CEnter of MAss

&

MOMENTUM!

Center of MassThe point at which

all of the mass of an object or system

may be considered to be concentrated. A system can be one or more objects that may or may not be connected in space.

CENTER OF MASS

CENTER OF MASS

CENTER OF MASS

CENTER OF MASS

Center of Mass for solid, symmetrical objects of uniform density

Located at the geometric center of the object

xcm

xcm

xcm xcm

xcmxcm

Center of Mass for more complicated situations

The bar above is not of uniform density. Center

of mass is not at the geometric center.

Towards which end is it?

Center of mass is not

inside the object

xcm

xcm

1. Hang a plumb line.2. Suspend the object

vertically from a point on the object so it can swing down freely.

3. Draw a vertical line where the object fell.

4. Suspend the object from a different point on the object. And let it swing down freely.

5. Draw another vertical line that intersects with the first vertical line that you drew. Where the two lines intersect is the CM.

xcm

Experimental Determination of CM

Even if the other parts

of the object have

other accelerations

Toppling Rule of Thumb

• If the CG of the object is above the area of support, the object will remain upright.

• If the CG is outside the area of support the object will topple.

Another look at Stability• Stable equilibrium: when for a

balanced object a displacement raises the CG (to higher U so it tends to go back to the lower U).

• Unstable equilibrium: when for a balanced object a displacement lowers the CG (lower U).

• Neutral equilibrium: when the height of the CG does not change with displacement.

Momentum• A measure of how hard it is to stop an

object which is moving.

• Related to both mass and velocity.

Momentum = mass x velocity

p = m v (in kg m/s)

Why is momentum important?

….and explosions!

The same momentum exists before and after a collision.Momentum lets us predict

collisions…

Elastic Collisions▪ Linear momentum is conserved

▪ Total energy is conserved

▪ Total kinetic energy is conserved

Inelastic Collisions▪ Linear momentum is conserved

▪ Total energy is conserved

▪ Kinetic energy is not conserved

Momentum vs.

Inertia

• Inertia is a property of mass that resists changes in velocity;

however, inertia depends only on mass. Inertia is a scalarquantity (no direction).

• Momentum is a property of moving mass that resists changes in a moving object’s velocity. Momentum is a vector quantity that

depends on both mass and velocity (has direction).

EXAMPLE: Momentum practice problem.

If a football player’s momentum is 400 kg m/s, and he has a mass of 120 kg, what is his velocity

(in m/s)?

Ans. Givens: p = 400 kg m/s, m = 120 kg

Unknown: v = ?Equation: p = m v

Solve: 𝐩

𝐦= 𝐯

Substitute: 𝟒𝟎𝟎 𝐤𝐠

𝐦

𝐬

𝟏𝟐𝟎 𝐤𝐠= 𝐯

Ans. 3.33 m/s = v

Bus: m = 19800 lbs. (2.2 lbs = 1 kg); v = 16 m /s

Train: m = 3.6 104 kg; v = 4 m /s

Car: m = 1800 kg; v = 80 m /s

P.O.D. 1: Find the momentum of the following.

2.2 lbs = 1 kg

• Kinetic energy and momentum are different quantities, even though both depend on mass and velocity.

• Kinetic energy (TKE = ½mv2) is a scalar quantity. Mass is always

+ and when you square the velocity you always get a + answer.

Kinetic Energy doesn’t depend on direction.

Momentum vs. Kinetic Energy

BA

Kinetic Energy Momentum

A = ½(4 kg)(1 m/s)2 = 2 J

B = ½(4 kg)(1 m/s)2 = 2 J

Momentum (p = mv)is a vector, so it always depends on direction.

Sometimes momentum is + if velocity is in the + direction and

sometimes momentum is if the velocity is in the direction.

Two balls with the same mass and speed have the same

kinetic energy but opposite momentum.

Momentum vs. Kinetic Energy

BA

Kinetic Energy Momentum

A = ½(4 kg)(1 m/s)2 = 2 J = 4 kg 1 m/s = 4 kgm/s

B = ½(4 kg)(1 m/s)2 = 2 J = 4 kg 1 m/s = 4 kg m/s

Conservation of MomentumThe law of conservation of momentum states

when a system of interacting objects is not influenced by outside forces (like friction), the total momentum of the system cannot change.

Collisions

m1 m2m1 m2

v1 v2 v3 v4

m1 m2

v1 v2 v3

m1 m2

EXAMPLE: Elastic collisions

Two 0.165 kg billiard balls roll toward each other and collide

head-on. Initially, the 5-ball has a velocity of 0.5 m/s. The 10-ball

has an initial velocity of -0.7 m/s. The collision is elastic and the

10-ball rebounds with a velocity of 0.4 m/s, reversing its direction.

What is the velocity of the 5-ball after the collision?

Ans. G.U.E.S.S. G. ivens: m1 = m2 = 0.165 kg, v1 = 0.5 m/s, v2 = 0.7 m/s, v4 = 0.4 m/sU. known: v3

E. quation: m1v1 + m2v2 = m1v3 + m2v4

S. olve: m1v1 + m2v2 = m1v3 + m2v4

m2v4 m2v4

m1v1 + m2v2 m2v4 = m1v3𝐦𝟏𝐯𝟏+𝐦𝟐𝐯𝟐−𝐦𝟐𝐯𝟒

𝐦𝟏= 𝐯𝟑

S. ubstitute: (𝟎.𝟏𝟔𝟓 𝒌𝒈)(𝟎.𝟓

𝒎

𝒔) +(𝟎.𝟏𝟔𝟓 𝒌𝒈)(−𝟎.𝟕

𝒎

𝒔) −(𝟎.𝟏𝟔𝟓 𝒌𝒈)(𝟎.𝟒

𝒎

𝒔)

𝟎.𝟏𝟔𝟓 𝐤𝐠= 𝐯𝟑

0.6 m/s = v3

P.O.D. 2: A 200 kg football player moves at 5 m/s towards a 150 kg player moving at 7 m/s. They collide and bounce off

each other in opposite directions. If the 200 kg player is moving at 3 m/s after the impact, how fast (in m/s) is the

150 kg player moving?

A train car moving to the right at 10 m/s collides with a

parked train car. They stick together and roll along the track.

If the moving car has a mass of 8,000 kg and the parked car

has a mass of 2,000 kg, what is their combined

velocity after the collision?

EXAMPLE: Inelastic collisions

Ans. G.U.E.S.S.

G. ivens: m1 = 8000 kg m2 = 2000 kg, v1 = 10 m/s, v2 = 0 m/s

U. known: v3

E. quation: m1v1 + m2v2 = (m1 + m2)v3

S. olve: 𝐦𝟏𝐯𝟏+𝐦𝟐𝐯𝟐

𝐦𝟏+𝐦𝟐= 𝐯𝟑

S. ubstitute:

(𝟖𝟎𝟎𝟎 𝒌𝒈)(𝟏𝟎𝒎

𝒔) +(𝟐𝟎𝟎𝟎 𝒌𝒈)(𝟎

𝒎

𝒔)

𝟖𝟎𝟎𝟎 𝐤𝐠 +𝟐𝟎𝟎𝟎 𝐤𝐠= 𝐯𝟑

8 m/s = v3

P.O.D. 3: A 2000 kg car rear ends a 2500 kg truck which is at rest. If the car was moving at 30 m/s initially, how fast (in m/s) would the car truck system move forward together?

Force is the Rate of Change of Momentum

• Momentum changes when a net force is applied.

• The inverse is also true: – If momentum changes,

forces are created.

• If momentum changes quickly, large forces are involved.

This means that force and momentum are

directly proportional

Force and Momentum ChangeThe relationship between force and motion follows directly

from Newton's Second Law.

Change in momentum

(kg m/sec)

Change in time (sec)

Force (N) F = Dp

Dt

F = ma F = m𝐯

𝐭 Ft = mv

The product of F t from the last slide is called Impulse.

The symbol for impulse is J. So, by definition:

J = F t

Impulse Defined

Example: A 50 N force is applied by a cattle

rancher to a 400 kg horse for 3 s to brand it. What

is the impulse of this force?

Ans. J = (50 N) (3 s) = 150 N·s.

Note that we didn’t need to know the mass of the object in

the above example.

P.O.D. 4

A sucker feels an impulse of 3000 kg m/s for 0.5 seconds

when he gets slapped by a buddy. How much force does he

feel (in N)?

Conservation of Momentum in 2-DSuppose a ball of mass m1 collides with another ball of mass m2. The

collision is not head on so the two balls will move off at angles to

each other after the collision.

Momentum before:

pin = m1v1+ m2v2 = m1v1+ m2 (0) = m1v1

pin = m1 v1

Momentum after:

pout = m1va + m2vb

m1

abm2

vbva

m1

v1

m2

By the law of

conservation of momentum:

pin = pout

So: m1 v1 = m1va + m2vb

m1

abm2

vbva

We can break our diagonal momentums after

the collision into x and y components as we

did with forces and velocities last semester:

Conservation of Momentum in 2-D

m1

v1

m2

Suppose a ball of mass m1 collides with another ball of mass m2. The

collision is not head on so the two balls will move off at angles to

each other after the collision.

b

m2

vb

a

m1

va

vax

vby

vbx

vay

m1

abm2

vbva

We can break our diagonal momentums after the

collision into x and y components as we did with

forces and velocities last semester:

Conservation of Momentum in 2-D

m1

v1

m2

Suppose a ball of mass m1 collides with another ball of mass m2. The

collision is not head on so the two balls will move off at angles to each other

after the collision.

a

m1

va

vax

vay

b

m2

vbvby

vbx

Horizontal things with horizontal things,

vertical things with vertical things:

m1vax = m2vbx

m1v1 = m1vay + m2vby

m1

abm2

vbva

We can break our diagonal momentums after the

collision into x and y components as we did with

forces and velocities last semester:

Conservation of Momentum in 2-D

m1

v1

m2

Suppose a ball of mass m1 collides with another ball of mass m2. The

collision is not head on so the two balls will move off at angles to each other

after the collision.

a

m1

va

vax

vay

b

m2

vbvby

vbx

The vertical components are given by vsin :

m1v1 = m1vay + m2vby

m1v1 = m1vasin a + m2vbsin b

3 kg3 kg

6 kg

EXAMPLE: Conservation of Momentum in 2-DSuppose a ball of mass 3 kg moving at a

velocity of 10 m/s collides with another ball of

mass 6 kg. The collision is not head on so the

two balls will move off at angles to each other

after the collision. The 3kg ball moves off at

an angle of 25 to the horizontal at a velocity

of 8 m/s while the 6 kg ball moves off at an

angle of 45. What is the final velocity of the 6

kg ball? pin = pout

m1v1 = m1vasin a + m2vbsin b

m1vasin a m1vasin a

25 456 kg

vb = ?8 m/s

10 m/s

m1v1 m1vasin a = m2vbsin b

m1v1 m1vasinam2sinb

= vb

(3 kg)(10 m/s) (3 kg)(8 m/s)sin 25(6 kg)sin 45

= vb 4.68= 𝐯𝐛

8 kg8 kg

4 kg

P.O.D. 5: Conservation of Momentum in 2-DSuppose a ball of mass 8 kg moving at a

velocity of 12 m/s collides with another ball of

mass 4 kg. The collision is not head on so the

two balls will move off at angles to each other

after the collision. The 8 kg ball moves off at

an angle of 40 to the horizontal at a velocity

of 6 m/s while the 4 kg ball moves off at an

angle of 45. What is the final velocity of the

4 kg ball?

40 454 kg

vb = ?6 m/s

12 m/s