center of mass & momentum!...force and momentum change the relationship between force and motion...
TRANSCRIPT
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CEnter of MAss
&
MOMENTUM!
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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
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Center of Mass for solid, symmetrical objects of uniform density
Located at the geometric center of the object
xcm
xcm
xcm xcm
xcmxcm
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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
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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
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Even if the other parts
of the object have
other accelerations
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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.
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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.
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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)
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Why is momentum important?
….and explosions!
The same momentum exists before and after a collision.Momentum lets us predict
collisions…
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Elastic Collisions▪ Linear momentum is conserved
▪ Total energy is conserved
▪ Total kinetic energy is conserved
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Inelastic Collisions▪ Linear momentum is conserved
▪ Total energy is conserved
▪ Kinetic energy is not conserved
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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).
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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
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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
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• 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
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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
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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.
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Collisions
m1 m2m1 m2
v1 v2 v3 v4
m1 m2
v1 v2 v3
m1 m2
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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?
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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
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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?
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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
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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
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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?
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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
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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
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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
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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.
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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)?
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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
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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
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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
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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
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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
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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