIONS 3 7. (10 points) An amusement park ride consists of a rotating circular platform d=11.4 m in diameter from which 10.0 kg seats are sus- pended at the end of l=1.18 m massless chains. When the system rotates, the chains makes an angle of θ = 43:7◦ with the vertical. What is the speed of each seat? Solution: The radius of the circular motion is given by R = l sin θ + d=2 R = 6:52m From the conditions of the force equilibrium in the horizontal and vertical direction we obtain: v2 T sin θ = m R T cos θ = mg tan θ = v2=gR v = pgR tan θ v = 7:82m=s SOLUTIONS 4 8. (10 points) A curve of radius 80 m is banked at θ = 45◦. Suppose that an ice storm hits, and the curve is effectively frictionless. What is the safe speed with which to take the curve without either sliding up or down? Solution: The simplest approach to this problem is if we consider the force equilibrium in the horizontal and vertical direc- tions. v2 N sin θ = m R N cos θ = mg v = pgR tan θ v = 28m=s SOLUTIONS 5 9. (10 points) A car travels along a reversed banked turn. That is the road is tilted away from the center of curvature of the road. If the coefficient of static friction is µs = 0:5, the radius of the curvature is R = 15 m, and the banking angle is θ = 10◦, what is the maximum speed at which a car can safely navigate such a turn? Solution: The normal reaction force N is due to the sum of the components of the gravity and the centripetal force in the normal to the surface direction. v2 N = (mg cos θ − m sin θ) R v2 µN = mg sin θ + m cos θ R Solving for v we get v2 (µ sin θ + cos θ) = g(µ cos θ − sin θ) R µ cos θ − sin θ v2 = Rg µ sin θ + cos θ v = 6:61m/s v2 Where we used the expressions for the centripetal and friction force Fc = m R and f = µN respectively. SOLUTIONS 6.
Load more

Recommended publications
  • Circular Motion Dynamics
    Problem Solving Circular Motion Dynamics Problem 1: Double Star System Consider a double star system under the influence of gravitational force between the stars. Star 1 has mass m1 and star 2 has mass m2 . Assume that each star undergoes uniform circular motion about the center of mass of the system. If the stars are always a fixed distance s apart, what is the period of the orbit? Solution: Choose radial coordinates for each star with origin at center of mass. Let rˆ1 be a unit vector at Star 1 pointing radially away from the center of mass. Let rˆ2 be a unit vector at Star 2 pointing radially away from the center of mass. The force diagrams on the two stars are shown in the figure below. ! ! From Newton’s Second Law, F1 = m 1a 1 , for Star 1 in the radial direction is m m ˆ 1 2 2 r1 : "G 2 = "m1 r 1 ! . s We can solve this for r1 , m r = G 2 . 1 ! 2s 2 ! ! Newton’s Second Law, F2 = m 2a 2 , for Star 2 in the radial direction is m m ˆ 1 2 2 r2 : "G 2 = "m2 r 2 ! . s We can solve this for r2 , m r = G 1 . 2 ! 2s 2 Since s , the distance between the stars, is constant m2 m1 (m2 + m1 ) s = r + r = G + G = G 1 2 ! 2s 2! 2s 2 ! 2s 2 . Thus the angular velocity is 1 2 " (m2 + m1 ) # ! = $G 3 % & s ' and the period is then 1 2 2! # 4! 2s 3 $ T = = % & .
    [Show full text]
  • Assignment 2 Centripetal Force & Univ
    LAST NAME______________________________ FIRST NAME_____________________DATE______ CJ - Assignment 2 Centripetal Force & Universal Gravitation, 5.4 Banked Curves Draw the vectors for a free body diagram of a car in an unbanked turn and a banked turn. For this situation assume the cars are turning to the left side of the drawing. Car in Unbanked Turn Car in Banked turn Is this car in Equilibrium? Is this car in Equilibrium? Do the same for the airplane. Airplane in Unbanked Turn Airplane in Banked turn Let’s be careful to fully understand what is going on with this one. We have made an assumption in doing this for the airplane which are not valid. THE BANKED TURN- car can make it a round a turn even if friction is not present (or friction is zero). In order for this to happen the curved road must be “banked”. In a non-banked turn the centripetal force is created by the friction of the road on the tires. What creates the centripetal force in a banked turn? Imagine looking at a banked turn as shown to the right. Draw the forces that act on the car on the diagram of the car. Is the car in Equilibrium? YES , NO The car which has a mass of m is travelling at a velocity (or speed) v around a turn of radius r. How do these variables relate to optimize the banked turn. If the mass of the car is greater is a larger or smaller angle required? LARGER, SMALLER, NO CHANGE NECESSARY If the VELOCITY of the car is greater is a larger or smaller angle required? LARGER, SMALLER, NO CHANGE NECESSARY If the RADUS OF THE TURN is greater is a larger or smaller angle required? LARGER, SMALLER, NO CHANGE NECESSARY To properly show the relationship an equation must be created.
    [Show full text]
  • April 2018 the Privileged View Steve Beste, President
    Volume 18 – 04 www.FlyingClub1.org April 2018 The Privileged View Steve Beste, President Maker Faire. When something breaks, who you gonna call? If it’s your car or your Cessna, you put it into the shop and get out your checkbook. But if you fly an experimental or Part 103 aircraft, it’s not so simple. That’s why - when newbies ask me about our sport - I find out if they’re tinkerers and do-it-yourselfers. If all they bring to a problem is their checkbook, then they’re better off going with the Cessna. The tinkerers and builders, though, they’re my kind of people. That’s why I went with Dick Martin to the annual Maker Faire at George Mason in mid-March. I thought we’d find a modest event that focused on 3D printing for adult hobbyists, but I was so wrong. They had 140 exhibitors filling three buildings. With several thousand visitors, there were lines outside the buildings by mid-afternoon. Yes, they had 3D printers, but so much more. And so many kids! The focus of the event was clearly on bringing young people into the world of making stuff. The cleverest exhibit was the one shown below. They set up four tables with junk computer gear and told kids to just take things apart. Don’t worry about putting anything back together again. It’s all junk! Rip it apart! See how it’s made! Do it! It looked like a destruction derby, with laptops and printers and even a sewing machine being reduced to screws and plates.
    [Show full text]
  • FAA-H-8083-3A, Airplane Flying Handbook -- 3 of 7 Files
    Ch 04.qxd 5/7/04 6:46 AM Page 4-1 NTRODUCTION Maneuvering during slow flight should be performed I using both instrument indications and outside visual The maintenance of lift and control of an airplane in reference. Slow flight should be practiced from straight flight requires a certain minimum airspeed. This glides, straight-and-level flight, and from medium critical airspeed depends on certain factors, such as banked gliding and level flight turns. Slow flight at gross weight, load factors, and existing density altitude. approach speeds should include slowing the airplane The minimum speed below which further controlled smoothly and promptly from cruising to approach flight is impossible is called the stalling speed. An speeds without changes in altitude or heading, and important feature of pilot training is the development determining and using appropriate power and trim of the ability to estimate the margin of safety above the settings. Slow flight at approach speed should also stalling speed. Also, the ability to determine the include configuration changes, such as landing gear characteristic responses of any airplane at different and flaps, while maintaining heading and altitude. airspeeds is of great importance to the pilot. The student pilot, therefore, must develop this awareness in FLIGHT AT MINIMUM CONTROLLABLE order to safely avoid stalls and to operate an airplane AIRSPEED This maneuver demonstrates the flight characteristics correctly and safely at slow airspeeds. and degree of controllability of the airplane at its minimum flying speed. By definition, the term “flight SLOW FLIGHT at minimum controllable airspeed” means a speed at Slow flight could be thought of, by some, as a speed which any further increase in angle of attack or load that is less than cruise.
    [Show full text]
  • 42O0kg)(25Ky = 30,000N 25In
    Unit 4: Uniform Circular Motion Name_______Key_______________________ WS-3 Vertical Motion and Banked Turns Date_______________ Block___________ 1. A car travels through a valley at constant speed, though not at constant velocity. a. Explain how this is possible. thereto - its direction e The car is constantly changing it is accelerating b. Is the car accelerating? What direction is the car's acceleration? (Explain how you know.) towards the center of the circle yes , c. Construct a qualitative force diagram for the car at the moment it is at the bottom of the valley. Are the forces balanced? Justify the relative sizes of the forces. ^ F w F T ÷ 8 " mg d. If the car's speed is 25 m/s, its mass is 1200 kg and the radius of valley (r) is 25 meters, determine the magnitude of the centripetal force acting on the car. t.my?42o0kg)(25kY = 30,000N 25in e. Construct a quantitative force diagram for the car at the bottom of the valley. - a N = 42,000 n mg 30,000N = (10%2) ÷ mg 1200kg u 12,000N 2. A car travels over a hill at constant speed. ftw • Img a. Is the car accelerating? What direction is the car's acceleration? (Explain how you know.) The car is accelerating. At the top the acceleration is down. The acceleration is always perpendicular to the tangent at the location b. If the speed of the car is 43 km/h, its mass is 1200 kg and the radius of the hill (r) is 25m, determine the magnitude of the centripetal force acting on the car.
    [Show full text]
  • Department of Transportation Federal Aviation Administration
    Tuesday, November 26, 2002 Part II Department of Transportation Federal Aviation Administration 14 CFR Parts 1, 25, and 97 1-g Stall Speed as the Basis for Compliance With Part 25 of the Federal Aviation Regulations; Final Rule VerDate 0ct<31>2002 14:24 Nov 25, 2002 Jkt 200001 PO 00000 Frm 00001 Fmt 4717 Sfmt 4717 E:\FR\FM\26NOR2.SGM 26NOR2 70812 Federal Register / Vol. 67, No. 228 / Tuesday, November 26, 2002 / Rules and Regulations DEPARTMENT OF TRANSPORTATION Docket you selected, click on the For example, under part 25, V2 must be document number for the item you wish at least 1.2 times VS, the final takeoff Federal Aviation Administration to view. climb speed must be at least 1.25 times You can also get an electronic copy VS, and the landing approach speed 14 CFR Parts 1, 25, and 97 using the Internet through the Office of must be at least 1.3 times VS. [Docket No. 28404; Amendment Nos. 1–49, Rulemaking’s Web page at http:// The speed margin, or difference in 25–108, 97–1333] www.faa.gov/avr/armhome.htm or the speed, between VS and each minimum Federal Register’s Web page at http:// operating speed provides a safety RIN 2120–AD40 www.access.gpo.gov/su_docs/aces/ ‘‘cushion’’ to ensure that normal aces140.html. operating speeds are sufficiently higher 1-g Stall Speed as the Basis for You can also get a copy by submitting than the speed at which the airplane Compliance With Part 25 of the Federal a request to the Federal Aviation stalls.
    [Show full text]
  • 5 | Newton's Laws of Motion 207 5 | NEWTON's LAWS of MOTION
    Chapter 5 | Newton's Laws of Motion 207 5 | NEWTON'S LAWS OF MOTION Figure 5.1 The Golden Gate Bridge, one of the greatest works of modern engineering, was the longest suspension bridge in the world in the year it opened, 1937. It is still among the 10 longest suspension bridges as of this writing. In designing and building a bridge, what physics must we consider? What forces act on the bridge? What forces keep the bridge from falling? How do the towers, cables, and ground interact to maintain stability? Chapter Outline 5.1 Forces 5.2 Newton's First Law 5.3 Newton's Second Law 5.4 Mass and Weight 5.5 Newton’s Third Law 5.6 Common Forces 5.7 Drawing Free-Body Diagrams Introduction When you drive across a bridge, you expect it to remain stable. You also expect to speed up or slow your car in response to traffic changes. In both cases, you deal with forces. The forces on the bridge are in equilibrium, so it stays in place. In contrast, the force produced by your car engine causes a change in motion. Isaac Newton discovered the laws of motion that describe these situations. Forces affect every moment of your life. Your body is held to Earth by force and held together by the forces of charged particles. When you open a door, walk down a street, lift your fork, or touch a baby’s face, you are applying forces. Zooming in deeper, your body’s atoms are held together by electrical forces, and the core of the atom, called the nucleus, is held together by the strongest force we know—strong nuclear force.
    [Show full text]
  • Chapter 2 – Aerodynamics: the Wing Is the Thing
    2-1 Chapter 2 – Aerodynamics: The Wing Is the Thing May the Four Forces Be With You May the Four Forces Be With You 3. [2-2/2/2] When are the four forces that act on an airplane in 1. [2-1/3/2] equilibrium? The four forces acting on an airplane in flight are A. During unaccelerated flight. A. lift, weight, thrust, and drag. B. When the aircraft is accelerating. B. lift, weight, gravity, and thrust. C. When the aircraft is in a stall. C. lift, gravity, power, and friction. 4. [2-2/2/2 & 2-2/3/2] What is the relationship of lift, drag, thrust, and weight 2. [2-1/Figure 1 ] Fill in the four forces: when the airplane is in straight-and-level flight? A. Lift equals weight and thrust equals drag. B. Lift, drag, and weight equal thrust. C. Lift and weight equal thrust and drag. 5. [2-3/2/1] Airplanes climb because of _____ A. excess lift. B. excess thrust. C. reduced weight. 6. [2-4/Figure 5] Lift acts at a _______degree angle to the relative wind. A. 180 B. 360 C. 90 2-2 Rod Machado’s Sport Pilot Workbook Climbs Descents 7. [2-4/Figure 5] Fill in the blanks for the forces in a 10. [2-6/Figure 9] Fill in the blanks for the forces in a climb: descent 8. [2-4/3/3] The minimum forward speed of the airplane is called the _____ speed. A. certified B. stall C. best rate of climb The Wing and Its Things 9.
    [Show full text]
  • Circular Motion
    6.1 – Circular Motion A-Level Physics: 0. Introduction 1. Measurements and their Errors 2. Particles and Radiation 3. Waves and Optics 4. Mechanics and Materials 5. Electricity 6. Further Mechanics and Thermal Physics 6.1 Circular Motion 6.2 Simple Harmonic Motion 6.3 Thermal Physics 7. Fields and their Consequences 8. Nuclear Physics 12. Turning Points in Physics Objects moving in a curved path is just a special case of motion along curved path. The main reasons that we focus on circular motion is that the maths works out nicely and that there are countless examples of objects moving in paths which, if not exactly circular, can be approximated as such. The planets orbiting the Sun and the electron orbiting the atom are two examples. There are plenty of examples of circular motion in our everyday life: a CD spinning in a CD player, food in a microwave, a spinning football, a loop-the-loop at the fairground, a car going around a roundabout – these are all examples of circular motion. There are three main parts to this booklet. In the first five sections, we develop the necessary tools to understand circular motion. Sections 6 to 8 then look at applying these tools to three different types of situation. There is only a small amount of new Physics here, but it can be applied to a vast array of different circumstances. My aim is that it will be easier to see the links between all the different types of problems if we break it down like this. Section 10 onwards develops our ideas of circular motion beyond the A-Level course and is intended for anyone looking for more to quench thirst for more circular motion.
    [Show full text]
  • Single-Engine Failure After Takeoff: the Anatomy of a Turnback Maneuver Les Glatt, Phd., ATP, CFI-AI July 25, 2020
    Single-Engine Failure after Takeoff: The Anatomy of a Turnback Maneuver Les Glatt, PhD., ATP, CFI-AI July 25, 2020 Summary In the latest Advisory Circular AC-83J, Appendix A.11.4 provides guidance to all CFI’s relating to the subject: “Return to Field/Single-Engine Failure on Takeoff”, i.e., (1) “Flight instructors should demonstrate and teach trainees when and how to make a safe 180-degree turnback to the field after an engine failure”. (2) “Flight instructors should also teach the typical altitude loss for the given make and model flown during a 180-degree turn, while also teaching the pilot how to make a safe, coordinated turn with a sufficient bank. These elements should give the pilot the ability to determine quickly whether a turnback will have a successful outcome. During the before-takeoff check, the expected loss of altitude in a turnback, plus a sufficient safety factor, should be briefed and related to the altitude at which this maneuver can be conducted safely. In addition, the effect of existing winds on the preferred direction and the viability of a turnback should be considered as part of the briefing”. The question that is most asked by Pilots about the turnback maneuver is “How high above the runway do I need to be before attempting a turnback maneuver?” In a recent NAFI webinar Capt Brian Schiff came up with a Rule-of-Thumb (ROT) regarding this altitude. Schiff’s ROT is based on a 13-step flight experiment that the Pilot was required to perform. It required the Pilot to determine the altitude loss during a 45-degree banked gliding turn, at a speed no faster than best glide speed (or slightly lower), and after turning 360 degrees, the aircraft is rolled to a wings level attitude, and then flaring the aircraft to the point where the vertical speed went to zero.
    [Show full text]
  • Flight Dynamics-I Prof. EG Tulapurkara Chapter-9
    Flight dynamics-I Prof. E.G. Tulapurkara Chapter-9 Chapter- 9 Performance analysis – V- Manoeuvres (Lectures 28 to 31) Keywords : Flights along curved path in vertical plane – loop and pull out ; load factor ; steady level co-ordinated-turn - minimum radius of turn, maximum rate of turn; flight limitations ; operating envelop; V-n diagram Topics 9.1 Introduction 9.2 Flight along a circular path in a vertical plane (simplified loop) 9.2.1 Equation of motion in a simplified loop 9.2.2 Implications of lift required in a simplified loop 9.2.3 Load factor 9.2.4 Pull out 9.3 Turning flight 9.3.1 Steady, level, co-ordinated-turn 9.3.2 Equation of motion in steady, level, co-ordinated-turn 9.3.3 Factors limiting radius of turn and rate of turn 9.3.4 Determination of minimum radius of turn and maximum rate of turn at a chosen altitude 9.3.5 Parameters influencing turning performance of a jet airplane 9.3.6 Sustained turn rate and instantaneous turn rate 9.4 Miscellaneous topics – flight limitations, operating envelop and V-n diagram 9.4.1 Flight limitations 9.4.2 Operating envelop 9.4.3 V-n diagram Exercises Dept. of Aerospace Engg., Indian Institute of Technology, Madras 1 Flight dynamics-I Prof. E.G. Tulapurkara Chapter-9 Chapter 9 Lecture 28 Performance analysis V – Manoeuvres – 1 Topics 9.1 Introduction 9.2 Flight along a circular path in a vertical plane (simplified loop) 9.2.1 Equation of motion in a simplified loop 9.2.2 Implications of lift required in a simplified loop 9.2.3 Load factor 9.2.4 Pull out 9.3 Turning flight 9.3.1 Steady, level, co-ordinated-turn 9.3.2 Equation of motion in steady, level, co-ordinated-turn 9.1 Introduction Flight along a curved path is known as a manoeuvre.
    [Show full text]
  • Dynamics Applying Newton's Laws Rotating Frames
    Dynamics Applying Newton's Laws Rotating Frames Lana Sheridan De Anza College Feb 5, 2020 Last time • accelerated frames: linear motion Overview • accelerated frames: rotating frames Rotating Frames 6.3 Motion in Accelerated Frames 159 Consider a car making a turn to the left. force has acted on the puck to cause it to accelerate. We call an apparent force such as this one a fictitious force because it is not a real force and is due only to observa- tions made in an accelerated reference frame. A fictitious force appears to act on an object in the same way as a real force. Real forces are always interactions between two objects, however, and you cannot identify a second object for a fictitious force. (What second object is interacting with the puck to cause it to accelerate?) In gen- eral, simple fictitious forces appear to act in the direction opposite that of the acceler- ation of the noninertial frame. For example, the train accelerates forward and there appears to be a fictitious force causing the puck to slide toward the back of the train. a The train example describes a fictitious force due to a change in theWe train’s know that there is a force (the centripetal force) towards the speed. Another fictitious force is due to the change in the direction of thecenter veloc of the- circle. ity vector. To understand the motion of a system that is noninertial because of a From the passenger’s frame of change in direction, consider a car traveling along a highway at a high speed and reference, a force appears to push her toward the right door, but it is approaching a curved exit ramp on the left as shown in Figure 6.10a.
    [Show full text]