Physics

[Topher]

Hey, I lost my book for the subject and test is on Monday, these are the problems I'm having trouble with, I need to know the formulas to use as they were in the book which I've since lost..

 

 

FIRST QUESTION

What is the torque required to accelerate a turbine undergoing a dynamic balancing test from 0 to 15000 r/min in 80 seconds, it's mass moment of inertia is 11.5kg.m²

 

Torque = X (unknown)

Time = 80

Acceleration = y (unknown)

Mass moment of inertia = 11.5x9.8 = 112.7N.m²

15000x60 = 900000r/s (divide by time in seconds to find acceleration)

900000/80 = 11250r increase per second .. I can tell I'm lost

 

 

SECOND QUESTION

I can't find a question I copied down to cover this, so any help would be appreciated, I need to be able to do questions revolving around angular motion, all I remember is it's units, rad/s

[Schrödinger's hat]

Hey, I lost my book for the subject and test is on Monday, these are the problems I'm having trouble with, I need to know the formulas to use as they were in the book which I've since lost..

 

 

FIRST QUESTION

What is the torque required to accelerate a turbine undergoing a dynamic balancing test from 0 to 15000 r/min in 80 seconds, it's mass moment of inertia is 11.5kg.m²

 

Torque = X (unknown)

Time = 80

Acceleration = y (unknown)

Mass moment of inertia = 11.5x9.8 = 112.7N.m²

15000x60 = 900000r/s (divide by time in seconds to find acceleration)

900000/80 = 11250r increase per second .. I can tell I'm lost

Why did you multiply by g?

Also you're going to want everything in the same units. What is one revolution in radians? What is one minute in seconds? Given that, what's the conversion between revs/minute and rads/second?

 

The equation you want is [math] \tau = I \alpha [/math] where I is the moment of inertia in SI units, tau is the torque and alpha is the angular acceleration.

 

You're on roughly the right track for finding the angular acceleration. Formally (given constant torque) the equation is:

[math]\alpha = \frac{\Delta \omega}{\Delta t} [/math]

But your general logic is sound. You should be careful with unit conversions though. A good way of getting them right is along these lines:

Let's say I want to convert 150 km/h to m/s

[math] 1km = 1000m\; \rightarrow \; 1 = \frac{1000m}{1 km} = 1000 {m}{km} [/math]

[math]60 s = 1min[/math]

[math]60 min = 1h[/math]

[math]\rightarrow 60\times60 s = 1h \rightarrow 1 = \frac{1h}{3600s} = \frac{1}{3600}\frac{h}{s}[/math]

 

This means I can multiply or divide anything by [math] \frac{1}{3600}\frac{h}{s}[/math] without changing it, because this quantity is equal to 1.

 

So I get my original figure 150km/h and multiply/divide to cancel out the units (treating them as if they were algebraic quantities.

[math]150 km/h \left(\frac{1}{3600}\frac{h}{s}\right) \left(1000\frac{m}{km}\right)[/math]

Rearranging:

[math]150 km/h\frac{m}{km}\frac{h}{s} \frac{1000}{3600}[/math]

Then you can cancel things out to get:

[math] 150 \frac{1}{3.6} m/s = 41.67 m/s[/math]

 

This is a bit laborious and abstract, but guarantees you won't mix up dividing/multiplying the ratios between your units.

SECOND QUESTION

I can't find a question I copied down to cover this, so any help would be appreciated, I need to be able to do questions revolving around angular motion, all I remember is it's units, rad/s

 

Here's a good source for explanations and equations related to rotational motion http://hyperphysics....hbase/circ.html

 

Here's another thread with some questions for testing/improving your understanding of torque: http://www.sciencefo...__1#entry637202

They're fairly conceptual rather than rote application of the formulae and some of them require calculus, so don't be disappointed if you don't understand them all.

 

Perhaps you could tell us what level you are at mathematically (high school, college, some calculus/no calculus etc -- I'm guessing high school from that question but cannot be sure) and we could come up with some more appropriate questions. Also have a read of that hyperphysics site and perhaps ask questions based on more specific concepts.

Edited November 26, 2011 by Schrödinger's hat

[Topher]

I multiplied by gravity to turn the units from kg.m to N.m .. obviously didn't need to do that :/ ... Ok so, alpha = change in w/change in time .. so alpha = (11250r x 57.3)/80 = 8057.8125 ... which means Torque = I x alpha = 11.5 x 8057.8125 = 92664.84375 .. is that correct?

 

Just heading to work will look at the angular motion web sites you gave me when I get back, thanks for the help

[Schrödinger's hat]

I multiplied by gravity to turn the units from kg.m to N.m .. obviously didn't need to do that :/ ... Ok so, alpha = change in w/change in time .. so alpha = (11250r x 57.3)/80 = 8057.8125 ... which means Torque = I x alpha = 11.5 x 8057.8125 = 92664.84375 .. is that correct?

 

Just heading to work will look at the angular motion web sites you gave me when I get back, thanks for the help

 

Without scrutinizing your calculations in detail (including units makes this a lot easier). You didn't put units on your answer (put units on your answers). If it was supposed to be SI, I can say 93 kN is highly unlikely. Did you read my section about unit conversion?

 

 

 

Just in case the message wasn't clear.

Put units on all your quantities as you define them.

Put units in your calculations for unit conversion.

Keep track of units in your calculations. Yes, write them in even if it's a bit tedious.

Units

UNITS

UNITS

UNITS

UNITS

[Topher]

alpha = (11250r/s/s x 57.3)/80 = 8057.8125rad/s² ... which means Torque = I x alpha = 11.5kg.m² x 8057.8125rad/s² = 92664.84375kg.m² rad/s² .. how do I change the units into just the N lol .. obviously done something wrong..

the acceleration in revolutions per second per second needed (on average) to achieve 15000 revolutions per minute in 80 seconds ... LOL how do I convert that to N?...

[Schrödinger's hat]

Sometimes units involving rotation can be a little tricky and disguise themselves as something else, but what I was mostly getting at is the whole conversion step and justifying the numbers you were using.

Maybe go back a few steps and lay it out a bit more formally.

What is the torque required to accelerate a turbine undergoing a dynamic balancing test from 0 to 15000 r/min in 80 seconds, it's mass moment of inertia is 11.5kg.m²

 

So

[math]\omega_1 = 15000 \frac{r}{min} = 15000 \frac{rev}{min} \times \frac{2\pi rad}{1 rev} \times \frac{1min}{60s} = 15000\times\frac{2\pi}{60} \frac{rads}{s} = 1571 \frac{rad}{s}[/math]

(keeping a couple extra sig figs around for now as guard digits even though 15000 only has two.)

[math] \alpha = \frac{\omega_1-\omega_0}{\Delta t} = \frac{1571 rad\, s^{-1} - 0 rad\, s^{-1}}{80 s} = 19.63 rad\,s^{-2} [/math]

 

Multiply by moment of inertia.

[math] \tau = \alpha I = 19.63 rad\,s^{-2} \times 11.5 kg m^2 = 225.8 rad\,kg\,m^2s^{-2} [/math]

 

Our units are a bit cumbersome here, but this is correct.

To find the conversion between Newtons and kg m/s^2, F=ma can be used.

Put in 1 for each of the quantities and you recover:

[math] 1N = 1kg\,ms^{-2} [/math]

 

So we can switch this with the kgms^(-2) with Newtons in our equation to get

[math] 225.8 rad\,kg\,m^2s^{-2} = 225.8 rad\,Nm [/math]

Or, 230 rad Nm to 2 sig figs.

 

Why are there some pesky radians hanging around, I here you ask?

Well they're here to derail my point about the importance of units somewhat.

 

 

There is actually a suppressed factor of rad^(-2) in the mass moment of inertia that the problem gave us, but radians are a bit odd.

 

 

They're somewhat of a fundamental unit of angle, and they are the only angle-like quantity.

If you use them in all your calculations you can safely ignore them -- as long as you are careful about one or two other points.

For example, energy has units of Joules = Nm

Torque also has units of Nm which should also equal Joules

But these are very different concepts. Energy is torque multiplied by an angle. So there is a hidden rad^(-1) in the units for torque.

 

 

Putting in our hidden radians gives us:

[math] 230 Nm/rad [/math]

Which is the units of torque (without the radians suppressed).

 

I prefer to keep the radians in when I think about it (which isn't always), and this is good practice for someone who is learning.

Unfortunately the convention in many textbooks is just to drop them, so you may be forced to just let them slide unless you know how to spot all of the missing ones.

 

TLDR you can safely ignore units of radians as long as you know what your equation represents.

 

Edit: I found an article backing all this radian nonsense up and going into more detail if anyone is skeptical/interested:

pdfserv.aip.org/PHTEAH/vol_30/iss_3/170_1.pdf

 

Edit: Flipped incorrect fraction, see below

Edited November 27, 2011 by Schrödinger's hat

[Topher]

I don't get this part .. still .. it's the only part I'm not understanding right now I'm unsure of how you managed to get that answer can you please expand it so I can see what your doing at each point .. this is how my working goes when I do what you did .. lol ... 15000 x rev/min x 2pi rad/rev x 60s/1min = 15000 x 376.99rev rad/min rev min (divide number by 60 2 times to turn the minutes to seconds) 1570.7rad/s x s .. why do I end up with the extra second multiplied below the divide? .. and also why are you using those numbers to multiply .. are they from a formula/what do each of them stand for .. obviously15000rev/min is the change in speed but the other 2 I'm not understanding..

 

EDIT: the hyperlink was just your last posts first equation.

Edited November 27, 2011 by Topher

[Schrödinger's hat]

Apologies, I made a typo (or rather copy and pasted the wrong fraction without paying attention).

Also if you click on the equations it will give you the code used to generate them

 

[math] \omega_1 = 15000 \frac{r}{min} = 15000 \frac{rev}{min} \times \frac{2\pi rad}{1 rev} \times \frac{1min}{60s} = 15000\times\frac{2\pi}{60} \frac{rads}{s} = 1571 \frac{rad}{s} [/math]

 

Basically I'm multiplying by 1 which in this case is disguised as [math]\frac{2\pi rad}{1 rev}[/math]. Then multiplying by 1 again [math] \frac{1min}{60s}[/math]

This is one way of changing units so that you don't lose track of things.

Basically I can write [math] 60s=1min [/math] and then divide both sides by 60s to get the fraction. This allows you to multiply out the minutes cancelling them as you did (min/min=1) and introduces the seconds.

I accidentally flipped the fraction, it's fixed now.

You can also just go [math] 1min = 60s[/math] and replace min with 60s everywhere you see it (and similarly rev with 2pi rad). this sometimes leads to nested fractions though and I've found people more frequently get confused.

 

Dividing by 60 (twice) was the correct thing to do to fix the mistake I made, the extra s unit came from that same mistake. (if you follow the revised working you see that the units work too.

 

This is all just a formal way of doing unit conversions. Ie. your dividing by 60 to convert minutes to seconds is just compressing the multiplying and cancelling of units into a single step.

I went into detail on this because you were doing many things in your conversion that were not mathematically consistent. If you follow the scheme of finding the correct '1' and multiplying by it (this does not change the value of the equation), then cancelling units you won't get strange answers. On top of this, if you make a mistake (like I did :/ ) the units will tell you there is something wrong.

If you're in the habit of just randomly multiplying/dividing by 60 when you see minutes, then replacing it with seconds; you won't be any the wiser if you did the wrong operation.

In the same vein, you can't randomly multiply by g to try and get the right units. This changes the equation to one that does not represent torque.

[Topher]

Thanks, great help!