Problem Set 5: Universal Law of Gravitation; Circular Planetary Orbits

Similar documents
Physics 111 Lecture 5 Circular Motion

Physics 4A Chapter 8: Dynamics II Motion in a Plane

Physics 101 Lecture 6 Circular Motion


Physics Courseware Electromagnetism

- 5 - TEST 1R. This is the repeat version of TEST 1, which was held during Session.

Uniform Circular Motion

Phys 201A. Homework 5 Solutions

Circular Motion & Torque Test Review. The period is the amount of time it takes for an object to travel around a circular path once.

Quiz 6--Work, Gravitation, Circular Motion, Torque. (60 pts available, 50 points possible)

Circular-Rotational Motion Mock Exam. Instructions: (92 points) Answer the following questions. SHOW ALL OF YOUR WORK.

Rotational Motion: Statics and Dynamics

Circular Motion. Mr. Velazquez AP/Honors Physics

Physics 201 Homework 4

The Laws of Motion ( ) N SOLUTIONS TO PROBLEMS ! F = ( 6.00) 2 + ( 15.0) 2 N = 16.2 N. Section 4.4. Newton s Second Law The Particle Under a Net Force

Chapter 4. Newton s Laws of Motion

Chapter 5. Uniform Circular Motion. a c =v 2 /r

Physics 201 Lecture 18

constant t [rad.s -1 ] v / r r [m.s -2 ] (direction: towards centre of circle / perpendicular to circle)

Dynamics of Rotational Motion

Physics 2001 Problem Set 5 Solutions

ω = θ θ o = θ θ = s r v = rω

Phys 201A. Homework 6 Solutions. F A and F r. B. According to Newton s second law, ( ) ( )2. j = ( 6.0 m / s 2 )ˆ i ( 10.4m / s 2 )ˆ j.

AY 7A - Fall 2010 Section Worksheet 2 - Solutions Energy and Kepler s Law

Describing Circular motion

Chapter 7-8 Rotational Motion

Spring 2001 Physics 2048 Test 3 solutions

06 - ROTATIONAL MOTION Page 1 ( Answers at the end of all questions )

Chap 5. Circular Motion: Gravitation

From Newton to Einstein. Mid-Term Test, 12a.m. Thur. 13 th Nov Duration: 50 minutes. There are 20 marks in Section A and 30 in Section B.

Physics 1114: Unit 5 Hand-out Homework (Answers)

Exam 3: Equation Summary

( ) ( ) Review of Force. Review of Force. r = =... Example 1. What is the dot product for F r. Solution: Example 2 ( )

Objective Notes Summary

Lab 10: Newton s Second Law in Rotation

Chapter 5: Uniform Circular Motion

Section 26 The Laws of Rotational Motion

Gravitation. Chapter 12. PowerPoint Lectures for University Physics, Twelfth Edition Hugh D. Young and Roger A. Freedman. Lectures by James Pazun


DYNAMICS OF UNIFORM CIRCULAR MOTION

b) (5) What average force magnitude was applied by the students working together?

Physics 181. Assignment 4

r cos, and y r sin with the origin of coordinate system located at

Name. Date. Period. Engage Examine the pictures on the left. 1. What is going on in these pictures?

to point uphill and to be equal to its maximum value, in which case f s, max = μsfn

= 4 3 π( m) 3 (5480 kg m 3 ) = kg.

Physics 107 TUTORIAL ASSIGNMENT #8

Understanding the Concepts

Circular Motion. x-y coordinate systems. Other coordinates... PHY circular-motion - J. Hedberg

Sections and Chapter 10

OSCILLATIONS AND GRAVITATION

PROJECTILE MOTION. At any given point in the motion, the velocity vector is always a tangent to the path.

Chapter 13 Gravitation

Physics 111. Lecture 14 (Walker: Ch. 6.5) Circular Motion Centripetal Acceleration Centripetal Force February 27, 2009

PHYS Summer Professor Caillault Homework Solutions. Chapter 5

15 B1 1. Figure 1. At what speed would the car have to travel for resonant oscillations to occur? Comment on your answer.

Physics Fall Mechanics, Thermodynamics, Waves, Fluids. Lecture 6: motion in two and three dimensions III. Slide 6-1

Conservation of Angular Momentum

Easy. P4.2 Since the car is moving with constant speed and in a straight line, the. resultant force on it must be regardless of whether it is moving

Extra notes for circular motion: Circular motion : v keeps changing, maybe both speed and

Centripetal Force. Lecture 11. Chapter 8. Course website:

PS113 Chapter 5 Dynamics of Uniform Circular Motion

Experiment 09: Angular momentum

Recap. Centripetal acceleration: v r. a = m/s 2 (towards center of curvature)

Tutorial on Strehl ratio, wavefront power series expansion, Zernike polynomials expansion in small aberrated optical systems By Sheng Yuan

Physics C Rotational Motion Name: ANSWER KEY_ AP Review Packet

Multiple choice questions [100 points] As shown in the figure, a mass M is hanging by three massless strings from the ceiling of a room.

Chapter 12. Kinetics of Particles: Newton s Second Law

c) (6) Assuming the tires do not skid, what coefficient of static friction between tires and pavement is needed?

Midterm Exam #2, Part A

Physics 11 Chapter 4: Forces and Newton s Laws of Motion. Problem Solving

Chapter 4: The laws of motion. Newton s first law

Chapter 5. really hard to start the object moving and then, once it starts moving, you don t have to push as hard to keep it moving.

Uniform Circular Motion

Circular Orbits. and g =

PHYSICS 220. Lecture 08. Textbook Sections Lecture 8 Purdue University, Physics 220 1

ESTIMATION MODELS USING MATHEMATICAL CONCEPTS AND NEWTON S LAWS FOR CONIC SECTION TRAJECTORIES ON EARTH S SURFACE

Rotational Motion. Every quantity that we have studied with translational motion has a rotational counterpart

Graphs of Sine and Cosine Functions

Exam 3: Equation Summary

Magnetic Field. Conference 6. Physics 102 General Physics II

10. Force is inversely proportional to distance between the centers squared. R 4 = F 16 E 11.

PHYS 1114, Lecture 21, March 6 Contents:

AH Mechanics Checklist (Unit 2) AH Mechanics Checklist (Unit 2) Circular Motion

b) (5) What is the magnitude of the force on the 6.0-kg block due to the contact with the 12.0-kg block?

Chapter. s r. check whether your calculator is in all other parts of the body. When a rigid body rotates through a given angle, all

Translation and Rotation Kinematics

Chapter 8. Accelerated Circular Motion

Chapter 5 Force and Motion

Chapter 5 Force and Motion

Chapter 1 Functions and Graphs

Circular motion. Objectives. Physics terms. Assessment. Equations 5/22/14. Describe the accelerated motion of objects moving in circles.

Chapter 2: Basic Physics and Math Supplements

SAMPLE QUIZ 3 - PHYSICS For a right triangle: sin θ = a c, cos θ = b c, tan θ = a b,

PHYS 1410, 11 Nov 2015, 12:30pm.

CIRCULAR MOTION. Particle moving in an arbitrary path. Particle moving in straight line

Between any two masses, there exists a mutual attractive force.

Force can be exerted by direct contact between bodies: Contact Force.

Unit 6 Practice Test. Which vector diagram correctly shows the change in velocity Δv of the mass during this time? (1) (1) A. Energy KE.

UNIVERSITY OF SASKATCHEWAN Department of Physics and Engineering Physics

Transcription:

Poblem Set 5: Univesal Law of Gavitation; Cicula Planetay Obits Design Engineeing Callenge: Te Big Dig.007 Contest Evaluation of Scoing Concepts: Spinne vs. Plowe PROMBLEM 1: Daw a fee-body-diagam of a spee in a ole, like is used to old te sot-puts and te balls. Figue 1: Diagam of a spee in a ole F N mg Nv Figue : FBD of spee in a ole. Assume no fiction ove line contact.

PROMBLEM : Te foce condition fo making a ball o sot-put to just ise up out of te ole. Summing te foces in te -diection fist wit igt as te positive diection: F = + 0 = F => F = N N (1) Summing te foces in te y-diection wit up as te positive diection: + Fy 0 mg N v () = = + => Nv = mg Lastly summing moments aound te contact point: M contact _ point = 0 = F ( ) + mg (3) => F = mg Note fo 0<<, te above moment am fo N v was obtained fom Figue 3, wic esults in equation (4). - Figue 3: Dimensions of spee. + = ( ) => = (4)

PROBLEM 3: Te foce condition fo making a ball sot-put to leave te ole and stat olling acoss te platte. Tis poblem is just like poblem ecept tee is an acceleation tem. Summing te foces in te -diection fist wit igt as te positive diection: F = + ma = F N => F = ma + N Summing te foces in te y-diection wit up as te positive diection: (5) + Fy 0 mg N v (6) = = + => Nv = mg Lastly summing moments aound te contact point: M = 0 = F ( ) + mg contact _ point => F = ma + mg wee te moment am fo N v was obtained fom Figue 3. (7) Note te diffeence between Equations (3) and (9) is te latte as an additional tem of ma.

PROBLEM 4: Angula velocity of te platte must be acieved in ode to meet te foce conditions in (3)? Fo foce conditions in poblem (3), we need te centifugal foce. v = = = (8) F Fcentifigual maadial m R but te linea velocity is equal to te angula velocity times te distance between te cente of platte to cente of ball wic is denoted by R, efe to Figue 4. Tus equation (8) becomes: ( ωr) Fcentifigual = m R F m R => = ω (9) Substituting fo F found in poblem 3 into equation (9) poduces: a + g = ω R a => ω = + g R R ( ) (10) R Figue 4: Distance between cente of platte to cente of sot-put.

PROBLEM 5: How ad would a lasso ave to pull (o a blade to pus) in ode to meet te foce conditions in (3)? Te amount of foce equied would be te same as equation (7): F = ma + mg (11) PROBLEM 6: Wat is a bette concept fo libeating te sot-puts o ockey balls, spinning te platte o pulling o pusing tem off? Compaing equations (3), (7), and (11), te diection acceleation needs to be consideed fo libeating te ball wile pusing o pulling it off wee wen otating, te moment of inetia needs to be known. Tus pusing te ball equies less foce, but as te platte is spun wit te pope angula velocity, moe balls fly off fo tat amount of foce. PROBLEM 7: How do consideations of macine design compleity and feasibility affect te oveall best concept? Spinning te Platte: a. Can te motos of te ca andle te amount of toque necessay to get te angula velocity needed? If sot-put balls can not be spun off, instead could te ockey balls be? Ten pus te sot-put off te platte? b. Wic is quicke wen emoving te balls fom te platte, spin o pus/pull? Pulling/pusing: a. Can te motos andle pusing te sot-put off te platte plus te ockey balls in one effot o ae multiple effots needed?

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Pysics Depatment Pysics 8.01T Fall Tem 003 Ball Coming Out of Socket Hee s ow solve poblem of a ball coming out of its socket on a otating table witout using non-intetial coodinate systems o knowing anyting about moments. Te figue below sows a ball just about to come out, in te sense tat it makes contact wit te table at only one point on te cicumfeence of te socket. Te ball as adius a and te socket as adius. a N N tan θ mg Te figue sows a fee body diagam wit all te foces acting on te ball sown in ed. Newton s nd law fo te vetical ais is N sin θ + N tan cos θ = mg wee θ = cos 1 ( a ) Te oizontal foce components can poduce a maimum centipetal foce F c = N cos θ N tan sin θ If R is te adius of te ball s cicula pat, te ball stats to come out wen mrω mrω ma = N cos θ N tan sin θ If tee is no fiction, N tan = 0 and we obtain ω ma = Te ole of fiction is discussed on te net page. g R tan θ J. D. Litste 1 Octobe 1, 003

Te ole of fiction can be undestood if we tink in a little moe detail about ow te ball escapes wen tee is no fiction. As soon as ω > ω ma te ball stats to slide up te edge of te socket. Tat inceases θ, and teefoe educes te maimum centipetal foce N cos θ. Tis is an unstable situation so te ball will escape. Because te ball slides up te edge of te socket as it escapes, any fictional foce must act in te opposite diection to N tan as dawn in te figue. Teefoe N (sin θ µ cos θ) = mg Rωma = N (cos θ + µ sin θ) m cos θ + µ sin θ = sin θ µ cos θ g = 1 + µ tan θ tan θ µ g d dµ (Rω ma) = 1 + tan θ (tan θ µ) g So you can see tat adding some fiction makes it ade to get te ball out, in te sense tat ω ma inceases. Tat is consistent wit ou intuitive epectation. Howeve, tee is anote possibility: tat te ball will oll ate tan slide ove te edge of te socket. In tat case, te value fo ω c is te same as fo te fictionless case. Tis means tat only effect of fiction is to make te ball oll out ate tan slide out of te socket. It does not cange te citical otation speed. If you ae not sue you believe tis, I found te easiest way to sow it was to go into te otating coodinate system and calculate te toques of te weigt and of te centifugal foce about te contact point at te edge of te socket. J. D. Litste Octobe 1, 003