Center of Mass

1
Center of Mass Linear Momentum

• Physics
• Montwood High School
• R. Casao

2
Center of Mass

• We have discussed parabolic trajectories using a
  particle as the model for objects.
• But clearly objects are not particles they are
  extended and may have complicated shapes mass
  distributions.
• So if we toss something like a baseball bat into
  the air (spinning and rotating in a complicated
  way), what can we really say about its
  trajectory?
• There is one special location in every object
  that provides us with the basis for our earlier
  model of a point particle.
• That special location is called the center of
  mass.
• The center of mass will follow a parabolic
  trajectory even if the rest of the bats motion
  is very complicated

3
Center of Mass

• To start, lets suppose that we have two masses
  m1 and m2, separated by some distance d.
• We have also arbitrarily aligned the origin of
  our coordinate system to be the center of mass
  m1.
• We define the center of mass for these two
  particles to be

4
Center of Mass

• From this we can see that if m2 0, then xcom
  0.
• Similarly, if m1 0, then xcom d.
• Finally, if m1 m2, then xcom ½d.
• So we can see that the center of mass in this
  case is constrained to be somewhere between x 0
  and x d.

5
Center of Mass

• Now lets shift the origin of the coordinate
  system a little.
• We now need a more general definition of the
  center of mass.
• The more general definition (for two particles)
  is

6
Center of Mass

• Now lets suppose that we have lots of particles
  all lined up nicely for us on the x axis
• The equation would now bewhere M m1 m2
  mn
• The collection of terms in the numerator can be
  rewritten as a sum resulting in

7
Center of Mass

• This result is only for one dimension however, so
  the more generalized result for 3 dimensions is
  shown here

8
Center of Mass

• You know by intuition that the center of mass of
  a sphere is at the center of the sphere for a
  rod it lies along the central axis of the rod
  for a flat plate it lies in the plane of the
  plate.
• Note however that the center of mass doesnt
  necessarily have to lie within the object or have
  any mass at that point
• The COM of a horseshoe is somewhere in the middle
  along the axis of symmetry.
• The COM of a doughnut is at its geographic
  center, but there is no mass there either.

9
Center of Mass

• Going back to our baseball bat, the COM will lie
  along the central axis (the axis of symmetry).
• And it is the COM that faithfully follows the
  line of a parabola.

10
Newtons 2nd Law for a System of Particles

• We know from experience that if you roll the cue
  ball into another billiard ball that is at rest,
  that the two ball system will continue on away
  from you after the impact.
• You would be very surprised if one or the other
  of the balls came back to you (without putting
  English on the ball).
• What continued to move away from you was the COM
  of the two ball system!
• Remember that the COM is a point that acts as
  though all of the mass in the system were located
  there.
• So even though we may have a large number of
  particles possibly of different masses, we can
  treat the assembly as having all of its mass at
  the point of its COM.

11
Newtons 2nd Law for a System of Particles

• So we can assign that point a position, a
  velocity and an acceleration.
• And it turns out that Newtons 2nd law holds for
  that point
• where Fnet is the sum of all the external forces
  acting on the mass M is the total mass and acom
  is the acceleration of the center of mass.

12
Newtons 2nd Law for a System of Particles

• Circling back around to where we started from, we
  can break the one vector equation down into an
  equation for each dimension

13
Newtons 2nd Law for a System of Particles

• Now lets re-examine what is going on when we
  have the collision between the two billiard
  balls
• Once the cue ball has been set in motion, there
  are no external forces acting on the system.
• Thus, Fnet 0 and as a consequence, acom 0.
• This means that the velocity of the COM of the
  system must be constant.
• So when the two balls collide, the forces must be
  internal to the system (the balls exert forces on
  each other no external forces act on the balls)
  which doesnt affect Fnet.
• Thus the COM of the system continues to move
  forward unchanged by the collision.

14
Newtons 2nd Law for a System of Particles

• When a fireworks rocket explodes, the COM of the
  system does not change while the fragments all
  fan out, their COM continues to move along the
  original path of the rocket.
• This is also how a ballerina seems to defy the
  laws of physics and float across the stage when
  doing a grand jeté.

15

• Suppose a cannon shell traveling in a parabolic
  trajectory (neglecting air friction) explodes in
  flight, splitting into two fragments of equal
  mass.
• The fragments follow new parabolic paths, but the
  center of mass continues on the original
  parabolic path as if all the mass were still
  concentrated at that point.

16
Linear Momentum

• Linear momentum of an object is the mass of the
  object multiplied by its velocity.
• Momentum p mvcom
• Unit kgm/s or Ns
• Newton expressed his 2nd law of motion in terms
  of momentum
• The time rate of change of the momentum of a
  particle is equal to the net force acting on the
  particle and is in the direction of that force.
• Both momentum and kinetic energy describe the
  motion of an object and any change in mass and/or
  velocity will change both the momentum and
  kinetic energy of the object.

17
Linear Momentum

• Momentum refers to inertia in motion.
• Momentum is a measure of how difficult it is to
  stop an object a measure of how much motion an
  object has.
• More force is needed to stop a baseball thrown at
  95 mph than to stop a baseball thrown at 45 mph,
  even though they both have the same mass.
• More force is needed to stop a train moving at 45
  mph than to stop a car moving at 45 mph, even
  though they both have the same speed.
• Both mass and velocity are important factors when
  considering the force needed to change the motion
  of an object.

18
Impulse

• Impulse (J) forcetime
• Equation J Ft Unit Ns
• The impulse of a force is equal to the change in
  momentum of the body to which the force is
  applied. This usually means a change in
  velocity.
• Ft m?v where ?v vf - vi
• The same change in momentum can be accomplished
  by a small force acting for a long time or by a
  large force acting for a short time.

19
Impulse

• If your car runs into a brick wall and you come
  to rest along with the car, there is a
  significant change in momentum. If you are
  wearing a seat belt or if the car has an air bag,
  your change in momentum occurs over a relatively
  long time interval. If you stop because you hit
  the dashboard, your change in momentum occurs
  over a very short time interval.

• The area under the curve in a force vs. time
  graph represents the change in momentum (m?v).

20
(No Transcript)
21
Impulse

• If a seat belt or air bag brings you to a stop
  over a time interval that is five times as long
  as required to stop when you strike the
  dashboard, then the forces involved are reduced
  to one-fifth of the dashboard values. That is
  the purpose of seat belts, air bags, and padded
  dashboards. By extending the time during which
  you come to rest, these safety devices help
  reduce the forces exerted on you.
• If you want to increase the momentum of an object
  as much as possible, you apply the greatest force
  you can for as long a time as possible.

22

• A 1000 kg car moving at 30 m/s
• (p 30,000 kg m/s) can be stopped by 30,000 N
  of force acting for 1.0 s (a crash!)
• or by 3000 N of force acting for 10.0 s (normal
  stop)

23
Impulse and Bouncing

• Impulses are greater when bouncing takes place.
• The impulse required to bring an object to a stop
  and then throw it back again is greater than the
  impulse required to bring an object to a stop.

24
(No Transcript)
25
Conservation of Linear Momentum

• In a closed system of objects, linear momentum is
  conserved as the objects interact or collide.
  The total vector momentum of the system remains
  constant.
• p before interaction p after interaction

26
Collisions

• A collision is an isolated event in which two or
  more bodies (the colliding bodies) exert
  relatively strong forces on each other for a
  relatively short time.
• An interesting point to note is that the
  definition doesnt necessarily require the bodies
  to actually make contact a near miss (near
  enough so that there are relatively strong
  forces involved) will suffice

27
Collisions

• To analyze a collision we have to pay attention
  to the 3 parts of a collision
• Before
• During
• After

28
Momentum Kinetic Energy in Collisions

• If we have a system where two bodies collide,
  then there must be some kinetic energy (and
  therefore linear momentum) present.
• During the collision, the kinetic energy and
  linear momentum of each body is changed by the
  impulse from the other body.
• We define two different kinds of collisions
• An elastic collision is one where the total
  kinetic energy of the system is conserved.
• An inelastic collision is one where the total
  kinetic energy is not conserved.
• A collision is not necessarily completely elastic
  or completely inelastic most are in fact
  somewhere in between

29
Momentum Kinetic Energy in Collisions

• For any kind of collision however, linear
  momentum must be preserved.
• This is because in a closed, isolated system, the
  total linear moment cannot change without the
  presence of an external force (and the forces
  involved in a collision are internal to the
  system not external).
• This does not mean that the linear momentum of
  the various colliding bodies cannot change.
• The linear momentum of each body involved in the
  collision may indeed change, but the total linear
  momentum of the system cannot change.

30
Perfectly Inelastic Collisions

• Perfectly inelastic collisions are those in which
  the colliding objects stick together and move
  with the same velocity.
• Kinetic energy is lost to other forms of energy
  in an inelastic collision.

31
Inelastic Collision Example

• Cart 1 and cart 2 collide and stick together
• Momentum equation
• Kinetic energy equation
• v1 and v2 velocities before collision
• v ? velocity after collision

32
Directions for Velocity

• Momentum is a vector, so direction is important.
• Velocities are positive or negative to indicate
  direction.
• Example bounce a ball off a wall

33
Inelastic Collisions

• Kinetic energy is lost when the objects are
  deformed during the collision.
• Momentum is conserved.

34
Elastic Collisions

• Momentum and kinetic energy are conserved in an
  elastic collision.
• The colliding objects rebound from each other
  with NO loss of kinetic energy.

35
Elastic Collision Example

• Example mass 1 and mass 2 collide and bounce
  off of each other
• Momentum equation
• Kinetic energy equation
• v1 and v2 velocities before collision
• v1? and v2? velocities after collision
• Velocities are or to indicate directions.

36
Elastic Collisions Involving an Angle

• Momentum is conserved in both the x-direction and
  in the y-direction.
• Before

37
Elastic Collisions Involving an Angle

• After

38
Elastic Collisions Involving an Angle

• Directions for the velocities before and after
  the collision must include the positive or
  negative sign.
• The direction of the x-components for v1 and v2
  do not change and therefore remain positive.
• The directions of the y-components for v1 and v2
  do change and therefore one velocity is positive
  and the other velocity is negative.

39
Elastic Collisions Involving an Angle

• px before px after
• py before py after
• Velocity after collision

40
Elastic Collisions

• Perfectly elastic collisions do not have to be
  head-on.
• Particles can divide or break apart.
• Example nuclear decay (nucleus of an element
  emits an alpha particle and becomes a different
  element with less mass)

41
Elastic Collisions

• mn mass of nucleus
• mp mass of alpha particle
• vn velocity of nucleus before event
• vn velocity of nucleus after event
• vp velocity of particle after event

42
Recoil

• Recoil is the term that describes the backward
  movement of an object that has propelled another
  object forward. In the nuclear decay example,
  the vn would be the recoil velocity.

43
Head-on and Glancing Collisions

• Head-on collisions occur when all of the motion,
  before and after the collision, is along one
  straight line.
• Glancing collisions involve an angle.
• A vector diagram can be used to represent the
  momentum for a glancing collision.

44
Vector Diagrams

• Use the three vectors and construct a triangle.

45
Vector Diagrams

• Use the appropriate expression to determine the
  unknown variable.

46
Vector Diagrams

• Total vector momentum is conserved. You could
  break each momentum vector into an x and y
  component.
• px before px after
• py before py after
• You would use the x and y components to determine
  the resultant momentum for the object in question
• Resultant momentum

47
Vector Diagrams

• Right triangle trigonometry can be used to solve
  this type of problem

48
Vector Diagrams

• Pythagorean theorem
• If the angle ? for the direction in which the
  cars go in after the collision is known, you can
  use sin, cos, or tan to determine the unknown
  quantity. Example determine final velocity vT
  if the angle is 25.

49
Vector Diagrams

• To determine the angle at which the cars go off
  together after the impact

50
Special Condition

• When a moving ball strikes a stationary ball of
  equal mass in a glancing collision, the two balls
  move away from each other at right angles.
• ma mb
• va 0 m/s

51
Special Condition

• Use the three vectors to construct a triangle.

52
Special Condition

• Use the appropriate expression to determine the
  unknown variable.

53
(No Transcript)
54
Ballistic Pendulum

• In the ballistic pendulum lab, a ball of known
  mass is shot into a pendulum arm. The arm swings
  upward and stops when its kinetic energy is
  exhausted.
• From the measurement of the height of the swing,
  one can determine the initial speed of the ball.
• This is an inelastic collision. As always, linear
  momentum is conserved.

55
Ballistic Pendulum
56
Ballistic Pendulum

• Potential energy of ball in gun

• Ball embeds in pendulum

57
Ballistic Pendulum

• Pendulum rises to a maximum height

• Solving for the initial speed of the projectile
  we get

58
Series of Collisions

• What happens when there are multiple collisions
  to consider?
• Consider an object that is securely bolted down
  so it cant move and is continuously pelted with
  a steady stream of projectiles.
• Each projectile has a mass of m and is moving at
  velocity v along the x axis.

59
Series of Collisions

• The linear momentum of each projectile is mv.
• Suppose that n projectiles arrive in an interval
  of ?t.
• As each projectile hits (and is absorbed by) the
  mass, the change in the projectiles linear
  momentum is ?p thus the total change in linear
  momentum during the interval ?t is n?p.

60
Series of Collisions

• The total impulse on the target is the same
  magnitude but opposite in direction to the change
  in linear momentum thus
• But we also know that
• This leads us to

61
Series of Collisions

• So now we have the average force in terms of the
  rate at which projectiles collide with the target
  (n/?t) and each projectiles change in velocity
  (?v).
• In the time interval ?t, we also know that an
  amount of mass ?m nm collides with the target.
• Thus our equation for the average force finally
  turns out to be

62
Series of Collisions

• Assuming that each projectile stops upon impact,
  we know that
• In this case the average force is

63
Series of Collisions

• Suppose instead that each projectile rebounds
  (bounces directly back) upon impact in this
  case we have
• In this case the average force is

64
Elastic Collision Example

• Example mass 1 and mass 2 collide and bounce
  off of each other
• Momentum equation
• Kinetic energy equation
• v1 and v2 velocities before collision
• v1? and v2? velocities after collision
• Velocities are or to indicate directions.

65
Elastic Collision Example

• Working with kinetic energy
• 0.5 cancels out.

66
Elastic Collision Example

• The velocity terms are perfect squares and can be
  factored
• a2-b2 (a b)(a b)
• We will use this equation later.

67
Elastic Collision Example

• Momentum equation

68
Elastic Collision Example

• Both the kinetic energy and momentum equations
  have been solved for the ratio of m1/m2.
• Set m1/m2 for kinetic energy equal to m1/m2 for
  momentum

69
Elastic Collision Example

• Get all the v1 terms together and all the v2
  terms together
• Cancel the like terms

70
Elastic Collision Example

• Rearrange to get the initial and final velocities
  back together on the same side of the equation
• This equation can be solved for one of the two
  unknowns (v1 or v2), then substituted back into
  the conservation of momentum equation.

71
Change in Momentum Example

• A 0.5 kg rubber ball is thrown towards a wall
  with a velocity of 14 m/s. It hits the wall,
  causing it to deflect in the opposite direction
  with a speed of 9 m/s. What is the change
  in the balls momentum?
• ?p F?t
• ?p mvf mvi
• ?p (0.5 kg)(-9 m/s) (0.5 kg)(14 m/s)
• ?p (-4.5 kgm/s) (7 kgm/s)
• ?p -11.5 kgm/s
• The change in the balls momentum is 11.5 kgm/s.