Halliday, Resnick & Walker Chapter 9

Centre of Mass and Linear Momentum Physics 1A – PHYS1121 Professor Michael Burton

MMM07VD1: Motion of the Centre of Mass

9-1 Centre of Mass ! 

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The motion of rotating objects can be complicated (imagine flipping a cricket bat?) But there is a special point on the object for which the motion is simple: ! 

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The centre of mass of the bat traces out a parabola, just as a tossed ball does All other points rotate around this point

9-1 Centre of Mass

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For two particles separated by a distance d, where the origin is chosen at the position of particle 1: Eq. (9-1)

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For two particles, with an arbitrary choice of origin:

Eq. (9-2)

9-1 Centre of Mass ! 

For many particles, we can generalize the equation, where M = m1 + m2 + . . . + mn:

Eq. (9-4) ! 

In three dimensions, we find the centre of mass along each axis separately:

Eq. (9-5)

9-1 Centre of Mass ! 

More concisely, we can write in terms of vectors: Eq. (9-8)

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For solid bodies, we take the limit of an infinite sum of infinitely small particles ! integration! Coordinate-by-coordinate, we write:

Here M is the mass of the object, dm an element of mass

9-1 Centre of Mass ! 

We limit ourselves to objects of uniform density, !, for the sake of simplicity Eq. (9-10)

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Substituting, we find the centre of mass simplifies:

Eq. (9-11)

Consider the Earth and Mars in their orbit around the Sun. Where is the centre of mass located for this three body system?

a) It is closer to the Earth, than it is to either the Sun or Mars. b) It is at the centre of a triangle that has the Sun at one apex, the Earth at another apex, and Mars at the third apex. c) It is half of the distance between the Sun and Mars. d) It is closer to the Sun, than it is to either the Earth or Mars. e) It is closer to Mars, than it is to either the Earth or the Sun.

9-2 Newton's Second Law for a System of Particles ! 

Centre of mass motion continues unaffected by forces internal to a system (e.g. collisions between billiard balls)

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Motion of a system's centre of mass:

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Reminders: 1.  Fnet is the sum of all external forces 2.  M is the total, constant, mass of the closed system 3.  acom is the centre of mass acceleration

6-2 Newton's Second Law for a System of Particles Exploding rocket: explosion forces are internal, so only the gravitational force acts on the system, and the c.o.m. follows a gravitational trajectory. [So long as air resistance can be ignored for the fragments….]

Two people sit in a large canoe on a still lake. The canoe is oriented with the front pointing due north. The person at the front of the canoe walks to sit by the person at the back and sits down. What effect does this event have on the canoe? a) The canoe will still be at rest, but it will be south of its original position. b) The canoe will still be at rest, but it will be north of its original position. c) The canoe will be moving toward the south. d) The canoe will be moving toward the north. e) The canoe will still be at rest at its original position. M1

MC

M2

MMM02VD1: Linear Momentum

21 seconds

9-3 Linear Momentum ! 

The linear momentum is defined as: Eq. (9-22)

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The momentum: o 

Points in the same direction as the velocity

o 

Can only be changed by a net external force

We can write Newton's second law thus: Eq. (9-23)

9-3 Linear Momentum ! 

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Linear Momentum for a system of particles equals product of the total mass M and the velocity of the centre of mass. Thus

The net external force on a system changes the linear momentum; i.e. ! 

Without a net external force, the total linear momentum of a system of particles cannot change

A stone of mass m is dropped and falls freely under the influence of gravity. Ignoring any effects of air resistance, which one of the following expressions gives the momentum of the stone as a function of time t? The local acceleration due to gravity is g.

a)

"mgt2

b)

gt

c)

"gt2

d)

mgt

e)

g dm/dt

Remember v=v0+at

9-4 Collision and Impulse

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In a collision, momentum of a particle can change

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We define the impulse J acting during a collision:

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This means that the applied impulse is equal to the change in momentum of the object during the collision:

9-4 Collision and Impulse

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Given the average force Favg and the time duration #t:

Only need to know the area under the force curve to obtain the impulse J

A skydiver whose parachute fails to open lands in snow; he is slightly hurt. Had he landed on bare ground his stopping time would have been 10 times shorter and the collision lethal. Does the presence of the snow increase, decrease or leave unchanged:

1. 

The skydiver’s change in momentum? Increase / decrease / unchanged

2.  The impulse stopping the skydiver? Increase / decrease / unchanged 3.  The force stopping the skydiver? Increase / decrease / unchanged

9-4 Collision and Impulse ! 

For a steady stream of n projectiles, each undergoes a momentum change #p

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The average force is:

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If the particles stop:

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If the particles bounce back with equal speed: !

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"v = v f # v i = 0 # v = #v. So Favg =

nmv "t

"v = v f # v i = #v # v = #2v. So Favg =

2nmv "t

The product nm is the total mass #m for n collisions so: Favg!= "

#m #m #m #v = v (stops) or 2 v (rebounds) #t #t #t

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MMA02VD1: Conservation of Momentum

36 seconds

9-5 Conservation of Linear Momentum ! 

Suppose the impulse is zero. Then: Eq. (9-42)

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This is the law of conservation of linear momentum

A rifle of mass M is initially at rest. A bullet of mass m is fired from the rifle with a velocity v relative to the ground. Which one of the following expressions gives the velocity of the rifle relative to the ground after the bullet is fired?

a) -mv b) mv c) Mv/m d) mv/M e) -mv/M

9-6 Momentum and Kinetic Energy in Collisions ! 

Types of collisions:

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Elastic collisions: o 

Total kinetic energy is unchanged (conserved)

o 

A useful approximation for common situations

o 

In real collisions, some energy is always transferred

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Inelastic collisions: some energy is transferred

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Completely inelastic collisions: o 

The objects stick together

o 

Greatest loss of kinetic energy

9-6 Momentum and Kinetic Energy in Collisions ! 

Momentum before collision = Momentum afterwards ! 

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In 1D:

For a completely inelastic collision, when target at rest:

9-6 Momentum and Kinetic Energy in Collisions ! 

The centre of mass velocity remains unchanged:

MMM06VD1: Elastic Collisions in One Dimension

9-7 Elastic Collisions in One Dimension ! 

Total kinetic energy is conserved in elastic collisions

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For a stationary target, conservation laws give:

9-7 Elastic Collisions in One Dimension ! 

With some algebra we get (exercise for the student!): Eq. (9-67) Eq. (9-68)

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Some deductions: o 

For equal masses: v1f = 0, v2f = v1i o 

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For massive target, m2 >> m1: v1f = -v1i , v2f=0 o 

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i.e. the first object stops

i.e. the first object just bounces back, speed mostly unchanged

For massive projectile, m1 >> m2: v1f " v1i, v2f " 2v1i o 

i.e. first object keeps going, target flies forward at about twice its speed

9-7 Elastic Collisions in One Dimension ! 

For a target that is also moving, we get (harder exercise!):

Worked Example

Consider an elastic collision between a Projectile and a stationary Target. Suppose the initial linear momentum of the Projectile is 6 kg m/s. What is the final linear momentum of the Target if the final linear momentum of the Projectile is: (a)  2 kg m/s? (b) -2 kg m/s? (c) What is final KE of Target if initial and final KE of Projectile are 5J & 2J, respectively?

Answers: (a) 4 kg m/s

(b) 8 kg m/s

(c) 3 J

9-8 Collisions in Two Dimensions ! 

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Apply the conservation of momentum along each axis Apply conservation of energy for elastic collisions

Example For a stationary target: o 

Along x:

o 

Along y:

o 

Energy:

MMM09VD1: Momentum Conservation in Rockets

60 seconds

9

Summary

Linear Momentum & Newton's 2nd Law ! 

Collision and Impulse ! 

Linear momentum defined as:

Eq. (9-30)

Eq. (9-25) ! 

Defined as:

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Write Newton's 2nd law:

Impulse causes changes in linear momentum

Eq. (9-27)

Conservation of Linear Momentum

Inelastic Collision in 1D ! 

Momentum conserved along that dimension

Eq. (9-42)

9

Summary

Motion of the Centre of Mass ! 

Eq. (9-51)

Unaffected by collisions/internal forces

Elastic Collisions in One Dimension ! 

K is also conserved Eq. (9-67) Eq. (9-68)

Collisions in Two Dimensions ! 

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Apply conservation of momentum along each axis individually Conserve K if elastic