Center of Gravity We have said that for rigid bodies, all of the forces act at the centre of mass. This is a normally a very good approximation, but strictly speaking, the forces act at the centre of gravity, not the centre of mass. The location of the center of gravity is defined by: x CG n i1 mgx i i i mg i i Page 7
Center of Gravity The center of gravity of an object coincides with its center of mass if the gravitational acceleration, g, is uniform over the entire object. This is the case for most objects. If gi g constant x gmx mx i i i i 1 mx CG M i i g m m i i x CM Page 8
APSC 111 Course Outline Chapters 1-1 1. Vectors. Kinematics (Linear and Rotational) 3. Dynamics (Newtons Laws, with friction, springs ) 4. Work and Energy (Conservation of energy, KE, PE) 5. Linear Momentum and collisions (elastic/inelastic) 6. Rotational Dynamics (rotations, torques, rolling) 7. Equilibrium (static equilibrium) Did not cover varying mass, angular momentum, elasticity Page 9
APSC 111 Physics I Grades: Online Assignments 7% Class Participation 3% Mechanics Background Quiz % By Dec 5,11PM Mid-Term Exam I 19% Mid-Term Exam II 19% Final Exam 50% All grades (other than final exam) will be posted to moodle next week. Please let me know ASAP of any problems. APSC APSC 111 111 Intro/Vectors Review Page 10
Advice for studying 1. This is a problem-based course. Practice solving problems, make sure you understand and can solve the assignment and tutorial problems. Looking at the solution is not nearly as useful as solving a problem on your own!. Old exams, course manual and end of chapter problems are good sources of sample problems to practice on. (Be careful, rotational dynamics was NOT part of APSC 111 last year). 3. The following slides summarize some of the main/important topics from the course. In general you are not expected to memorize equations or results (you should be familiar with the formula sheet on moodle, which will be provided with the exam), but you should understand how to apply concepts we have studied. Page 11
A set of 55 end-of-chapter problems have been added to moodle for you to practice on (in addition to tutorials and Mastering Physics). Approximate level for exam is or 3 dot problems. These are meant to be representative and span most of the topics we ve covered (but may not be complete!) Solutions to those will be posted next week, please attempt the problems before looking at the solutions! Solutions to all tutorial problems will be posted by end of this week. Midterm with answers are now on moodle. Page 1
New formula sheet is on moodle You should be familiar with It! Page 13
Formula sheet, page Page 14
Advice for writing the exam 1. Read the whole problem and make sure you understand it! This is very important!. Even if the problem appears too complex, all of the simple rules we learned still hold! Think about how to simplify the problem to be able to apply the rules you ve learned! 3. Draw a diagram and choose coordinate axes. 4. Decide what physics applies and then look at which equations are available to use (ie constant acceleration, conservation of energy, momentum, etc.) 5. Check your solution: estimate the answer if possible and make sure your solution is reasonable. If time permits you can sometimes cross-check your result with supporting calculations (ie total energy, etc.) Make sure the units are correct! 6. You don t need to solve the exam in order. You may want to start with easiest questions first, and work in order of increasing difficulty. Page 15
Q0 Bob is looking at Jane. Jane is looking at Larry. Bob is married. Larry is not. Is a married person looking at an unmarried person? 0% 10% 71% 1. Yes. No 3. Not enough information Page 16
Final exam is 3 hours long, with (approximately) 15 multiple-choice questions 30 marks 6 long-answer questions 60 marks The questions are comprehensive (all course material). Approximately ½ of the questions will be the new material since Midterm Answer ALL questions, including multiple-choice!! Draw diagrams, show your work! Page 17
(35% correct answers) Page 18
s FBD for B N B m B m A When not slipping: F m m a A B Both blocks have same a F m B mg B F f F 0 N m g y B B N B m g F N m g f s B s B F maf x B f a g s B Page 19
9. A cart loaded with sand slides forward along a horizontal frictionless track. As the cart moves, sand trickles out at a constant rate through a hole in the back of the cart. The acceleration of the cart is: A) constant and in the forward direction B) constant and in the backward direction C) variable and in the forward direction D) variable and in the backward direction E) zero v (1% correct answers) When grain falls out, horizontal velocity unchanged: v c Before this grain falls: p mvmv xt g c grain: cart + rest of sand: v pxt mv g mv c c v v a 0 c Page 0
y (a) Remember, for projectile motion we usually find t from the vertical motion: 1 1 v0 y v y0 v0 yt gt 0 t v0 y gt) 0 t g ( 0 sin g s range R v0xt v0 cost 0 m Page 1
range R 0 m (b) Long way: slick way: Find y and v y at x = 4.0 m, then start a new projectile problem there with speed v x,v y Realize that above will lead the mirror image of the original projectile motion Page
V, V x y V, V x y 4m R 0m Page 3
(text problem 8-7 dots ) Page 4
a), b) mg T y cos TMax T mg T 933N v Fy T mtgcos mt r v T mtgcos mt r v TMax mt gmt r When θ = 0 K 0 1 K mv U i i m gh U 0 T mv 1 T T v gh f f mgh T Page 5
c) Before collision: p mv m gh T T T After collision: p ( m m ) v T J ( m m ) v m v v T J T T mt v ( m m ) T J T 4.11 m/s 1 K ( ) i mt mj v Ui 0 K f 0 U ( m m ) gh f T J v gh h 0.86 m Page 6
d) T J y F T Mgcos F y y M m m v M r T Mgcos h cos 0 h cos 0.95 h T Mg Mg cos T 16N h Page 7
Collisions In a collision, the total kinetic energy may or may not be conserved (since KE may transformed into other forms). We divide collisions into two classes: Elastic collision: total kinetic energy is conserved Inelastic collision: total kinetic energy is NOT conserved Linear Momentum: always conserved when comparing just before and just after the collision. Page 8
We also learned that a system of particles always behaves as though all forces acts on the Centre of Mass, even if the particles break apart Page 9
We introduced the basic variables for rotations: Summary of Rotational variablee,, angle angular velocity angular acceleration, same every where in object!!! Which are related to instantaneous linear variables as: Depend on location s v a t r r r va, t r, s Page 30
Summary of Equations of motion (1) o t v v at o () t t o o 1 x x v t at o o 1 (3) ( ) 1 t o o 1 x xo vvo t ( ) (4) o ( o) v v a( xx ) o o Angular Rotation These equations will be used extensively: Refer to them often. Linear Motion Note that each one differs by being independent of a different variable, so solving problems will be easier if the correct equation is used. Page 31
Nonuniform Circular Motion If an object is moving in a circular path but at varying speeds, it must have a tangential component to its acceleration as well as the radial one. Page 3
r v r v C r r r r C r add r r C r r r Rotation only r v r v C Result View from Floor v C C v 0 v C Wheel seems to rotate about instantaneous point of rest Instantaneous point at rest. (No slipping) Page 33
Conservation of Energy for Rotating Bodies: General Form: ( K U) ( K U) W W final initial ext fric K I Mv 1 1 cm cm External work (done on body or system) Work done by friction (Always negative) Where Moment of Inertia I defines Mass distribution in rotations I I m r P body p p r dm Solid bodies Discreet Masses or I body r dv Parallel axis theorem I I Mh h cm Page 34
10.7: Calculating the Rotational Inertia If a rigid body consists of a great many adjacent particles (it is continuous, like a Frisbee), we consider an integral and define the rotational inertia of the body as Page 35
Torque The perpendicular distance from the axis of rotation to the line along which the force acts is called the lever arm. We define the magnitude of the torque in this case as: RF Page 36
Application of Newton s Second Law to rotational problems: Step 1: Draw a free body diagram for each body in the system. Define the coordinate system and direction of torque. Step : Apply second Law (in both forms if required): Step 3: Find the Kinematic relations. Step 4: Solve!! F ma cm I cm Icm For rotations with translation Note: This is an alternative approach to using conservation of energy technique we learned previously. Either technique should work. Which you use will depend on personal preference, ease of application. Page 37
Work and Rotational Kinetic Energy Work-kinetic energy theorem W K ( I Mv ) ( I Mv ) 1 1 1 1 Net cm cm f cm cm i Wext External Work Terms F d W ext f i Power P dw FV or About a fixed axis dt Page 38
10.10: Work and Rotational Kinetic Energy Page 39
Conditions for Equilibrium In order for an object to be in equilibrium: The vector sum of all external forces acting on the object must be zero. F 0 F 0 F 0 x y z The vector sum of all torques acting on the object must be zero. 0 0 0 x y z Page 40
Conditions for Equilibrium We will normally have problems in which all of the forces act in a plane. (Say x and y). Then F 0 F 0 x y In this case, all of the torques have to be along a z axis (They are always perpendicular to force and moment arm, which are in the x-y plane.) Therefore: z 0 The torque can be about any convenient axis! A clever choice can simplify the problem significantly. Page 41
Good Luck!! Page 4