First year college CALCULUS and PHYSICS, I can teach. Send PM

[IanB]

What is the power consumption, relative to time, of a freely falling body, vs. TIME ?

In reality there is no power consumption. The power is zero. This is similar to reactive "power" (VAR) in electrical systems, vs real power (watts).

In a free falling body, there is an exchange of energy between potential energy and kinetic energy, but if we consider the "system" to be the body in motion, then the total energy of the system remains constant. Power expresses a transfer of energy across a system boundary. Since there is no such transfer of energy, the power is zero.

[RJSV]

(Using diagram in previous post).
   I've got a pretty good idea of how to present a more full set of those three equations, using pre-existing Position, and Velocity as the starting condition, rather than simply dropping an object from rest.  This relates to my comments previous regarding putting (unknown) constants into an equation after Integration.  The next diagram reflects this, as the time variable, starts at zero, but the velocity can start at some existing value.  So the expression would contain (V0) for example, to delineate a starting velocity, (P0) as starting position, etc.
   So, for example, the current instantaneous Velocity will be (V0) + AxT rather than having V0 or 'starting'  value = 0.  The time spent dropping under acceleration then uses (T- T0) as the elapsed time.  That idea also applies to any starting position; where you have the usual equation, but provision for having any starting point (or height in this case).
   For example, a quiz question could say that your object has already been moving (falling), is now at 309 meters high, and right now moving at 19 meters per second.  Working those numbers out actually sounds like a more realistic real-world problem.

Thanks.

[IanB]

There are standard equations of rectilinear motion learned in high school physics class. For instance:

  v = u + at
  s = ut + ½at²
  s = ½(u + v)t
  v² − u² = 2as

(u = initial velocity, v = final velocity, a = acceleration, t = elapsed time, s = distance traveled)

These equations will apply to a body in free fall, where the acceleration "a" becomes the acceleration due to gravity, "g".

You can apply any of these equations according to need.

[RJSV]

   Yes, thanks, that's the sort of thing that I've been thinking.
   Never really encountered the more general and everyday type problem.  For example I'm assuming that the coordinate system (X,y,z) type, would start calculating at 309 meters high but movement (downward falling) would start involving negative values, as object loses position, and, of course the velocity would be non-negative, but result is a loss of height...hinting that it's numerically a 'negative' velocity (meters per second).
   But, It's close enough that someone more skilled might suggest corrections.

   As to those constants needed, of unknown value, I recall the general solution calls for having two equations, with two unknowns to be able to solve.  You manipulate one equation so that one of the unknowns drops out, when you've combined the two equations.
   
   So I'm going to get the fuller set of the three equations up (Position, Velocity, Acceleration.

[IanB]

Note that "rectilinear motion" is a posh way of saying "in a straight line". So there is only one distance coordinate, x.

In object in free fall is moving in a straight line, and the acceleration due to gravity is constant. So the equations I listed work perfectly for that scenario.

If an object is falling, then you simply say that downwards is the positive direction, and then velocity, distance and acceleration are all positive.

Let's do a thought experiment. You let go of an object that is 309 meters high, and you wish to know how fast it will be moving after it has fallen to 289 meters?

We pick the following equation as suitable:

  v² − u² = 2as

We set our datum to 309 meters, and when it has reached 289 meters high it has fallen through 20 meters. So distance traveled, s = 20 meters.

We let go of the object at at the beginning, so the initial velocity u is 0 m/s.

The acceleration due to gravity is a = g = 9.81 m/s².

We plug the numbers into the formula:

  v² − 0² = (2)(9.81)(20)
  v² = 392.4  m²/s²
  v = 19.81 m/s

So after the object has fallen 20 m from rest, it's speed will be 19.81 m/s.

[MrAl]

"lbl" is the symbol for poundal.  Thus, pound mass and poundal force are as "logical" as kilogram mass and Newton force.  I never thought of "lbl" as just a way to distinguish the two.

Hi,

What is "lbl" ?

[jpanhalt]

"lbl" is the unit for pound plus the "l" from poundal.  I pronounce it as pound "ell" or "el" (i.e, L spelled out).

Edit: Why is the unit for "pound" "lb" and not pd or something similar?  That's because of Latin.

[TimFox]

Latin "libra" means pound, or scales/balance.
A set of Latin currency values is libra, solidus, denarius;  whence pounds, shillings, pence abbreviated L, S, and D.

[newbrain]

Latin "libra" means pound, or scales/balance.
A set of Latin currency values is libra, solidus, denarius;  whence pounds, shillings, pence abbreviated L, S, and D.

And those three words gave origin in Italian to:
- "Lira" the Italian currency used before Euro (the symbol most commonly used was the same for pound £, though also L. and ₤ were used)
- "Soldo" used in the singular for coin, but more commonly in the plural form "Soldi" meaning money.
- "Denaro" another word for money

[IanB]

"lbl" is the symbol for poundal.

Hi,

What is "lbl" ?

"lbl" is the symbol for poundal.

Sorry. Keep seeing this, and the temptation is too hard to resist 

[jpanhalt]

"lbl" is the symbol for poundal.

Hi,

What is "lbl" ?

"lbl" is the symbol for poundal.

Sorry. Keep seeing this, and the temptation is too hard to resist 

You need your nystagmus checked.  Please correct the sequence to include the post (#32)  that follows , "What is "lbl" and tries to dissect its meaning.  Perhaps mistakenly, I assumed everyone knew the derivation of "lb", as there are many such examples in English, Pb, Au, K, Na ... are just a few. 

[watchmaker]

I want to say thank you for the intended purpose of this thread. 

I am 70 and reigniting my interest in understanding electronics from my 1960s Voc Ed days.  It is better than crossword puzzles or golf.

In following the Analog Real course (and and Dave's videos and MIT and NI and Analog Wiki) I realized I needed matrix algebra to solve complex ckts.  My kids (daughter (structural), her husband (EE/Patent Attn) started me off on the tangent of strengthening my math skills.  At the same time I am doing labs, I am using Kahn and Brilliant to relearn (in reality learn) all the stuff I was taught but failed to learn (including Diffi Q).

I failed to learn because at the time it was all taught in the abstract; with no relevance to application.  All theorems and proofs.  This is where STEM is a game changer.  It was the reason I left the gifted program in high school to join the Vocational Education program. 

I fell into the category of those who passed by not understanding but writing enough stuff to finally get an answer.

I am amazed at how "kids" are taught math today.  I am starting at middle school algebra and matrix operations.  Will need to also relearn trig function math.  But the focus on graphing is great!

To be honest, I hold a Ph.D. in social research which relied on all the higher order statistical analyses.  But I NEVER visualized y=x squared.  Nor did I use the distributive property to break large numbers into reasonable chunks for earlier multiplication.  Sad, I know.  But there it is.

It will be some time before I get to Diff Calc but I really appreciate the effort to provide math support on this forum.  I started but realized I really needed to back up to middle school.

It was because of this forum (and Dave's videos) that I found Wulfram (sic?) for solving the more complex linear ckts.

So, I would ask that you not let the "perfect be the enemy of the good".  At least for me, there is potential value in this thread.  And I am old enough to know I am not unique.

Hopefully, this post will encourage others to expose their confusion and ask for help.  Maybe there is a way to break this thread out so it becomes a sub-forum if demand is demonstrated?

Regards,

Dewey

[IanB]

"lbl" is the symbol for poundal.

Hi,

What is "lbl" ?

"lbl" is the symbol for poundal.

Sorry. Keep seeing this, and the temptation is too hard to resist 

You need your nystagmus checked.  Please correct the sequence to include the post (#32)  that follows , "What is "lbl" and tries to dissect its meaning.  Perhaps mistakenly, I assumed everyone knew the derivation of "lb", as there are many such examples in English, Pb, Au, K, Na ... are just a few.

Joking aside, the symbol for poundal is actually "pdl". So six poundals would be 6 pdl.

[RJSV]

Thanks, especially Dewey, for your thoughts.
   One benefit here, no matter how screwed up my personal organization is, has to do with being available for questions and criticism, unlike some more-perfected text book.  (Lol) I could end up benefiting from corrective commentary.

  Right now, I'm still focused on analogies in the basic mechanical physics, and attempting to get that into diagram form,...that being the three hierarchy related items of Position, Speed, and Acceleration.
   Looks like I might need to just think up a suitable form, meanwhile noting that it's NOT a correctly expressed formula, for the motion physics.  Others responding helped a lot with that frustration!

   So, if the dynamics of the 3 equations, involving differentiation and integrals isn't leading to heavy use, for numerical answers, I'm hoping to at least give a basis for why that understanding is still useful.
   For example, Newton, and others, used the calculus when describing the planet orbit problem, of whether an orbiting body is 'falling' by the math analysis, along with the analytic geometry involved in taking apart that (orbit) question.

Thanks...more soon

[RJSV]

   Now I've gotten things in a better shape, for illustrating the tiers or 'hierarchy' of the three parameters of motion; that being Position, Speed, Acceleration.
   I'm in deep water a bit, here, as I'm attempting to generalize for illustrating the calculus aspect, and so have tried to get a most general case.

[RJSV]

   Explanation goes like this:
   Starting at the top equation (actually just a constant to represent gravity field induced acceleration, AND, in this case I will be putting an additional little rocket motor, for an initial segment of time).

   I figure to replace my example of falling from 309 meters high, to using a backwards vertical notation system, where the start is up somewhere, like 340, and THAT is newly, the place notated as 'zero' position.
Moving downward then is a positive direction for position and velocity.
   Simply, otherwise it's a hassle in the math.

[TimFox]

An important nit-pick:  position, speed, and acceleration are variables, not parameters:  they all vary as a function of time in your example.
The strength of the gravitational field is a parameter.

[RJSV]

   In this aspect of my presentation, I think ianb has provided best explanations...please see those posts, a bit back.

   In the illustration here I've added an Xtra feature to try and get the most general case.
Shown us a little rocket motor, let's say the object (we want to characterize) is attached to a rocket motor, before the free-fall phase of downward motion (now considered as a positive direction just to simplify, as ianb has suggested).
   The reason I'm doing this is to get a general equation, rather than just the '1/2 AT squared' term.  Either we could have the rocket motor accelerate, say at 5 newtons per second per second, or just have it give a push at the beginning (looks better).
That way you've got all the terms expressed
Idea is to start the experiment at (t0), maybe move at a constant rate, to (t1) before doing the free-fall.... But that is wrong because gravity is still there in that initial time period.
   See how messy this gets ?  Hopefully I can at least convey what I want, if not in numerically correct fashion.  Sooo, even if incorrect, let's ignore gravity acceleration until t1....  (my bad).

   This approach let's me introduce the full calculus expressions, where each Integration operation introduces an unknown constant.
   In that sense, your period where there is acceleration would be from t1 to tx time.  For that, you would use (tx - t1) as the length of time that acceleration happens.

[RJSV]

   Here is that modified experiment, for figuring the variables.

[RJSV]

Uh shit, more upload problems...

[RJSV]

   Sorry about the picture upload glitch...I think it's errors in my cellphone memory (heat caused aging).
   This last diagram for today shows a more complete use of the TIME variable, where each place that uses time is literally a time period, helping to identify which segment of the little experiment is applicable.

   Whew, this is confusing for sure, but I think close enough that I can get back into the pure math aspects.  That is, if you want to take a constant, and integrate that (twice), as a function then you get that stack of three equations, shown a few illustrations back.

   Again, though, I'm not sure if the absolute position, in a purely free falling body, will need these fully expressed terms.  Part of the whole point, here, is to show the reversibility between integrals and derivatives...

Clear as mud...(I may have just convinced 27 students to QUIT Engineering, lol)

[IanB]

The two basic equations we have for linear motion are (1) velocity from acceleration:
$$\frac{\mathrm{d}v}{\mathrm{d}t}=a$$
And (2) distance from velocity:
$$\frac{\mathrm{d}s}{\mathrm{d}t}=v$$
If we define \$u\$ to be the initial velocity and \$v\$ to be the final velocity, we can integrate the first equation over a particular interval:
$$\int_{u}^{v}\mathrm{d}v=\int_{0}^{t}a\,\mathrm{d}t$$
If acceleration is constant it can move outside the integral, to give:
$$\int_{u}^{v}\mathrm{d}v=a\int_{0}^{t}\mathrm{d}t$$
Evaluating the integrals therefore gives:
$$v-u=at$$
Or
$$v=u+at$$
Now, we can integrate the distance equation in a similar way:
$$\int_{0}^{s}\mathrm{d}s=\int_{0}^{t}v\,\mathrm{d}t$$
Which tells us that:
$$s=\int_{0}^{t}v\,\mathrm{d}t$$
We can take this integral and substitute in the previous expression for \$v\$ in terms of \$u\$, \$a\$ and \$t\$:
$$s=\int_{0}^{t}v\,\mathrm{d}t=\int_{0}^{t}(u+at)\,\mathrm{d}t$$
Evaluating this latter integral, with the understanding that \$u\$ and \$a\$ are constants, yields:
$$s=ut+\textstyle{\frac12} at^2$$
This shows how my earlier posts are related to the fundamental derivatives and integrals that express motion in a straight line, and how they simplify when acceleration is constant.

[watchmaker]

The two basic equations we have for linear motion are (1) velocity from acceleration:
$$\frac{\mathrm{d}v}{\mathrm{d}t}=a$$
And (2) distance from velocity:
$$\frac{\mathrm{d}s}{\mathrm{d}t}=v$$
If we define \$u\$ to be the initial velocity and \$v\$ to be the final velocity, we can integrate the first equation over a particular interval:
$$\int_{u}^{v}\mathrm{d}v=\int_{0}^{t}a\,\mathrm{d}t$$
If acceleration is constant it can move outside the integral, to give:
$$\int_{u}^{v}\mathrm{d}v=a\int_{0}^{t}\mathrm{d}t$$
Evaluating the integrals therefore gives:
$$v-u=at$$
Or
$$v=u+at$$
Now, we can integrate the distance equation in a similar way:
$$\int_{0}^{s}\mathrm{d}s=\int_{0}^{t}v\,\mathrm{d}t$$
Which tells us that:
$$s=\int_{0}^{t}v\,\mathrm{d}t$$
We can take this integral and substitute in the previous expression for \$v\$ in terms of \$u\$, \$a\$ and \$t\$:
$$s=\int_{0}^{t}v\,\mathrm{d}t=\int_{0}^{t}(u+at)\,\mathrm{d}t$$
Evaluating this latter integral, with the understanding that \$u\$ and \$a\$ are constants, yields:
$$s=ut+\textstyle{\frac12} at^2$$
This shows how my earlier posts are related to the fundamental derivatives and integrals that express motion in a straight line, and how they simplify when acceleration is constant.

THAT and circumference/area/volume are the only things I remember from calculus.  Always thought it was a great "trick".  Those guys were pretty damned smart.

Regards,

Dewey

[RJSV]

     Creating and Solving Integrals, using Analytic Geometry:

 _______-----------_________-------------________------
   By dividing up into smaller areas, the classic solution, for obtaining the area under a curve, can be determined, if maybe only approximately.
   Nice thing is that, by design, those areas can be made smaller and smaller;  meanwhile the sheer quantity (of infinitesemals) of the partitioned areas grow (rapidly!).

   For example, (see diagram): you could have two pieces, or areas, and then go with four pieces.  Obviously you also have each piece at half the area as before.

[IanB]

   In the illustration here I've added an Xtra feature to try and get the most general case.
Shown us a little rocket motor, let's say the object (we want to characterize) is attached to a rocket motor, before the free-fall phase of downward motion (now considered as a positive direction just to simplify, as ianb has suggested).
   The reason I'm doing this is to get a general equation, rather than just the '1/2 AT squared' term.  Either we could have the rocket motor accelerate, say at 5 newtons per second per second, or just have it give a push at the beginning (looks better).
That way you've got all the terms expressed

But actually, the simple equation can still work, even in some cases that seem complicated.

Consider the illustration below. Someone fires a stone from a catapult up onto the roof of a building. If the launch velocity of the stone is 10 m/s, at an angle of 80°, starting 1 m above the ground, and if the building is 4 m high, how long does it take before the stone lands on the roof? Assume that air resistance can be neglected.

To work the solution, we can use the equation derived earlier:
$$s=ut+\textstyle{\frac12} at^2$$
To work out the time of rise and fall, we only need to consider the vertical component of velocity, acceleration and distance, and we can ignore the horizontal component. So first of all, we can find the vertical component of the initial velocity. This is found with trigonometry:
$$u_y=u\sin(80^{\circ})=(10)(0.985)=9.85\,\mathrm{m/s}$$
Secondly, we have the total distance traveled:
$$s_y=4-1=3\,\textrm{m}$$
Since we are using up for the positive direction, then the acceleration due to gravity is negative:
$$a_y=-g=-9.81\,\textrm{m/s}^2$$
Now we can plug the numbers into the formula:
$$s_y=u_y t+\textstyle{\frac12} a_y t^2$$ $$3=(9.85)t+\textstyle{\frac12}(-9.81)t^2$$
This is a quadratic equation we can solve for \$t\$. We get two solutions:
$$t=0.37\,\textrm{s}\;\;\textrm{or}\;\;t=1.63\,\textrm{s}$$
The first time will be on the way up, and the second time will be on the way down. So the answer we are looking for is the second one, and the time to land on the roof is 1.63 s.