Download Impulse, momentum, and center of mass

Impulse, momentum, and center of mass
Suppose a crate has a mass of m and begins at rest on a frictionless surface. You push on it with a
force F over a time of Δt.
By Newton’s second law, F = m·a
By definition, a =
𝛥𝑣
𝛥𝑡
, therefore F = m·
𝛥𝑣
𝛥𝑡
Rearranging this equation yields
F·Δt = m·Δv
If we define impulse (J) as the product of force and time span and also define momentum (p) as mass
times velocity, then this equation becomes:
J = Δmv = Δp
read as “Impulse equals change in momentum”
If there was a graph of force versus time, the area under that curve would equal the impulse.
For instance, the impulse from 0s to 7s would be the area trapped to the x-axis, which is -12N·s. This
𝑚
would cause a change in momentum of -12kg· 𝑠 . These two quantities are the same by the equivalence
of impulse and change in momentum. Even the units are the same if you expand 1N out into 1kg·m/s2.
Question 1: What is the impulse in the graph above from 15s to 18s?
The simple situation described at the beginning had motion in only one dimension, but the little
derivation works just as well in two and three dimensions, making the impulse equation a vector
equation.
J = Δp
where J = F·Δt
and p = m·v
And if the force applied varies with time, you just use calculus notation: J = ∫ 𝐅 · dt
Momentum (p) is often seen as a vector in two dimensions.
𝑚
For example, if the above diagram had units of kg· 𝑠 , the little blue puck would have a momentum of
𝑚
𝑚
4kg· 𝑠 in the x-dimension and 5kg· 𝑠 in the y-dimension. In polar coordinates, this would be a
𝑚
momentum of (6.4kg· 𝑠 , 51.34º).
Suppose you have two very large magnets sitting on a frictionless surface, each beginning at rest.
In this orientation, the two magnets will be attracted to each other, each experiencing the same
magnitude of force from the attraction. Therefore, they will begin sliding together.
Let’s say we look at the amount they both slide during a time Δt. If we call F just the magnitude of the
force, the left-hand magnet will experience an impulse of +F·Δt. The right-hand magnet will
experience an impulse of -F·Δt.
Impulse equals change in momentum, so the left-hand magnet will change momentum by +F·Δt the
right-hand magnet with change momentum by -F·Δt.
Clearly, the total change in momentum for the system is zero, and this derivation works regardless of
whether or not the magnets began at rest.
So internal forces like the one shown above cannot change the momentum of a system. If the sum of
the external impulses also adds to zero, then the system cannot change momentum at all.
ΣF·Δt = F1·Δt + F2·Δt + …..
If the right-hand side of the equation is zero, so is the left-hand side, so is 𝛴𝐹, the net external force on
the system.
Altogether, this principle is stated as the conservation of momentum: If the net external force acting on
a system is zero, the total momentum of the system remains constant.
Clearly, if there is a net external force acting on a system, it can change momentum. Kicking a ball is
a simple example.
Question 2: What is another example of where a net external force causes a change in momentum in a
system?
As with the concepts of impulse and momentum, the derivation above holds just as true in two and
three dimensions. It is a vector equation.
Σpi = Σpf
if ΣFexternal = 0
𝑚
𝑚
If the diagram above has units of kg· 𝑠 , the little blue puck would have a momentum of < 4, 5> kg· 𝑠
𝑚
the little green puck would have a momentum of <-1, 4> kg· ,
𝑠
𝑚
making the total momentum of the system < 3, 9 > kg· 𝑠 .
The two pucks may collide and stick together or collide and bounce off each other, but whatever
𝑚
happens, the total momentum of this closed system will remain < 3, 9 > kg· 𝑠 .
This idea of conservation of momentum can also be reached through the idea of center of mass.
Conceptually, the center of mass of a system is the weighted average of the component’s positions.
Let me illustrate that with an example.
Above is a simple system of three masses. Where is the center of mass of the system? Well, we need
to weight each position with the mass at that position. The equation looks like this:
Rcm =
𝑚1 ·𝑅1 + 𝑚2 ·𝑅2 +𝑚3 ·𝑅3
𝑀
=
(5𝑘𝑔)(−3𝑚)+(2𝑘𝑔)(1𝑚)+(8𝑘𝑔)(4𝑚)
15𝑘𝑔
M is the total mass of the system, so here the center of mass, Rcm = 1.26̅m
Question 3: What is the center of mass in the system above if the 8kg sphere moves 1m right?
This concept can easily be extended into two and three dimensions. Suppose you had a square
with sides of 2m. Relative to the bottom left corner, where is the center of mass?
In the x-dimension: Rcm =
In the y-dimension, Rcm =
(3𝑘𝑔)(0𝑚)+(5𝑘𝑔)(0𝑚)+(18𝑘𝑔)(2𝑚)+(2𝑘𝑔)(2𝑚)
28𝑘𝑔
(3𝑘𝑔)(0𝑚)+(5𝑘𝑔)(2𝑚)+(18𝑘𝑔)(2𝑚)+(2𝑘𝑔)(0𝑚)
28𝑘𝑔
= 1.43m
= 1.64m
So Rcm = < 1.43, 1.64 >m. That puts the center of mass about where the little x is:
Incidentally, the center of mass is the point on which a system would balance. This is because both
torque and center of mass are linear functions of position. It’s also the point around which the system
naturally spins because centripetal force is also a linear function of position (FC = m·ω2·R).
So how is this concept related to momentum? Well, go back to the original equation:
Rcm =
𝑚1 ·𝑅1 + 𝑚2 ·𝑅2 +𝑚3 ·𝑅3
𝑀
If you differentiate both sides with respect to time, you get:
vcm =
𝑚1 ·𝑣1 + 𝑚2 ·𝑣2 +𝑚3 ·𝑣3
or
𝑀
(vcm)(M) = 𝑚1 · 𝑣1 + 𝑚2 · 𝑣2 + 𝑚3 · 𝑣3
Conceptually, the total momentum of a system can be equated to the total mass of the system times the
velocity of the center of mass of the system
Differentiate again and you get:
acm =
𝑚1 ·𝑎1 + 𝑚2 ·𝑎2 +𝑚3 ·𝑎3
or
𝑀
(acm)(M) = 𝑚1 · 𝑎1 + 𝑚2 · 𝑎2 + 𝑚3 · 𝑎3
Now m1·a1 is the net force on particle one, m2·a2 is the net force on particle two, etc.
All internal forces within the system will add to zero, so the right hand side of the equation becomes
the sum of all external forces on the system.
ΣFext = (M)(acm)
Suppose, for instance, you throw a tennis racket across a football field. The racket is likely to wobble
around itself in a very complex pattern with many internal forces. But the total external force on the
racket is just -M·g,
so -M·g = (M)(acm)
and
acm = -g.
The center of mass of the racket is going to move with a nice, simple acceleration of –g, in a parabolic
path, just like a ball tossed across the field.
Question 4: What is another example of something that would wobble around itself if thrown through
the air?
Lastly, it is very easy to use these equations to derive conservation of momentum:
If ΣFext = 0, then acm = 0, so
And if
𝑑𝑣𝑐𝑚
𝑑𝑡
𝑑𝑣𝑐𝑚
𝑑𝑡
= 0.
𝑑
= 0, then, from above, 𝑑𝑡(𝑚1 · 𝑣1 + 𝑚2 · 𝑣2 + 𝑚3 · 𝑣3 ) = 0
or
𝑑
𝑑𝑡
So, if ΣFext = 0, then the net momentum of the system does not change with time
(Σp) = 0
To review:
J = F·Δt
Impulse equals force applied over a time span
J = ∫ 𝐅 · dt
if the force is not constant
The area under a force-time function is also the impulse imparted.
J = Δp
Impulse equals change in momentum
p = m·v
Momentum equals mass times velocity
Because force and velocity are vectors, impulse and momentum are also vectors, quantities often
analyzed in two dimensions.
If the net external force acting on a system is zero, the total momentum of the system remains constant.
Symbolically, if ΣFexternal = 0
For the center of mass, Rcm =
Σpi = Σpf
𝑚1 ·𝑅1 + 𝑚2 ·𝑅2 +𝑚3 ·𝑅3
𝑀