NEWTON’S SECOND LAW: LINEAR MOMENTUM

NEWTON’S SECOND LAW: LINEAR MOMENTUM

1. What is Momentum?

87-351 Fluid Mechanics and Hydraulics

any object with mass that moves has momentum momentum = mass * velocity

[ physical interpretation: what are we doing today? ]

2. How are Newton’s second law and momentum related? Newton’s second law can be expressed to say that:

“The rate of change of momentum for an object is proportional to the imposed force on that object”

3. Who cares !? understanding momentum conservation is fundamental to our

understanding of the motion of anything with mass that moves

applying Newton’s second law empowers us to relate how forces affect the motion of an object and vice-versa


1. The basis for the derivation is really just a mathematical description of Newton’s second law, we can specifically say:

[ derivation of the linear momentum equation ]

where this represents a summation of time rate of change of momentum for the entire system

OR

[EQN1]



2. For a fixed and non-deforming CV that is coincident with the system, the linear momentum equation can be expressed by our old friend the “Reynolds Transport Theorem”

[ derivation of the linear momentum equation (cont’d) ]

where in this case, the velocity is b (the intrinsic property), and the system momentum is B (the extensive property)

OR

in this way, we can think of velocity as momentum per unit mass

[EQN2]



3. Thus, for a non-deforming and fixed CV, we can express the linear momentum equation as

[ derivation of the linear momentum equation (cont’d) ]

here the CV form of the RTT reminds us that the flow of momentum in and out of the CV should behave no differently than mass flow

this statement tells us that the net force on the contents of the CV is equal to the momentum time rate of change of the contents of the CV + the net flow of momentum across the CV boundaries

[EQN2]

[EQN3]


THE LINEAR MOMENTUM EQUATION: APPLICATION[ example 1 ]

Let’s look at a simple example application of the linear momentum equation

GIVEN:

REQD: What anchoring force do we require to hold the deflection block in place? (assume viscous and gravitational forces negligible)

= 10 fps


THE LINEAR MOMENTUM EQUATION: APPLICATION[ example 1 (cont’d) ]

SOLU: 1. We will apply a CV approach, we select the CV to contain the deflector and a the portion of flow in contact with it



SOLU: 2. We know the linear momentum equation is a vector equation, so we identify all the angles and relevant forces acting on the CV



SOLU:

at the entrance and exit of the CV, the water acts as a “free” jet (exposed to the atmosphere) i.e., the same atmospheric pressure acts at each boundary, thus net pressure forces are zero

we have neglected gravity, so the weight forces of the deflector and water are also out

all that remains are the horizontal, FAx, and vertical, FAz, components of the force that anchors the deflector



SOLU:

we further recognize that the flow enters the CV at (1) and leaves it at (2), at an angle, q



SOLU: 3. Now let us write the x and z component of the momentum eqn

- (1)

- (2)

with a steady flow, (momentum flow in = momentum flow out), we may neglect the unsteady term in our eqn, as the net rate of momentum accumulation in the volume is zero



SOLU:

- (3)

- (4)

here we pay particular attention to the unit normal vector directions

further, we acknowledge that equal pressure at entrance and exit, and negligible gravity and friction will not slow the flow from (1) to (2), *(remember Bernoulli Eqn)

thus we can write (1) and (2) as:

we eliminate the necessity to integrate by assuming the velocity profiles entering and leaving the CV are uniform



SOLU:

- (6)

- (7)

and we know V1 = V2, thus A1 must equal A2 s.t.

Q1 = V1A1 = Q2 = V2A2

3. We simplify (3) and (4) further by applying the conservation of mass

- (5)

4. Now we plug GIVEN data into (6) and (7), and solve

- (ans1)



SOLU:- (ans2)

what does this solution tell us about the momentum relationship between the deflector geometry and momentum imposed forces?

- (ans1)


THE LINEAR MOMENTUM EQUATION: APPLICATION[ example 2 ]

Let’s look at another example application of the linear momentum equation

GIVEN: REQD: Determine the anchoring force required to hold the nozzle on the sink faucet

Q = 0.6 l/s

mnozzle= 0.1 kg

p1= 464 kPa



SOLU: 1. We draw the CV with all applicable forces and flow delineations

where:

FA = anchoring force holding nozzle in place Wn = weight of nozzle Ww = weight of water in nozzle p1 = gauge pressure at 1 A1 = area of nozzle at 1 p2 = gauge pressure at 2 A2 = area of nozzle at 2 w1 = z direction velocity at CV entrance w2 = z direction velocity at CV exit



SOLU:2. We write the linear momentum eqn

- (1)



SOLU:NB: further, we recall that the simplicity of this expression is available (no integrals) because we have assumed uniform velocity distributions

- (2)

then

where mdot = rAV (mass flowrate)



SOLU:

- (4)

but we know from conservation of mass that mdot1 = mdot2, so

We note here that (4) is an interesting illustration of how Newton’s second law relates forces to matter in motion

3. Now we solve (3) for the anchoring force

- (3)

i.e., look at how each force and mass flux impacts the anchoring force FA



SOLU:

- (6)

we start with the mass flux

4. Now let us solve for some numbers

- (5)

then it follows that

- (7)

and

similarly simple

- (8)



SOLU:

- (10)

the weight of the water is obtained by virtue of a little geometry (see expression for the volume of a truncated cone)

- (9)or

finally, we can compute the anchoring force



SOLU:

- (10)

finally, we can compute the anchoring force

- (ans)

or


NEWTON’S SECOND LAW: LINEAR MOMENTUM

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