angular motion

[steveB]

Well, there is nothing fictitious or virtual about the force caused by a centrifuge. And, I don't hear too much discussion about its noninertial frame of reference, either. So, perhaps the physicists should explain themselves better.

Ratch

Ratch, I would say, perhaps you should learn physics better, but I'm contributing here to help PG answer his question, not to teach you physics. So, I understand that you don't get it, but I think physicists do explain themselves very well generally. So, go look at what they say, and I'm sure you'll get it.

PG, and Ratch,

As far as the question. Just to clarify, I should describe what a reaction centrifugal force is. First, the term centrifugal means a force directed out from the center, as opposed to centripetal which means in towards the center. The pseudo-force typically called centrifugal force is an outward force that exists in the moving non-inertial reference frame. This differs from the reaction centrifugal force, which is a real Newtonian force that is the reaction to centripetal force. Examples include the internal stress on a rotating turbine, motor or string, that keeps things from flying apart. In the case of a satellite orbiting a planet, the reaction centrifugal force is the force that the satellite places on the planet. Such forces are outward forces and hence are suitable to be called centrifugal. Again, physicists frown on this terminology, particularly for pedagogical reasons. They will insist all centrifugal force is virtual, fictitious or pseudo, and would prefer that you just say "reaction to centripetal force" to describe "reaction centrifugal force".

In my view, this is what makes the situation confusing. The physics is easy, but the terminology and established pedagogy make it confusing.

 

[Ratchit]

Ratch, I would say, perhaps you should learn physics better, but I'm contributing here to help PG answer his question, not to teach you physics. So, I understand that you don't get it, but I think physicists do explain themselves very well generally. So, go look at what they say, and I'm sure you'll get it.

I doubt very much that I will get it. I have learned this aspect of physics as well as I could and can. I believe I explained centrifugal and centripetal forces correctly, and any other extensions of the definitions which include changing frames of reference, relativity, tensors, or whatever, are just an obfuscation of what the OP wanted to know.

PG, and Ratch,

As far as the question. Just to clarify, I should describe what a reaction centrifugal force is. First, the term centrifugal means a force directed out from the center, as opposed to centripetal which means in towards the center. The pseudo-force typically called centrifugal force is an outward force that exists in the moving non-inertial reference frame. This differs from the reaction centrifugal force, which is a real Newtonian force that is the reaction to centripetal force. Examples include the internal stress on a rotating turbine, motor or string, that keeps things from flying apart. In the case of a satellite orbiting a planet, the reaction centrifugal force is the force that the satellite places on the planet. Such forces are outward forces and hence are suitable to be called centrifugal. Again, physicists frown on this terminology, particularly for pedagogical reasons. They will insist all centrifugal force is virtual, fictitious or pseudo, and would prefer that you just say "reaction to centripetal force" to describe "reaction centrifugal force".

In my view, this is what makes the situation confusing. The physics is easy, but the terminology and established pedagogy make it confusing.

That explanation is clear as mud. I read it over several times and I still not sure what you mean. You seem to say one thing and contradict yourself later. I am sticking to what I averred earlier. That has always worked for me.

Ratch

 

[steveB]

Ratch, I know you don't get it, and I didn't expect that you would get it. I'll wait to see if PG understands or if he is also confused by what I'm trying to say. If he is also confused, then I'll try to explain better.

 

[PG1995]

Thank you, Ratch, Steve.

I'll wait to see if PG understands or if he is also confused by what I'm trying to say. If he is also confused, then I'll try to explain better.

I believe that I understand it now but I need to go through it more carefully and see if I have any follow-on queries.

Regards
PG

 

[BR-549]

I think everyone missed the point. The skater example is to show the Law Of Conservation of Angular Momentum.

In the skater example, there is no energy exchanged.

The skater example shows that because of the Law, one can trade distance for speed and visa versa.

You are demonstrating the relationship between R and V in the angular momentum formula.

Angular Momentum, L = M x R x V. If the L and M remain constant, then R and V are inversely related to maintain balance.

Mathematicians give angular momentum a force/direction magnitude. However it is not. Angular momentum is an alignment force or magnitude, not translational.

 

[Ratchit]

I think everyone missed the point. The skater example is to show the Law Of Conservation of Angular Momentum.

In the skater example, there is no energy exchanged.

No external energy is exchanged except for ice and air friction, but there is an energy exchange from the skater's muscles to the rotational energy.

The skater example shows that because of the Law, one can trade distance for speed and visa versa.

You are demonstrating the relationship between R and V in the angular momentum formula.

Both the angular momentum and the rotational energy have to be conserved. Just like in translational movement.

Angular Momentum, L = M x R x V. If the L and M remain constant, then R and V are inversely related to maintain balance.

Yes.

Mathematicians give angular momentum a force/direction magnitude. However it is not. Angular momentum is an alignment force or magnitude, not translational.

No, physicists define angular momentum as a vector, which means it has magnitude and direction. It is proportional to the vector cross product of the radius and velocity. That means angular momentum is orthogonal to both the radius and velocity.

Ratch

 

[PG1995]

Hi

Please have a look here. I'm not able to see how the following expression constitute the moment of inertia on a round nut and also I can't see the reasoning behind its derivation.

J = (1/3)rho (R^3 - r^3)

Could you please help me? Thank you.

Regards
PG

 

[Ratchit]

Hi

Please have a look here. I'm not able to see how the following expression constitute the moment of inertia on a round nut and also I can't see the reasoning behind its derivation.

J = (1/3)rho (R^3 - r^3)

Could you please help me? Thank you.

Regards
PG

What are you trying to do? Accelerate spin of the nut on the threads, or figure out how much torque is needed? Anyway, the nut represents a hollow cylinder, so you should be able to look that up here. https://en.wikipedia.org/wiki/List_of_moments_of_inertia. I don't see to what that formula for J above relates.

Ratch

 

[PG1995]

Thanks, Ratch.

Please don't mind my saying this but are you really sure that that formula for J doesn't really useful to approximate the moment of inertia of nut?

 

[Ratchit]

Thanks, Ratch.

Please don't mind my saying this but are you really sure that that formula for J doesn't really useful to approximate the moment of inertia of nut?

In the list of moments of inertia link I sent you, the moment of inertia of a hollow cylinder is listed exactly. As far as I can see, a nut is a hollow cylinder.

Ratch

 

[steveB]

PG,

I'm not sure i understand your question about inertia for a nut, but are you questioning the formula from the reference you quoted? Because I cant make sense of that formula either. Maybe i'm misunderstanding what they wrote, but it doesnt look correct to me.

 

[PG1995]

Thanks, Ratch, Steve.

Let's forget that formula then.

The torque for Toyota Camry is given 170 lb.-ft. @ 4,100RPM [source]. I interpret it to mean that when the wheels are rotating on a flat, hard surface with maximum friction at the rate of 4,100 revolution per minute, the torque exerted by the tires is 170 Ib.-ft or 230 Nm. Is my interpretation correct? Thanks.

Regards
PG

 

[steveB]

I'll have to let someone more knowledgeable about cars answer that with certainty. I'm not sure. However, I believe that the RPMs refers to engine (crank-shaft or other component) RPMs, and not necessarily tire RPMs. As to whether the torque is related to the wheel torque or crank-shaft, I'm not sure, but I suspect it is the latter.

Also, remember that power is force times speed, or torque times rotational speed for rotary systems. So this is really telling you the power capability of the engine at that rotational speed.

 

[Ratchit]

Thanks, Ratch, Steve.

Let's forget that formula then.

The torque for Toyota Camry is given 170 lb.-ft. @ 4,100RPM [source]. I interpret it to mean that when the wheels are rotating on a flat, hard surface with maximum friction at the rate of 4,100 revolution per minute, the torque exerted by the tires is 170 Ib.-ft or 230 Nm. Is my interpretation correct? Thanks.

Regards
PG

Forget about the Toyota for a minute. Suppose I am trying to loosen a stubborn nut with a 1 foot wrench. Since I am a big guy, I can brace myself and probably exert 170 lbs at the end of the 1 foot handle. So I am exerting 170 foot-lbs of torque even if the nut remains stuck, and no energy has been expended on the nut thus far. Now suppose the nut breaks loose, and I only have to exert 100 lbs at the end of the 1 foot wrench handle to unwind the nut. At the end of 1 rotation, I have expended 2*π*100 ft-lbs of energy unwinding the nut. If it takes me 10 seconds to unwind the nut one revolution at 100 ft-lbs of torque, then I have expended energy at the rate of 2*π*100/10 ft-lbs/sec of power. So you can see that power is proportional to the product of revolutions per time and torque. Now you can scale up my example and convert the English units into metric if you desire.

Ratch

 

[JimB]

I interpret it to mean that when the wheels are rotating on a flat, hard surface with maximum friction at the rate of 4,100 revolution per minute, the torque exerted by the tires is 170 Ib.-ft or 230 Nm. Is my interpretation correct?

No.
In vehicle specifications, the torque and BHP figures are those measured at the flywheel, ie where the engine meets the gearbox.

In use the torque available at the driving wheels will depend on which gear is selected.
For example a 4x4 (SUV) in low ratio 1st gear will have much more torque at the wheels to allow it to crawl slowly up a very steep hill, than it will in high ratio 5th gear for cruising along the motorway.
I both cases the engine can be producing the same torque at the same RPM, but the forces exerted on the road by the tyres are quite different.

JimB

 

[PG1995]

Thank you, everyone.

JimB, I have been through some articles and videos and I agree with what you have said.

Please note that I don't want to turn this thread into a discussion about automobile engines therefore we will keep things summarized and to the point. Thanks.

Videos and articles which explain torque, braking horsepower (BHP), and other important stuff:
1: http://mechanics.stackexchange.com/...he-meaning-of-bhp-and-torque-in-bikes-or-cars
2: http://physics.stackexchange.com/qu...ionship-and-difference-between-bhp-and-torque
3: **broken link removed**
4: http://www.carthrottle.com/post/the-difference-between-bhp-and-whp-explained/
5: This video is a very good one to understand engine torque and engine horsepower (BHP).
6: This is another good video which shows torque-hp-rpm graphs for actual cars.

Q1:
In video #5 above, from 1:06 to 1:15 it is said, "...we are going choose an engine such that the lever arm of the crankshaft is one foot meaning that if the engine produces a 100 ft-pound...".

I don't get that what's the purpose of his saying where he says "the lever arm of the crankshaft is one foot...". Could you please help me with it?

Q2:
In this video from the article at #4 above, from 1:33 to 2:13, it says something like 'so what other things they might exclude from the engine horsepower, well if it has power steering they are not going to account for power required in order to steer the car, they may also not include the power if you have air conditioner on, other things which drain power from your engine which might not be included in what the manufacturer claiming is the horsepower are alternator, water pump, restrictive exhaust'.

I do believe that wheel horsepower and wheel torque are affected to great extent by the operation of an air conditioner and I'm sure that other things also affect but I don't think that their effect is that much. Do you agree with me?

Q3:
I have never been through the manual of any automobile but I do think that the manufacturers should show torque-BHP-RMP curve for every model. In video #6 above, at 2:42 it is shown that 2011 Mercedes-Benz S63 AMG has maximum torque over an extended range of RMP which is a good thing, in my opinion. What do you think?



Example cars:
1: **broken link removed** (Suzuki Alto)
2: http://www.auto-data.net/en/?f=showCar&car_id=3349 (Toyota Corolla VII with 1.3 XLI engine)

Suzuki Alto has:
Maximum output kW/rpm: 50/6,000 (50 kW = 67 hp)
Maximum torque N.m/rpm: 90/3,400
Engine displacement 1.0L

Toyota Corolla VII has:
Maximum power 75 hp @ 5400 rpm.
Torque 115/4300 Nm
Volume of engine 1.3L

Q4:
I have always thought that the rating for volume of engine or engine displacement really tells you about how power an engine is. In my opinion, engine torque is mostly related to engine displacement. Do you agree? Thank you.



General information:
1: http://en.wikipedia.org/?title=Joule
2: http://en.wikipedia.org/wiki/Dynamometer (device to measure toque, horsepower, rpm, etc. of motors, engines etc.)
3: This video explains horsepower.
4: This video might be good; didn't watch it carefully.
5: http://en.wikipedia.org/wiki/Horsepower#Brake_horsepower
6: speed-torque curve for AC or induction motor: http://raise.spd.louisville.edu/ECE252/images/L19-18.gif
7: speed-torque curve for DC motor: http://lancet.mit.edu/motors/colorTS1.jpg


Regards
PG

 

[JimB]

Q1:
In video #5 above, ...
he says "the lever arm of the crankshaft is one foot...". Could you please help me with it?

That video had my head spinning and I know what he is talking about!

All that he is trying to say, in a very round about way, is
that he has a pulley on the end of the crankshaft,
the pulley is 1ft in diameter,
there is a rope over the pulley,
so, the distance between the rope and the centre of the crankshaft is 1 foot.

Q2:
In this video from the article at #4 above, from 1:33 to 2:13, it says something like 'so what other things they might exclude from the engine horsepower, well if it has power steering they are not going to account for power required in order to steer the car, they may also not include the power if you have air conditioner on, other things which drain power from your engine which might not be included in what the manufacturer claiming is the horsepower are alternator, water pump, restrictive exhaust'.

I do believe that wheel horsepower and wheel torque are affected to great extent by the operation of an air conditioner and I'm sure that other things also affect but I don't think that their effect is that much. Do you agree with me?

All to do with making specifications look better to help salesmen.
It used to be that some manufacturers would quote the "Gross BHP" figure for an engine. This would be the power at the flywheel for the bare engine.
No cooling fan
No water pump
No Alternator.
No air filter
No silencer (muffler)
No anything else that you can think of which uses power from the engine.

The "Net BHP" figure would include the losses for everything needed for a practical vehicle.

The aircon can be a big load on the engine, especially for a small car with a small engine.

Q3:
I have never been through the manual of any automobile but I do think that the manufacturers should show torque-BHP-RMP curve for every model. In video #6 above, at 2:42 it is shown that 2011 Mercedes-Benz S63 AMG has maximum torque over an extended range of RMP which is a good thing, in my opinion. What do you think?

Less than 1% of owners would have any idea what it was telling them.

A broad torque range is a good thing.
An example form my own experience.
At one time I owned a Mk2 VW Golf GTI, quite a fast little car with a 1.8 litre engine.
There was good low-down torque, it would pull well from about 1500rpm.

I replace that car with a Mk3 VW Golf GTI which had a 2 litre engine and was more powerfull 150 BHP instead of 120BHP for the Mk2.
But on my daily commute to/from work it was slower!
Why, because below 3000RPM there was not as much torque as there was with the older car with the smaller engine.
It would be a bad isea to try and drive at 80MPH on the little back roads which I used on my way to work.
I would have been in 2nd/3rd gear all the way! No good for the fuel economy, not good for the nerves!

Q4:
I have always thought that the rating for volume of engine or engine displacement really tells you about how power an engine is. In my opinion, engine torque is mostly related to engine displacement. Do you agree? Thank you.

Engine displacement has a big effect on power and torque.
But there are many factors in the design of an engine which affect the power torque curves.
Cylinder bore and stroke.
Valve timing.
Ignition timing.
Length and diameter of inlet manifold.
Length and diameter of exhaust manifold.

For any basic engine, you can get good low down torque but poor (?) peak power, or, good high top end power and poor low down torque by varying these factors.
Example, I once had a Suzuki Vitara, a cheap almost crude 4x4, with a 1.6 litre engine. But it did exactly what it said on the tin!
If I drove the thing at 60MPH it scared me stiff!
But, put it in low ratio 1st gear and it would go anywhere. If you could get suitable tyres, it would climb the North Face of the Eiger* at about 2000RPM.

* A steep mountain somewhere in Europe, the basis for one of Clint Eastwoods worst films.

JimB

 

[PG1995]

Thank you, JimB. Your reply was very good and to the point.

All that he is trying to say, in a very round about way, is
that he has a pulley on the end of the crankshaft,
the pulley is 1ft in diameter,
there is a rope over the pulley,
so, the distance between the rope and the centre of the crankshaft is 1 foot.

The engine is producing 100 pound-feet torque. How would torque be affected if the diameter of pulley is 3 feet instead of 1 foot? I believe that the torque is to be measured along the circumference of pulley. Thanks.

Best regards
PG

 

[JimB]

The torque would not be affected.
Look at the units 100 POUNDS . FEET force x distance

Torque is the same, the force changes.

At 1 foot, force = 100/1 pounds

At 3 feet, force = 100/3 pounds

Be aware that there is oversimplification here.
I am on holiday, the sun is shining and I do not wish to spend the day on the computer! I can do that at home!
Talk to you later.

JimB

 

[PG1995]

Thank you, JimB.

The torque would not be affected.
Look at the units 100 POUNDS . FEET force x distance

Torque is the same, the force changes.

At 1 foot, force = 100/1 pounds

At 3 feet, force = 100/3 pounds

Be aware that there is oversimplification here.
I am on holiday, the sun is shining and I do not wish to spend the day on the computer! I can do that at home!
Talk to you later.

JimB

I would like to expand on my previous post because I'm little confused about JimB's above post.

Please have a look here. Suppose at first we have a pulley whose diameter is 0.5 foot. Then, we put up on a new pulley made up of similar material as 0.5 foot one; the only difference is the diameter.

T=Iα where T=Fr and I is moment of inertia of pulley and r is the distance from center of pulley. I'm ignoring the moment of inertia of crankshaft or rod connected to the pulley. We have F=(Iα)/r=T/r.

The 1 foot pulley will have more moment of inertia compared to 0.5 foot one (perhaps double?).

According to you, torque will remain the same which means that angular acceleration should go down because moment of inertia is going to be larger for 1-foot pulley.

What is that 'oversimplification' you were talking about? Thank you.

Regards
PG