Roger's Equations

Hello Roger - good summary, only thing I'd query is your rotational units. Should be in radians, not revs. Angular velocity radians/sec, etc. Otherwise answer for energy = ¹/₂(Iω²) (for example) comes out wrong, though it's still OK dimensionally as in your Lagrangian calculation since angle is dimensionless.

Can check by comparing KE of a point mass moving in a circle, when ¹/₂(Iω²) and

¹/₂(mv²) should be the same.

Cheers

You Wrote: "Hello Roger - good summary, only thing I'd query is your rotational units. Should be in radians, not revs. Angular velocity radians/sec, etc. Otherwise answer for energy = ¹/₂(Iω²) (for example) comes out wrong, though it's still OK dimensionally as in your Lagrangian calculation since angle is dimensionless."

I called the units of Angular Velocity Hertz, which is a cycle a second, which is 360°/s or 2π/s. A radian is just 57.2957 degrees or .15915 rotation. So a radian/sec is just .15915 rotations/sec. The two are just different ways of measuring the same thing. Just like an inch is .0254 meters, or 2.54 cm. All can be used as a measure of distance. The SI unit for distance is meter. The SI unit for angular velocity (degrees/sec) is Hertz. Radian is considered a derived unit in the SI system (see link below).

http://en.wikipedia.org/wiki/Radian

Roger - reply to #6

Maybe so, but you still have to put ω in rad/sec to get the right answer. An object of MoI 1 kg.m² rotating at 1 rev/sec has KE ¹/₂(2.π)² joule, not ¹/₂ joule.

If it consists of a mass 1 kg at radius 1 m its linear velocity is 2.π m/s

You Wrote: "Maybe so, but you still have to put ω in rad/sec to get the right answer. An object of MoI 1 kg.m² rotating at 1 rev/sec has KE ¹/₂(2.π)² joule, not ¹/₂ joule."

I see your point. From a calculation standpoint we would convert the rotations/s to radians/s in order to calculate the energy. An interesting question is why? I've been trying to think of a reason why this should be the case but all I can think of is that one method is base 360 and the other is base 10, but I'm not sure. I'd appreciate your insight on this if you have any.

Roger - reply to #11

Apart from energy considerations discussed, only thing I can think of is it's the unit of angle θ that gives direct relation between radius r and arc length L, L = r.θ, as well as v = r.ω.

Ref wikipedia, I suppose it's a derived unit in the sense that it's defined from arc length and radius. Unlike revolution, it's not an intuitive unit.

Yes, but you were correct when you said that 1/2 J would be incorrect, just as 360 J would be incorrect. Why should this be? Again, I wonder if one being base 10 and the other being base 360 plays a role. Or maybe it has something to do with unitless quantities. I'm not sure.

An excellent and valuable thread. Thank you.

As Roger said, angular measure is dimensionless, so is independent of the basic concept of dimensional analysis. On the other hand, it is sometimes useful to consider a hybrid between dimensional and numerical analysis, and this is where angular measure can become relevant.

Base 10 and base 360 have no basic physics or mathematical origin - they are entirely artificial and historic divisions with accounting and packaging convenience.

Therefore, for rotational and frequency units, the choice would be between full circles (cycles) and radian measure. Radians are said to be a derived unit - but I believe that this is as opposed to being a fundamental dimension, rather than in comparison with full cycles - and degrees are not generally considered in this context.

Anywhere that differentiation is also relevant (e.g. angular position versus velocity), the radian would become the unit of choice - because in any measure other than radian you acquire a pi multiplier between angular position and velocity, etc.

There isn't really a full answer, but my personal preference (as I know of no similar simplification from using cycles) would be to choose radians over cycles wherever possible.

Fyz

More on derived units:

My previous comments were perhaps putting the cart before the horse, so...

First, a definition: a derived unit is a unit that can be expressed in terms of the "fundamental" units* used for dimensional analysis. Use of derived units can extend the power of dimensional analyses; however, unlike the absolute requirement for matching dimensions, the additional implications can improve insight but do not directly lead to any results.

*dimensions = mass,length,time,mdot
or in MKS units = kg,m,s,kgdot

The advantage of sometimes using derived units is that they can give an idea as to the origins of the dimensions, so they can give an idea of some of the multiplying constants to watch for. If a derived unit has time in the denominator, and the expression includes this unit "to the power n", we should not be surprised to see a factor n! in the equation. But this is by no means universal - consider for example standard kinetic energy E=m.v^2/2 and intrinsic (mass) energy E=m.c^2.

Be that as it may, the clear intrinsic feature of a derived unit is that it implies no added constants. Now, I think, we are ready to return to the radian.

The radian measures circumferential_extent/radius of an angle. It's units are length/length - it vanishes in terms of basic dimensional analysis. (Other measures of angle, such as the cycle, do not have a dimensional basis in the same sense; that is to say that the radian is the only known 'dimensional' unit of angle, albeit a derived one). Generally, radians do not accrue any standard numerical multipliers.

Fyz

You Wrote: "The radian measures circumferential_extent/radius of an angle. It's units are length/length - it vanishes in terms of basic dimensional analysis.

So where I was treating degree as being on equal footing with meter, kilogram, time, that isn't necessary. It's more efficient (as I'm seeing it now) to express rotations in terms of radians, which are merely derived units of distance (m/m) and fit without effort into the SI system.

You Wrote: "Generally, radians do not accrue any standard numerical multipliers."

I'm sorry, what do you mean by "standard numerical multipliers"? Do you mean they can be treated like other SI quantities in equations (unlike degrees which must be converted to provide real results.

You Wrote: "If a derived unit has time in the denominator, and the expression includes this unit "to the power n", we should not be surprised to see a factor n! in the equation."

I'm a little lost. Can you provide an example of where this is true so I can get your meaning.

Between Codemaster's and Fyz's comments, I'm quite convinced about radians. I might even say "converted" because now I'm wondering why we even bother to learn Trig in terms of degrees (I know the "it divides easily" argument, but that seems outdated now). It seems counter productive to learn in degrees and spend the rest of your life converting degrees to radians. It would be like a French man being taught only English for his first 15 years, so that for the rest of his life he has to convert his thoughts from English into his native language. Wouldn't it just make more sense to learn French first so he could think and speak in the same language (if you get my analogy)? Anyway, I'm now wondering that if radians is unitless because its really meters/meters, are degrees unitless for a similar reason? Say because they are really seconds/seconds?

Hi Roger, this is Fyz

(I'm not at my desk, and can never remember passwords, so not logged in)

As regards acquiring 1/N! terms, this is related to differentials and integrals. So kinetic energy, which undergoes a single integration from force to force*distance aquires the 1/2! (which just looks like 1/2). You will may areas in dynamics where 1/time^3 appears, and this is sometimes associated with 1/3! As I said, it's only of use as a "watch-out for" indicator, as the factor is associated with the detailed mechanism, and cannot in an way be relied upon. The reason that I commented that Radians do not generally exhibit this sort of behaviour is that radian^N usually comes about as a result of differentiating sinusoidal or rotational behaviout, where each level arises through its association with a different variable inside a trig function.

Regarding degrees etc - unfortunately, the straight line (and right angle) have importance in geometry, and pi (pi/2) radians is/are not convenient. So it's horses for courses (again).

Fyz

Fyz,

Yeah, I can never remember my passwords either. I look forward to the day the whole password thing is simplified (maybe never, but I can hope).

Anyway, your explanation has cleared it up for me. I find so many things can go wrong when I'm calculating that its always nice to have a few tricks to make sure on track. My favorite trick is "ok, I've got an answer, but does it make sense(f(0)=?; f(∞)=?)" closely followed by "I was looking for force but I have units of m⁴/s⁴·kg ... that can't be good"

I'll keep an eye out for this new one and see if it helps sometimes.

Roger

Roger - reply to #17

You wrote "now I'm wondering why we even bother to learn Trig in terms of degrees (I know the "it divides easily" argument, but that seems outdated now). It seems counter productive to learn in degrees and spend the rest of your life converting degrees to radians."

You're dead right that in Maths radian is the appropriate unit, but scrapping ° would be going a bit far I think. Angle measure is used in a lot of other more practical fields, engineering, DIY etc and I wouldn't like the idea of doing home woodwork using radians! And if you went into a plumbers' merchant and asked for a Π/2 elbow you'd get a funny look.

Cheers

pi/2 elbow certainly made me chuckle. Imagine air traffic control: "N1348, turn right to a heading of 1.308 radians."

Something nobody has pointed out is that if you multiply an angle measured in radians by the radius of the circle you are actually calculating the distance round the circumference of the circle. For example if you look at what happens when you swing through a full circle you have

Distance = 2π r = Circumference

Put another way if you multiply a rotational velocity in radians by the radius of the circle you are calculating the linear distance that object has moved through which is the definition of a linear velocity.

As for using the SI system of weights an measures there are now only 3 countries left in the world that are sticking to the old imperial system, So how come the USA, the most technically advanced nation on earth, is still wallowing around in the imperial system with a couple underdeveloped countries like Burma and Liberia?

PS Great thread Roger.

Perhaps the US still regrets its early involvement with Bonaparte?

BTW, 2.54-mm and its sub-multiples still seem to be much used in many metric countries

You Wrote "Perhaps the US still regrets its early involvement with Bonaparte?"

The US "support" of Bonaparte consisted of buying land. Considering Bonaparte's main enemy was the British, who the US were to war with in 1812 as they did in 1776, I think a little of "the enemy of my enemy is my friend" makes sense.

Nobody seems to be using metric time, why is that?

"Nobody seems to be using metric time, why is that?"

Some applications do in use a sort of metric time. For example astronomical date/times are in days after a particular date, for example a UTC date/time of Tuesday 16^th January 2007 at 04:06:59 has a UT Julian date of JT 2 454 116.67156 and Julian Ephemeris date/time of JDE 2 454 116.672412. PCs store dates in a similar manner except that they start counting from a different point and use the local time so it comes out 39 098.171516.

An interesting point to note is that if you add 12 hours, or 0.5 days, to the PC time you get 39 098.671516 and the digits that represent the portion of the day are now identical to the JT date/time.

Then of course there is the infamous stardate of Star Trek fame that I have never been able to come to grips with. Are there any trekies out there that can enlighten me and explain the vagaries of faster than light travel and stardate to me? I am sure there is an explaination.

Physicists and scientists for the most part you use seconds and this is a true decimal metric unit.

There are a few imperial measurements that have been carried across to the metric system. For example the nautical mile has be redefined as 1,852 metres and is still used navigationally since it corresponds to one minute of arc at the equator. Another carry over is, that for the most part, the altitude of an aircraft which is still measured in feet. While some of the eastern block countries do have altimeters calibrated in meters they are in the minority and you will find that feet is the predominant unit. I believe the reason for the continued use of these units is that any change over would result in absolute pandemonium in the air. To convert to metric safely you would need to simultaneously world wide convert every aircraft, air traffic control instrument and map, then somehow transient everybody to new flight plans without causing a collision. Maybe some day in the future when all aircraft instruments are digital it will be possible but I can't see it happening too soon.

Hello Masu reply to #23

You wrote "Something nobody has pointed out is that if you multiply an angle measured in radians by the radius of the circle you are actually calculating the distance round the circumference of the circle."

I made the same point in post #12. Having thought about it a bit more (ref Roger's post #11) I can't improve on it as the reason for definition of radian i.e. to have a direct relation between radius, angle and distance round the circumference. Though that's not to say nobody else can.

Cheers

So you did Codemaster. I must apologize for that, some how I mist your post.

kindly send the same way the british units too fir easy refernece

regards

ennappan

Hi Roger, nice, useful summary!

Just one point, how is "free-fall acceleration" a constant? Do you mean "g" (9.81 m/s^2)? It is however not a constant in the sense of the other fundamental ones.

Don't want to take words out of Roger's mouth, but isn't this just to give the units of these quantities? After all, no values are given for any of them.

Your Wrote: "Just one point, how is "free-fall acceleration" a constant? Do you mean "g" (9.81 m/s^2)? It is however not a constant in the sense of the other fundamental ones."

I did mean "g". It's true its not a fundamental constant but it shows up a lot in physics so I included it.

Hi Roger: units its a strong point. We (in Argentina) use the metric system, but when I was a student teachers mess gravitational system (basic unit Kgr force) with absolute system (basic unit Kg mass). So, always where a "g" bothering the equations. Undoubtely the best way is using the SI and no more units like "calories", Kgr force and the like.

Now, a question: I don´t understand the third term of this formula:

T=¹/₂(Iω²⁾ + ¹/₂m((dr/dt)² + r²(dφ/dt)²)

what represents? Also, a mass unit is missing.

Regards,

Gabriel, Buenos Aires, Argentina

You Wrote: "

Now, a question: I don´t understand the third term of this formula:

T=¹/₂(Iω²⁾ + ¹/₂m((dr/dt)² + r²(dφ/dt)²)

what represents?"

The equation above represesnts the kinetic energy for the ball or radius R rolling off a log of radius R. The first term represents the rotational kinetic energy of the system. The second term, which itself contains two terms represents linear motion through a plane. Think of it this way, when the ball rolls of the log, lets say, to the left, then the ball is moving both to the left and downward at the same time. The linear motion can be expressed as a combination of these motions. We can right this as:

we can convert this to polar coordinates (which is more convenient when dealing with rotational motion with the relations;

which after a bit of math gives us the equation;

you can see that you should multiply the mass through to both terms, so I wasn't missing a mass unit.

I hope that helps.

Hi Roger:many thanks for your response. I miss a parentheses, so I didn´t understand the formula.

Regards.

Comment on 12 and 13

As I understand it, it is the dimensionless constant linking angular to linear as noted in 12. ie. one revoultion = 2pi radians. I believe this was why the SI system now regards it as a derived unit.

KenS

Hi Roger:

Another great thread, which I have only glanced through. The dimensionless angle is something I hadn't really thought about, so I will have to come back here soon. Thanks, Ken