### It’s Obvious. Not! A Few Answers and More Questions

This installment of “It’s Obvious. Not!” looks at:

Book: “Classical Dynamics of Particles and Systems”

Edition: third

Authors: Jerry B. Marion and Stephen T. Thornton

Publisher: Harcourt Brace Jovanovich

Page: 52

This installation of the series provides a few clarifications into the example presented in the textbook and asks even more questions. I have a feeling that readers steeped in differential equations will immediately follow the reasoning of the example as it is written in the textbook. Please, if you have answers to the remaining questions below, or even ideas, please comment. Thanks!

The first example of the chapter titled “Newtonian Mechanics” asks the reader to find the velocity of an object sliding down a ramp. The solution for the acceleration, (second derivative of the position x), has already been derived as:

The process for deriving the velocity as a function of position illustrated by the author starts with the above equation for acceleration. I’ll simplify by first briefly stating the strategy of the authors solution.

Basic Strategy:

Working from the expression for acceleration, the authors want to demonstrate how to derive velocity as a function of position. They use the following steps that I’ll elaborate on here.

1. Introduce a velocity term to each side of the equation to form a differential equation that can be integrated on each side to solve for velocity in terms of position.
2. Use the product rule of differentiation to re-write the left hand side of the expression so that it only contains velocity.
3. Multiply both sides of the expression by dt so that the integration on the right handside will only integrate over dx producing an x.

We want an equation that relates velocity, (the first derivative of position), to the position of the object on the ramp. Ultimately, the example will manipulate the original expression above so that the left side only contains a differential of velocity and the right side contains a differential of position. When both sides of the expression are integrated, we’ll be left with an expression that relates velocity to position.

Clarifications and Questions:

Now that we know where we’re headed, let’s see how the authors take us there.

The authors multiply both sides of the originalequation by 2 times the velocity, (first derivative of position with respect to time), to give:

and then, in one giant step for physics students everywhere, the authors state on the very next line:

Ummm what… !? OK, let’s start at the start. First, why did the author mysteriously decide to multiply both sides of the original equation for acceleration by two times the velocity?

My first thought was why not just integrate the acceleration to get the velocity and be done with it? That would work if we wanted the velocity as a function of time, but we have to solve for it as a function of position. Speaking of time, let’s rewrite the first equation above as:

We can do this because acceleration is the derivative of velocity with respect to time. Why we did it is to show explicitly that we have velocity and time so far, but not position. So, we need to somehow introduce position.

Now, returning to the authors original multiplication written in terms of position and velocity:

Now, dx is available in the equation. By integrating both sides, we can possibly come up with an answer that provides velocity in terms of position.

This presents the second problem. At present, dx is on both sides of the equation. We’d like to have velocity on one side of the equation and position on the other.

This brings us to the authors’ ‘clever trick’ and another question. The question: Is there a simple mathematical operation to get from

as the book seems to imply by performing this ‘simplification’ in one step? As far as I can tell, there isn’t. Please comment below if you know the way!

That brings us to the authors’ ‘clever trick’. By starting at the end of the simplification and writing

as

and then using the product rule, (use the first dx/dt as u and the second dx/dt as v), to carry out the derivative with respect to time on the contents of the parenthesis, we get:

So, by ‘cleverly’ choosing to multiply both sides of the original equation by 2 times the velocity, (first derivative of x or position), the authors were left with a quantity on the left-hand-side that is recognizable as the time based derivative of the velocity squared. With that recognition and simplification, the left-hand-side is now written only in terms of velocity.

So, after the ‘clever trick’ that introduced position into the equation and then the ‘clever’ simplification, we’re left at the second step of the example:

The authors then multiply both sides by dt and then integrate both sides using the boundary values of the example to arrive at, (the authors’ third step):

This step brings us to the last question. Is there a typo in the third step of the authors’ example? In the text, the upper limit of integration for the left hand side is:

the boundary velocity of the object when it is at position x-naught.

Hopefully, I’ve helped clarify how the authors’ moved from step to step. The following three questions still remain:

1. How did the author decide to multiply both sides of the equation for acceleration by 2 times the velocity?
2. Is there a simple mathematical operation that makes the following equality clear?
3. Is there a typo in the third step of the authors’ example?

If you have an idea or answer, please help everyone out and comment below. Thanks!

Handy Stuff

Anonymous said…
1) It's a fairly standard thing to do in differential equations...you multiply the whole thing by something that reduces the LHS to a recognisable derivative via product rule.

2. Yes. It's often known as product rule. You wrote it out yourself.

3. No. You missed a point here. The LHS is integrating with respect to velocity squared. When x = x0, v^2 = v0^2, making that the upper limit (and the value of the LHS (+C))

Hope I helped.
Hamilton said…
This comment has been removed by the author.
Hamilton said…
Thanks!

Your explanations cleared up everything! Once the reader is aware of the technique you mentioned in your first answer, then it becomes common-place to look for it.

Thanks for the correction on the third question and the confirmation that there isn't a typo.

### Cool Math Tricks: Deriving the Divergence, (Del or Nabla) into New (Cylindrical) Coordinate Systems

The following is a pretty lengthy procedure, but converting the divergence, (nabla, del) operator between coordinate systems comes up pretty often. While there are tables for converting between common coordinate systems, there seem to be fewer explanations of the procedure for deriving the conversion, so here goes!

What do we actually want?

To convert the Cartesian nabla

to the nabla for another coordinate system, say… cylindrical coordinates.

What we’ll need:

1. The Cartesian Nabla:

2. A set of equations relating the Cartesian coordinates to cylindrical coordinates:

3. A set of equations relating the Cartesian basis vectors to the basis vectors of the new coordinate system:

How to do it:

Use the chain rule for differentiation to convert the derivatives with respect to the Cartesian variables to derivatives with respect to the cylindrical variables.

The chain rule can be used to convert a differential operator in terms of one variable into a series of differential operators in terms of othe…

### Lost Phone

We were incredibly lucky to have both been in university settings when our kids were born.  When No. 1 arrived, we were both still grad students.  Not long after No. 2 arrived, (about 10 days to be exact), mom-person defended her dissertation and gained the appellation prependage Dr.

While there are lots of perks attendant to grad school, not the least of them phenomenal health insurance, that’s not the one that’s come to mind for me just now.  The one I’m most grateful for at the moment with respect to our kids was the opportunities for sheer independence.  Most days, we’d meet for lunch on the quad of whatever university we were hanging out at at the time, (physics research requires a bit of travel), to eat lunch.  During those lunches, the kids could crawl, toddle, or jog off into the distance.  There were no roads, and therefore no cars.  And, I realize now with a certain wistful bliss I had no knowledge of at the time, there were also very few people at hand that new what a baby…

### Lab Book 2014_07_10 More NaI Characterization

Summary: Much more plunking around with the NaI detector and sources today.  A Pb shield was built to eliminate cosmic ray muons as well as potassium 40 radiation from the concreted building.  The spectra are much cleaner, but still don't have the count rates or distinctive peaks that are expected.
New to the experiment?  Scroll to the bottom to see background and get caught up.
Lab Book Threshold for the QVT is currently set at -1.49 volts.  Remember to divide this by 100 to get the actual threshold voltage. A new spectrum recording the lines of all three sources, Cs 137, Co 60, and Sr 90, was started at approximately 10:55. Took data for about an hour.
Started the Cs 137 only spectrum at about 11:55 AM

Here’s the no-source background from yesterday
In comparison, here’s the 3 source spectrum from this morning.

The three source spectrum shows peak structure not exhibited by the background alone. I forgot to take scope pictures of the Cs137 run. I do however, have the printout, and…