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Physics 111: Lecture 13 Today’s Agenda

Physics 111: Lecture 13 Today’s Agenda. Potential Energy & Force Systems of Particles Center of mass Velocity and acceleration of the center of mass Dynamics of the center of mass Linear Momentum Example problems. Potential Energy & Force.

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Physics 111: Lecture 13 Today’s Agenda

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  1. Physics 111: Lecture 13Today’s Agenda • Potential Energy & Force • Systems of Particles • Center of mass • Velocity and acceleration of the center of mass • Dynamics of the center of mass • Linear Momentum • Example problems

  2. Potential Energy & Force • For a conservative force we define the potential energy function: • Therefore: • Consider some potential energy functions we know, and find the forces: Spring: Gravity near earth: Newton’s Gravity: It’s true!!

  3. Potential Energy Diagrams • Consider a block sliding on a frictionless surface, attached to an ideal spring. m x U x 0

  4. m Potential Energy Diagrams • Consider a block sliding on a frictionless surface, attached to an ideal spring. • F = -dU/dx = -slope F x U F x x 0

  5. Potential Energy Diagrams • The potential energy of the block is the same as that of an object sliding in a frictionless “bowl”: Ug = mgy = 1/2 kx2= Us m U x is the height of an object in the bowl at position x 0

  6. Equilibrium • F = -dU/dx = -slope • So F = 0 if slope = 0. • This is the case at the minimumormaximum of U(x). • This is called an equilibrium position. • If we place the block at rest at x = 0, it won’t move. m x U x 0

  7. Equilibrium • If small displacements from the equilibrium position result in a force that tends to move the system back to its equilibrium position, the equilibrium is said to be stable. • This is the case if U is a minimum at the equilibrium position. • In calculus language, the equilibrium is stable if the curvature (second derivative) is positive. F m x U F x 0

  8. Equilibrium Balance cone Birds U • Suppose U(x) looked like this: • This has two equilibrium positions, one is stable (+ curvature) and one is unstable (- curvature). • Think of a small object sliding on the U(x) surface: • If it wants to keep sliding when you give it a little push, the equilibrium is unstable. • If it returns to the equilibrium position when you give it a little push, the equilibrium is stable. • If the curvature is zero (flat line) the equilibrium is neutral. unstable neutral stable x 0

  9. System of Particles • Until now, we have considered the behavior of very simple systems (one or two masses). • But real life is usually much more interesting! • For example, consider a simple rotating disk. • An extended solid object (like a disk) can be thought of as a collection of parts. The motion of each little part depends on where it is in the object!

  10. m2 m1 r2 r1 RCM m3 y r3 x m4 r4 (In this case, N = 4) System of Particles: Center of Mass Ice table • How do we describe the “position” of a system made up of many parts? • Define the Center of Mass (average position): • For a collection of N individual pointlike particles whose masses and positions we know:

  11. r2 - r1 m2 m1 RCM r2 r1 So: y where M = m1 + m2 x System of Particles: Center of Mass • If the system is made up of only two particles:

  12. If m1 = m2 the CM is halfway between the masses. System of Particles: Center of Mass • If the system is made up of only two particles: where M = m1 + m2 r2 - r1 + m2 m1 RCM r2 r1 y x

  13. System of Particles: Center of Mass • If the system is made up of only two particles: where M = m1 + m2 If m1 = 3m2 r2 - r1 + m2 m1 RCM r2 the CM is now closer to the heavy mass. r1 y x

  14. + + m2 m1 m1 m2 System of Particles: Center of Mass Baton • The center of mass is where the system is balanced! • Building a mobile is an exercise in finding centers of mass.

  15. m2 m1 r2 r1 RCM m3 y r3 x m4 r4 (In this case, N = 4) System of Particles: Center of Mass • We can consider the components of RCM separately:

  16. Example Calculation: • Consider the following mass distribution: 2m (12,12) m m (0,0) (24,0) RCM = (12,6)

  17. ò ò r dm r dm = = R CM ò dm M System of Particles: Center of Mass • For a continuous solid, we have to do an integral. dm r y where dm is an infinitesimal mass element. x

  18. System of Particles: Center of Mass • We find that the Center of Mass is at the “center” of the object. y RCM x

  19. y x System of Particles: Center of Mass • We find that the Center of Mass is at the “center” of the object. The location of the center of mass is an intrinsic property of the object!! (it does not depend on where you choose the origin or coordinates when calculating it). RCM

  20. System of Particles: Center of Mass • We can use intuition to find the location of the center of mass for symmetric objects that have uniform density: • It will simply be at the geometrical center ! + + CM + + + +

  21. + + System of Particles: Center of Mass Pisa Bottle • The center of mass for a combination of objects is the average center of mass location of the objects: m2 R2 - R1 + R2 so if we have two objects: RCM R1 m1 y x

  22. Lecture 13, Act 1Center of Mass • The disk shown below (1) clearly has its CM at the center. • Suppose the disk is cut in half and the pieces arranged as shown in (2): • Where is the CM of (2) as compared to (1)? (a)higher(b)lower(c)same XCM (1) (2)

  23. The CM of the compound object will be halfway between theCMs of the two halves. • This is higher than the CM of the disk X XCM Lecture 13, Act 1Solution • The CM of each half-disk will be closer to the fat end than to the thin end (think of where it would balance). X X (1) (2)

  24. System of Particles: Center of Mass Double cone • The center of mass (CM) of an object is where we can freely pivot that object. • Gravity acts on the CM of an object (show later) • If we pivot the objectsomewhere else, it willorient itself so that theCM is directly below the pivot. • This fact can be used to findthe CM of odd-shaped objects. pivot + CM pivot pivot + + CM CM mg

  25. System of Particles: Center of Mass Odd shapes • Hang the object from several pivots and see where the vertical lines through each pivot intersect! pivot pivot pivot + CM • The intersection point must be at the CM.

  26. 3 pronged object Lecture 13, Act 2Center of Mass Fork, spoon, and match • An object with three prongs of equal mass is balanced on a wire (equal angles between prongs). What kind of equilibrium is this position? a) stable b) neutral c) unstable

  27. Lecture 13, Act 2Solution If the object is pushed slightly to the left or right, its center of mass will not be above the wire and gravity will make the object fall off The center of mass of the object is at its center and is initially directly over the wire CM CM + + mg mg (front view)

  28. Lecture 13, Act 2Solution • Consider also the case in which the two lower prongs have balls of equal mass attached to them: + CM + CM mg mg In this case, the center of mass of the object is below the wire When the object is pushed slightly, gravity provides a restoring force, creating a stable equilibrium

  29. Velocity and Accelerationof the Center of Mass • If its particles are moving, the CM of a system can also move. • Suppose we know the position riof every particle in the system as a function of time. So: And: • The velocity and acceleration of the CM is just the weighted average velocity and acceleration of all the particles.

  30. p v Linear Momentum: • Definition: For a single particle, the momentump is defined as: • (p is a vector since v is a vector). p= mv • So px = mvxetc. • Newton’s 2nd Law: F = ma dv • Units of linear momentum are kg m/s.

  31. Linear Momentum: • For a system of particles the total momentum P is the vector sum of the individual particle momenta: But we just showed that So

  32. Linear Momentum: • So the total momentum of a system of particles is just the total mass times the velocity of the center of mass. • Observe: • We are interested in so we need to figure out

  33. Linear Momentum: • Suppose we have a system of three particles as shown. Each particle interacts with every other, and in addition there is an external force pushing on particle 1. m3 F31 F32 F13 F23 F12 m1 F21 (since the other forces cancel in pairs...Newton’s 3rd Law) m2 F1,EXT All of the “internal” forces cancel !! Only the “external” force matters !!

  34. Linear Momentum: • Only the total external force matters! m3 Which is the same as: m1 m2 F1,EXT Newton’s 2nd law applied to systems!

  35. Center of Mass Motion: Recap Pork chop • We have the following law for CM motion: • This has several interesting implications: • It tells us that the CM of an extended object behaves like a simple point mass under the influence of external forces: • We can use it to relate F and A like we are used to doing. • It tells us that if FEXT = 0, the total momentum of the system can not change. • The total momentum of a system is conserved if there are no external forces acting. Pendulum

  36. Example: Astronauts & Rope • Two astronauts at rest in outer space are connected by a light rope. They begin to pull towards each other. Where do they meet? m M = 1.5m

  37. Example: Astronauts & Rope... m M = 1.5m • They start at rest, so VCM = 0. • VCM remains zero because • there are no external forces. • So, the CM does not move! • They will meet at the CM. CM L x=L x=0 Finding the CM: If we take the astronaut on the left to be at x = 0:

  38. Lecture 13, Act 3Center of Mass Motion • A man weighs exactly as much as his 20 foot long canoe. • Initially he stands in the center of the motionless canoe, a distance of 20 feet from shore. Next he walks toward the shore until he gets to the end of the canoe. • What is his new distance from the shore. (There no horizontal force on the canoe by the water). 20 ft (a)10 ft(b)15 ft(c)16.7 ft before 20 ft ? ft after

  39. Since the man and the canoe have the same mass, the CM of the man-canoe system will be halfway between the CM of the man and the CM of the canoe. • Initially the CM of the system is 20 ft from shore. X X 20 ft CM of system Lecture 13, Act 3Solution x

  40. Therefore, the man ends up 5 ft to the left of the system CM, and the center of the canoe ends up 5 ft to the right. • He ends up moving 5 ft toward the shore (15 ft away). 10 ft 15 ft 5 ft Lecture 13, Act 3Solution • Since there is no force acting on the canoe in the x-direction, thelocation of the CM of the system can’t change! X X x 20 ft CM of system

  41. Recap of today’s lecture • Systems of particles (Text: 8-1) • Center of mass (Text: 8-1 & 12-6) • Velocity and acceleration of the center of mass (Text:8-3) • Dynamics of the center of mass (Text: 8-3 to 8-4) • Linear Momentum • Example problems • Look at textbook problems Chapter 8: # 3, 7, 17, 29, 35, 77, 111

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