# Homework Help: Maximum velocity of circular motion

1. Apr 22, 2010

### lemon

1. The problem statement, all variables and given/known data

A mass of 0.30kg is attached to one end of a string of length 0.50m is whirled in a horizontal circle with gradually increasing speed.

a) Determine the maximum linear speed reached if the string breaks when the tension reaches a value of 25N.

b) What is the corresponding rotational speed in rad/s?

2. Relevant equations

Vmax=2(pi)fA
Tcos(theta)=mg

3. The attempt at a solution

Not sure what to do
I need only the frequency to find max velocity but how?
f=1/T
360 degrees = 2pi
So is my frequency 1/2pi=0.1591549431?

2. Apr 22, 2010

### housemartin

well, don't know if my advice will be of any help, but try to picture the situation - what forces act on the mass and how are they related? One useful expression might be F=mv2/R

3. Apr 22, 2010

### lemon

There are only 2 forces acting on the ball
The centripetal force of Tension acting toward the centre and the weight acting downward.
I just read that the tension provides component forces to balance the weight of the mass and provide centripetal force.
It said 'forces'. So do I need to break this into horizontal and vertical components? One vertically upwards?

Or just simply rearrange the equation you gave for velocity? I thought about this but couldn't see where to go next. I couldn't make the step from velocity to...

Ahh! So stupid. - maximum linear speed IS velocity, right?

V=6.45m/s

yes? - I'm totally gonna kick myself if that is correct

4. Apr 22, 2010

### clombard1973

Yup

Yup,
The equation that states F = mV2/r is the equation for the centripetal force acting upon the rotating mass. The V is the tangential velocity, or the linear velocity as you put it. There is another equation that relates centripetal force to angular velocity, which is the velocity you are looking for, it is F = mw2r where w is the radial velocity, sometimes called the radial frequency (how many times per second it rotates around). You know the force applied, the mass, and the length of the string, you should be able to solve for the unknown values now.

Also since V2/r = w2r then V2 = w2r2 which is just V = wr, which should make sense because w has units of radians per second and r has units of meters, so the product of the two yields meter-radians per second. Now radians are a unit-less concept used to describe the angle some radius is at within that circle; for instance 180 degrees around a circle is equivalent to pi radians around the circle, so you can drop them from the unit analysis leaving V with units of meters per second, which is correct for velocity, so that's another way to find the angular velocity/frequency from the tangential velocity (V = wr). Or you can use the F = mw2r to find the angular velocity/frequency of the rotating mass and then also use F = mV2/r to find the tangential velocity and then use V = wr, plug in the values of w and V that you got and solve for r and see if it is the correct r, if it is, then you more than likely have the correct answers.

Now I haven't given you an exact answer, all I did was confirm your notion that the V in the function is the correct velocity and I showed you another form of the same equation using angular velocity/frequency and the radius and not the tangential velocity and the radius. I'm only saying this because there's a good chance this reply will be seen as giving away too much information and will be removed by the moderator. So this is for him/her more than it is for you. I didn't answer any question directly, I only confirmed what this person had thought and showed him/her a different form of the centripetal force applied upon a rotating mass, which happens to have one of the values he/she is looking for in it, but so did the F = mV2/r reply and that wasn't removed.

Now back to lemon. I hope this has helped you some.

Many Smiles,
Craig

5. Apr 23, 2010

### housemartin

well your answer is almost correct (numerically, i got 6.41 m/s for linear speed), but i am not sure that the way you get it was correct ;] Tension of the string is not acting horizontally - its has vertical and horizontal components. Thats because it has to compensate force of gravity (which acts downward) and provide centripetal force (which act horizontally). Also i would say that there are three forces acting on the mass - gravity, string tension, and the name of the third i do not know ;] well if you were hanging on that string and rotating, you could pretty much tell that there is something that push you outward - this forces has magnitude F=mv2/R, and strings tension horizontal component has to provide same force in magnitude, but acting toward the center of circle of rotation.
Velocity is a vector (you have to know its direction) and speed is just modulus of this vector - so they are not the same.

6. Apr 23, 2010

### clombard1973

I don't understand your point. Centripetal acceleration acts to induce a centripetal force that acts inward towards the center of the circle. If the string breaks due to the mass spinning around at too great of an angular speed, it is the centripetal force that is too great upon the string that breaks it. It has nothing to do with gravity on the string nor mass. The tangential velocity fond using F = mV2/r is the tangential velocity at any point around the circle, and if the string breaks that mass moves away from that circle at an angle that is tangent to the point where the mass was when the string broke; making the mass move off in a single straight line direction. So no velocity is not speed as velocity is a vector and speed is a scalar, but the linear velocity he is looking for is the velocity of the mass when the string has broken at which point the mass is no longer changing direction and is moving away from the circle in one direction and with a constant velocity (ignoring any frictional effects from air that slows it down and gravity that pulls it toward the ground, not that it affects the speed at which the mass moves away from the circle at).

So once the string has broken it really is just moving away at some speed and constant direction, so it can be viewed as a scalar speed only since the direction will not and does not change.

Angular velocity is how many times per second the mass is rotating around some circle. It too is found directly from the centripetal force by F = mw2r. Some people call it the angular speed, or angular velocity or angular frequency, since the one value describes all three. This function basically states the same thing as the other, and that is, the faster the mass is spinning the greater the centripetal force acting upon it. Hence the reason the string breaks at some point.

You do not even have to take gravity into account for this problem. The questions asked were for the tangential velocity (although it wasn't stated as such, but that is what they want) and the angular velocity, which is the angular frequency. Both of these values can be found without gravity affecting their values. That's because gravity has no effect on the speed at which the mass leaves tangentially from a point along the circumference of the circle it is rotating through, other than to pull it down wards, but they want to know what the speed of the mass is when the string first breaks freeing the mass to move tangentially away from the point along the circle it was at when the string broke. That's just the tangential velocity at that point in time.

The angular velocity too is independent of gravity. How many times in a second the mass moves in a circle has nothing to do with the downward pull gravity has on the mass, it's a function of the applied force that spins the mass in a circle.

And no there are not three forces acting on the mass there are only two. Gravity which acts to pull the mass down wards and the centripetal force which acts to pull the mass inward toward the center of the circle it moves in. Think about it, what are the forces acting on the mass. There's the string being held onto that also holds that mass in a circle and gravity that acts to pull the mass down wards. That's it, no other forces. You may be thinking of centrifugal force. That force is not real, but can be viewed as a force acting on the mass, but mathematically one doesn't use it in analyzing this problem, nor any periodic circular motion problems, other than for "insight" into how the forces act on a rotating body of matter.

So I don't understand what your point is. The tension on the string that has anything to do with the tangential velocity is solely due to the centripetal force that acts to pull the mass in wards, so using F = mV2/r and then solving for V is the correct method for finding the speed at which the mass leaves the circle in a direction that is tangent to the point where that mass was when the string broke.

You can leave gravity out of this entire problem. The two questions asked are independent of any vertical forces and are only dependent on the horizontal forces, namely the centripetal force. If you do not believe me try it out. I know when I was in college we actually setup, predicted the angular velocity, the tangential speed the mass would move away at when the string we were using broke, and the centripetal force that broke the string, and we got great agreement between the theoretical values and the measured ones. That's the scientific method in action and it is hard to argue with it, and seem sane to everyone else. The tension in the string is only important due to it breaking at some pull on it from the spinning mass, which is directly related to the masses tangential velocity and angular velocity; having nothing to do with gravity.

I can ensure you everything I wrote is true. I have four advanced degrees in electrical engineering including my Doctorate and have been a full professor at our states university for over a decade now, so I know something about this stuff and I teach it daily. Leave gravity out of the problem and just use the equations for the centripetal force to find the tangential velocity and the angular speed (velocity/ frequency/ whatever you want to call it) and you'll have the two answers you are looking for, if you are looking for the same two answers the original poster is.

Many Smiles,
Craig

7. Apr 23, 2010

### housemartin

about three forces, yeap - i am wrong - you don't need that to solve this problem;] but
http://www.ipix.lt/desc/74523253 [Broken]
as far as i can see it, in absence of gravity (g=0): Ft = Fn, that is angle theta = 0. But when you take gravity into account, Fn is not simply Ft, but Ft+mg (as vector sum). But i don't hold any degree of education - so i might be wrong ;]

Last edited by a moderator: May 4, 2017
8. Apr 23, 2010

### inutard

I find the above post very difficult to swallow. I believe that the force of gravity on the mass must be included in the 25 N of breaking tension (of course I do not have "four advanced degrees in electrical engineering and a doctorate" as clombard1973 claims).

Thus, the breaking tension is actually equal to the resultant force of gravity + maximum centripetal force. Drawing a vector diagram of forces and a simple application of the pythagorean theorem will solve this problem easily.

Last edited: Apr 23, 2010
9. Apr 23, 2010

### AimAce

Excuse me, but I feel that I should reply, to inutard and housemartin's chagrin, that gravity would indeed not play a role in the speed of the mass once it has broken from the string, (or more specifically, once the tension in the string reaches maximum). The reason for this is because although gravity pulls downwards, it pulls downwards evenly from all direction, meaning that the forces cancel out, thereby not affecting the actual velocity of the mass.

I don't understand how anyone that registers to these physics forums can make such a rudimentary mistake. I hope that the original inquirer of the question has figured out the correct solution through Clombard1973's post. Gravity will not play a role.

10. Apr 23, 2010

### ehild

Yes, The vertical component of the tension balances the weight and the horizontal component provides the centripetal force.

*1 Horizontal component: Tsin{alpha}=mv^2/R

*2 Vertical component: Tcos{alpha}=mg

*3 From the geometry of the set-up, it comes that sin{alpha}=R/L,

therefore

T^2cos^2{alpha} = T^2(1-(R/L)^2)=(mg)^2.

You know the maximum allowed tension, so you can calculate the radius of the circle from this last equation.
Knowing R, you can determine sin(alpha), and with that, the linear speed from equation *1.

Edit: I made a mistake in the picture before: alpha should be the angle of the string with respect to the vertical. I replaced the figure

ehild

Last edited: Jun 29, 2010
11. Apr 23, 2010

### clombard1973

I like this, but a small problem

Hi there,
The only thing I think that may be incorrect with your reply is the sine and cosine terms. The tension vector is pointing somewhat upwards and towards the origin in the diagram you provided, and the two component forces are the weight, mg, and the horizontal centripetal force labeled FCP in your attached diagram. So the vertical forces are just the gravity and sin(alpha) = mg/T, but you have cos(alpha) = mg/T. Cosine is the side adjacent over the hypotenuse, and sine alpha is the side opposite over the hypotenuse and the side opposite to the angle alpha in your diagram is the weight mg, so I think you just accidentally mixed the sine with the cosine. So it should be according to your diagram:

Vertical forces = mg = Tsin(alpha) (which should make sense because if the angle is 0, the forces only act in a horizontal direction)

Horizontal forces = FCP = Tcos(alpha) (which should also make sense because again if the angle is 0 then the entire tension along the string acts inward only)

and cos(alpha) = R/L where R is the radius and hence is the adjacent side to the angle alpha in your diagram, so you want cos(alpha) = R/L. This too should make sense because if the angle is 0 then the mass is moving in a plane with the origin and the length of the string is the radius of the circle the mass is moving around (cos(0) = 1 = R/L, therefore R = L).

Other than this, your analysis is perfect. Your diagram does show a mass hanging from a fixed point moving in circles with not enough speed to bring the mass up so that alpha approaches 0, but that's what happens when the mass is rotated faster and faster. The mass approaches circular motion in a plane and alpha approaches 0, therefore the centripetal force approaches the maximum tension before the string breaks.

So we have:
FCP = Tcos(alpha) = TR/L now as the mass is spun faster and faster at some point the mass moves nearly in a plane so alpha is nearly zero leaving almost the entire tension along the string to be caused by the centripetal force only, so Tcos(0) = TR/L or T = TR/L, R = L, so the radius is the string length if the mass is moving in circles in a plane.

Now for "inutard", first I don't appreciate your condescending reply given that I know what I'm talking about and you are simply guessing, so why not read what I have here and maybe you will learn something. Tangential and angular velocity functions were derived by Newton and Huygens in the latter part of the 17th century. It has been known for all of this time that F = mV2/r = mw2r is the centripetal force and is a inward acting force only. We are given that when the Tension reaches some value the string breaks and furthermore that when this happens the mass is moving in circles in a near horizontal plane (this isn't explicitly stated but the weight of the mass is 2.94 N and the tension the string breaks at is 25 N, so there's more than enough centripetal force to spin the mass in a near plane), so the tension that breaks the string is virtually the centripetal force acting on the mass through the string. If the mass were rotating slowly enough so that it hung kind of low while moving then gravity would apply a force on the mass that would also act along the string to the center of the circle (like in the attached diagram echild has on his reply), but if the mass is moving in circles in a near horizontal plane the 25N force that breaks the string is almost entirely the centripetal force, as the weight acts to pull the mass down but has virtually no effect on how the mass is pulling on the origin with the centripetal force it has; because the weight of the mass is virtually canceled from the lift it gets from the vertical component of the tension acting along the string. That is why gravity really is not an issue in finding the tangential or angular velocity.

The mass itself only weighs 2.94 N, so if the tension in the string reaches 25 Newtons, there is no doubt that is enough force to keep a 2.94 N mass moving in a near circular plane while rotating. The centripetal force need only overcome the weight of the mass to lift it nearly into a circular plane therefore if the tension reached 25 Newtons the origin feels that force just before the string breaks. The force upon the origin is due almost entirely to the centripetal force so 25 N = mV2/r = mw2r. Gravity will always pull down on the mass so the inward force should always be a function of the angle between the mass and the plane the origin sits in. As the speed of the mass rotating is increased it acts to "lift" the mass upwards. When it is first starting to rotate the mass hangs down quite a bit and therefore the tension applied upon the origin from the mass has much to do with the weight of the mass, but as the mass is spun faster the centripetal force acts to pull the mass inward, but since its hanging a little low that inward force can be separated into a horizontal and vertical component, the vertical component acts to lift the mass so that it spins more or less around the origin in the same plane the origin is in.

It cannot lift it to a perfect none zero angle alpha because the closer alpha gets to 0 the greater the vertical component of the string tension needs to be in magnitude to keep the mass lifted and spinning within a plane the origin lies within, but again it cannot lift it entirely into that plane as the angle alpha then is zero and that means there is no vertical lift on the rotating mass, which would cause it to drop a little where there is a vertical component to the tension along the string that acts to keep it nearly in the same plane as the origin; it comes very close to spinning in the same plane as the origin as the centripetal force is bout 10 times the weight of the mass.

So is the angular velocity and tangential velocity independent of gravity? In the strictest sense, no, but for all practical purposes gravity can be ignored and the results are almost identical.

Most periodic circular motion problems do not involve gravity, typical examples may be the moon orbiting the Earth. Of course this is a perfect example of the gravitational force between the two bodies of matter being equal to the centripetal force between them, but the point is there are no other differently directed forces acting on the moon, so the analysis is simply the force between the two bodies is the centripetal force and from that and knowing the distance between the Earth and the moon one can find how fast the moon is spinning around the Earth and the tangential speed of the moon at some location around its orbit at some point in time.

So in the strictest sense, gravity does have an effect on the values desired, but the effect is negligible, particularly at high rotational speeds. Even for this problem if one accounts for gravity in finding the centripetal force acting on the origin from the spinning mass, and someone else disregards gravity and assumes the entire 25 N is the centripetal force, the difference between the answers is one is 0.993 times the other, or there a 0.7% difference between not using gravity and using gravity to find the centripetal force that yields the values desired. A 0.7% difference is well within any margin of error for actual measured data, and the mathematics we do in physics is for predicting the outcome of an event given a set of initial conditions, so what's important is the ability to use the math to predict the physical outcome of some event. If one answer is 0.7% different from another, but is a far easier method for finding that answer, it more than justifies that tiny difference in the answers.

I had said that gravity was not needed for finding the values desired, and for the most part that is true, but in the strictest sense there is a factor of 0.007 difference between using gravity and not using gravity in finding the values desired. That is so small of a difference that I thought it would be better to tell the original poster of the question that he/she does not need gravity to solve the problem. Why make it more complicated than it has to be when the two answers are virtually identical. If you go through and do the math you'll find there is less than a 7 degree difference between the mass and the plane the origin of the circle lies within (alpha = arcsine(2.94N/25N) = 6.75 degrees) so the centripetal force is:

Horizontal force = centripetal force = FCP = Tcos(alpha) = T*0.993, so including gravitational effects reduces the centripetal force by a factor of 0.007 times the full 25 N tension the string can handle. Those two answers are so close, it is more than safe to assume gravity has no play in finding the tangential or angular velocity, I mean:

Centripetal force including gravity = 24.83 N
Centripetal force not including gravity = 25 N

Horizontal Radius including gravity = R = cos(alpha)L = 0.993*0.5m = 0.497
Horizontal Radius not including gravity = L = 0.5 m

Centripetal force including gravity = 24.83 = 0.3kg*V2/0.497
re-arranging and solving for V yields:

V = [(24.83*0.497)/0.3kg]^0.5 = 6.41 m/s

Centripetal force not including gravity = 25 N = 0.3kg*V2/0.5
re-arranging and solving for V yields:

V = [(25*0.5)/0.3kg]^0.5 = 6.45 m/s

Now does a 4 cm/sec difference in velocity justify going through and having to find and properly use the effects from gravity upon the mass? I guess that depends on who is solving the problem and what accuracy is desired, but personally a 4 cm/sec difference in velocity is so small that it is well within any measured variance one would find if they decided to solve this through experimental means, so I think not.

That's my "4 advanced degrees including my Doctorate" speaking to your condescending mind (I'm reffering to inutard not you echild, in fact I like your reply, just the little boo-boo with the sine and cosine operators was the only thing that stuck out).

Craig

12. Apr 23, 2010

### housemartin

Sure thing you can almost ignore effect of gravity in this problem, but where is the line between reasonable to neglect and not reasonable. What if mass in lemon's problem would have been 1 kg, does it not fit "practical purpose" then? Cause you would get v = 3.16 (without gravity) and v = 2.76 (with gravity). I guess before ignoring something one should know what is he ignoring. And there is nothing wrong with inutard post, and saying that everything you write is true, cause you have "4 advanced degrees including my Doctorate" is kinda strange thing to do, looks a bit silly ;]
many smiles ;P

13. Apr 23, 2010

### AimAce

Okay, no more trolling. Lemon, don't listen to Clombard, he is honestly full of poop. Gravity plays a large role in this. Maybe if it was an electron on a string, then gravity would play no role. But in anything that has visible mass, gravity is *always* one of the forces on a body diagram. Think about it like this, if gravity didn't exist, swinging that thing would be so much easier, it would take LESS force. Gravity plays a very crucial role, so you need to have gravity as the vertical force, and a cetripedal force from the swinging motion in the horizontal.

14. Apr 23, 2010

### clombard1973

The line between when to neglect gravity and when not to is dependent on the size of the mass on the end of the string and how much centripetal force is applied upon the origin from that rotating mass, typically when the weight of the mass on the end of the string is about 10 times smaller than the force applied by the centripetal acceleration plus the horizontal tension added by the weight of the mass, one can ignore gravitational effects. If the mass is increased to 1 kg then make sure the force applied at the origin is at least 98 N by spinning the mass faster around the circle. In fact the 10 times smaller rule of thumb actually improves the accuracy between the values including gravity and those not including gravity compared to the closeness of the answers given here, for instance with 1 kg of mass it weighs 9.8 N, so ten times that is 98 N, which means the separation between the mass and the plane the origin rest in is only 5.74 degrees and with a 6.75 degree difference using the weight of 2.94 N and the maximum tension before the string breaks of 25 N yielding answers of 6.41 m/s using gravity and 6.45 m/s neglecting gravity, the little over 1 degree difference by making the tensional force 10 times the weight attached to the end of the string would actually bring those two answers even closer to each other.

I had already solved the problem before even posting my first reply and that is why I was saying it was fine to ignore gravity in this case because I knew the effect it had on the final answers was negligible.

For instance, if the weight is 2.94 N and we could apply a 29.4 N force without the string breaking the new answers are 6.999 m/s including gravity and 7 m/s not including gravity. So simply look at the weight attached to the end of the string, find the total force on the string, take the ratio of the weight to the total force on the string and then take the inverse sine of that fraction and it will tell you the degrees that separate the mass from the plane the origin lies within. The difference between 6.999 m/s and 7 m/s is definitely justifiable in neglecting gravity.

As far as my comments to inutard about "being the guy with 4 advanced degrees including my Doctorate" I made because I found his reference to my credentials and years of teaching these concepts to others to be more of a "jab" at me than a recognition of the time and work I have put in earning those degrees, and teaching those students. He has a right to what he thinks is true in Physics, but if he isn't certain than he shouldn't take shots at people who are here to genuinely help others, but rather state that he believes this to be the answer and perhaps ask me why I think differently; as that is how I would have reacted to someone disagreeing with something I had posted. I wouldn't make an off-handed remark about somebody else's education and thoughts on a subject, I would state my point of view if I could prove it mathematically and then ask the person what is wrong with my method and right with theirs.

So yes, you are correct, I didn't need to take a "jab" back at inutard. For that I apologize, but if he doesn't know the answer he shouldn't say I'm incorrect when he doesn't even know if he is correct. Nonetheless, you are correct in that I didn't need to take a "jab" at him.

So inutard if you read this I apologize for being curt with you. I deal with a number of students on a daily basis and sometimes I get frustrated with some of them and take it out on others when they did nothing to frustrate me, so again I apologize. I do hope what I wrote provides some insight into this problem for you.

Many Smiles,
Craig

Last edited: Apr 23, 2010
15. Apr 23, 2010

### ehild

In my previous attachment, one "alpha" was in the wrong place. I corrected it.

As for the role of gravity: it can not be ignored. With this data, it is much less then the maximum tension, so the radius of the circle is almost the same as the length of the rope. With some other critical tension, closer to the weight of the body, R would be considerable shorter than L.

ehild

16. Apr 23, 2010

### clombard1973

Are you nuts?

I just read your reply a few above this one and at that time you seemed convinced that gravity has no effect on the mass, now it is the biggest thing to consider? Where's your mathematical justification for such a statement. Why am I "full of poop" but not you for stating the same thing in your previous reply?

I gave you a mathematical derivation of the velocities found both including and neglecting gravity and they are different by 0.7%; gravity 4.61 m/s, no gravity 4.65 m/s, a difference of 4 cm/sec. How can you call that important?

How about backing your statement up with a little mathematical rigor and not just a thought argument. Perhaps because if you do you'll get the same answers and then realize that gravity does apply a fair downwards force, but the horizontal force (the centripetal force)due to gravity acting on the mass is hardly affected. Did you not read the mathematical derivation I gave, or can you just not follow it. If you disagree with it then prove it wrong. It's pretty straightforward mathematics, good luck in showing any of it to be incorrect.

Craig

17. Apr 23, 2010

### clombard1973

That much is true and is the reason why I stated that a good rule of thumb for ignoring gravity playing on the centripetal force is to be sure that the mass attached at the end of the string weighs at least ten times less than the tension upon the string, or upon the origin if the string is inelastic in nature. If the tensional force is 10 times the weight of the mass the difference between using the the centripetal force taking gravity into consideration and not taking gravity into consideration is 0.5%. That hardly seems like a difference that needs to be accounted for.

I know one needs the gravity to find the weight, and if the total tension on the string is also known, than the centripetal force is simply Tcos(alpha) = mV2/r = mw2r. When the weight is at least ten times less than the total tension on the string, the angle alpha is 5.74 degrees or smaller (5.74 are the degrees when the mass weighs exactly 10 times less than the total tension on the string), and the cosine of 5.74 degrees is about 0.995, so the centripetal force using gravity is 0.995T = mV2/r, or T = 1.005*(mV2/r), which implies that gravity acts to increase the tension on the string by a factor of 1.005 when the weight of the mass is ten times less than the tensional force felt at the origin; why do some of you think this is needed for an accurate answer to be derived? We're talking about a difference in the total centripetal force of 0.5% if one includes gravity in finding the centripetal force compared to not using gravity to find the centripetal force.

I'm not saying gravity doesn't play a role in this problem, as one, we need it to find the radius if we are going to include gravity in the tension along the string (and it doesn't have to be an electron on the end because the mass is moving virtually in a horizontal plane so gravity has almost no effect on horizontal forces, and no AimAce, the horizontal force acting on the mass is only slightly affected by gravity, next time you find yourself in outer-space try spinning a mass around your head attached to a string, the string will still break at 25 Newtons and that force will be reached when the mass is spinning around at virtually the same rate it does in a gravitational field; BECAUSE CENTRIPETAL FORCE IS 90 DEGREES OUT OF PHASE WITH A GRAVITATIONAL FORCE, THEREFORE IT HAS ONLY A DOWNWARD AFFECT ON THE MASS, BUT IF THE MASS IS SPUN FAST ENOUGH IT LIFTS AND THE FORCE FROM GRAVITY ACTS TO PULL THE MASS DOWN BUT IT IS BUSY SPINNING AROUND IN CIRCLES HELD IN THIS PATTERN BY THE CENTRIPETAL FORCE THAT ACTS INWARDS; THIS MEANS GRAVITY ONLY DISPLACES THE STRING FROM THE PLANE THE ORIGIN IS IN BY THE SMALLEST OF ANGLES, MAKING THE CENTRIPETAL FORCE NEARLY THE SAME TAKING GRAVITY INTO ACCOUNT AS IT IS IF GRAVITY IS IGNORED.

The only difference between the two functions if gravity is used or not is this:

Using gravity the centripetal force is:

FCP = Tcos(alpha) = mV2/r

Not using gravity the centripetal force is:

FCP = T = mV2/r

where the two r's are not the same and T is the total tension felt at the origin due to the string having to constantly move the mass around in circles and the effect gravity has on the mass, while it is spinning, which will be small because the mass is moving nearly in a horizontal plane, the only thing that stops it from moving entirely in the horizontal plane the origin lies within is gravity pulling the mass downwards creating a small angle, alpha, between the mass and the plane which the origin resides in; the cosine of a small angle is nearly 1, so the centripetal force is hardly affected in this situation. The only time gravity becomes important is if the weight of the mass is much closer to the total tension applied by the string on the origin, then the two centripetal forces found will differ, perhaps significantly; but that is not the case for this problem.

I don't care, you guys believe what you want. I went through and showed mathematically the effect gravity has on the velocities that were desired and it is negligible. If you don't think a 0.7% difference between the velocities is negligible then you haven't done much lab work. If anybody disagrees with what I have said then either prove my mathematical derivation as false, or derive one of your own and show me the difference between the velocities using and not using gravity that shows a large difference justifying the need to use it in finding the centripetal force. AimAce, I thought maybe you were bright enough to see that, but apparently you either cannot follow a simple mathematical argument, or you care to ignore it and go with your "instincts" as to the importance of gravity. I don't believe people who tell me something based on their opinion, provide me with a mathematical proof like I did for you and then if I find nothing wrong with it I'll look back through mine and see if there's a mistake in it. You didn't even bother to point out the error in my mathematics (most likely because there is no problem with my derivation), you simply jumped ship from your previous reply and cut down my argument without reason and now state that gravity is so important. If I'm full of "poop", then you're just a f***ing idiot.

Again, if you disagree with me, try mathematically proving your point, or disproving mine and not simply take shots at me personally, which I find offensive coming from a dumbass like yourself (AimAce). I gave you equations and factual formulas that I used to prove my point of view, you did nothing but insult me and showed no proof of your ideas. Reactions like that from people show them to be hacks. You have plenty to say, but you provide no rigor to back up your argument, nor did you disprove anything I did; HACK!!

Craig

18. Apr 24, 2010

### inutard

Hello clombard1973, let me start off by saying something that many people should have said to you upon registering at this forum:

Welcome to physicsforums.com.

As for your reply to my post, I admit it was a rather snide remark on my end. However, doubt is part of the scientific method and it serves as a promoter of greater education rather than a deterrent. Thus, it is not up to you or I to censor critical information such as the role of gravity in the question. True, the effect could be small in the particular problem shown. Yet, I believe that we must always include the role of gravity to promote "good habits". Approximations may be extremely useful in the lab (and who can doubt the great approximations of Fermi and the likes?), but it us up to us as aids to provide the fullest information possible.

I apologize for my remarks but in turn I hope that you will provide more complete advice to whomever you choose to help in your future endeavors.

Regards,
PL

19. Apr 24, 2010

### clombard1973

I agree, but

Thank you for welcoming me to this forum, and for your gracious reply; as I was not so kind to you when I replied, I do apologize and will again before I finish.

There's no doubt that "doubt" plays a role in the scientific method, but doubt is not the same as insulting; that perhaps we both were, I certainly was. The only reason I do believe there to be no problem ignoring the gravitational effects on a spinning mass if it weighs at least 10 less than the tensional force along the string felt at the origin is because the difference in answers is almost always going to within a margin of error. You say we should try and be as precise as possible whenever we can, and usually I would agree, but the function including gravity is itself an approximation function. Newton was not entirely correct when he derived his laws of motion as he knew nothing of relativity.

So even Newton's equations including gravity are approximation functions, but certainly things like the air resistance on the spinning mass slowing therefore needing an additional bump in the force to keep the mass spinning a in vacuum like atmosphere that the equation you want to use assumes. It also assumes that no energy is lost along the string while it is spinning and most likely twisting some itself in the process which heats the string taking energy away from the final answer that should also be accounted for to come up with an accurate as possible formula. Even then I'm not sure if I've missed something, the relative effects, the air resistance, any other frictional losses that take useful energy and convert it to thermal energy (heat) is a wasteful process, and we are not accounting for these real life occurrences in this analysis.

So if a difference in a mass that weighs 9.8 N that spins with a centripetal force including gravity at a tangential velocity of 6.999 m/s and spins at a tangential velocity ignoring gravity of 7 m/s why not just say since I'm approximating the final answer either way this much of a difference adds so little variation to my final answer, that if I'm unsure if my derivation using gravity is correct, then why not just drop it and not use it.

I'm all for accuracy, but I can tell you from experience that no one model of a situation ever has the kind of accuracy that justifies putting in the additional work when the two results are going be be a factor of 1.005*one value = the other value; or perhaps even less of difference for a final answer. For periodic circular motion problems for certain. The work we do in electromagnetics often requires accuracy beyond this to be correct, but simple laws of motion from experiment do line up well with the classical models but they themselves have a variance that justifies making assumptions at times if it makes it easier on oneself and has no real effect on the final answer.

So let's just say it depends on the desired accuracy of the final result. If including gravity were not accurate enough one would have to include things left out like I have already brought up to help bring the final desired result closer to what they get for a measurement. It is more fun when working on paper to most often include as many terms as one can mathematically handle in working an equation to find a final answer, but if one just wants an answer and isn't looking to have some mathematical fun then the easiest way to do this is to take your approximate functions to begin with and then ask yourself if there's anything that can either for appropriate reasons, or for reasons of ease that have no real effect on the final answer, be removed from the equation making it even easier to solve for a desired result.

So it depends on what one wants for accuracy in their model for a predicted answer that should dictate the equations used to predict that outcome. There are so many situations where students give me answers to predicted results for things like force, distance, velocity, etc.... that have like 5 decimal places of accuracy and then beside their predicted results are the ones they measured in lab which are perhaps a full Newton or two away from their prediction, their measured lengths are off by cm's and their measured velocities are always never exactly what they predicted. There are really just a number of good and useful classical derivations that yield good results, but unless one can account for all possible inputs and outputs of energy in any form, and knows exactly how the energy state of everything is at any moment so they can exactly model the situation in a truly closed form analysis then no one formula can exactly answer a real life problem. If if we aren't using the mathematical predictions to do something useful for us, then we are just having mathematical fun.

Certainly it is good to know what and how many terms a function should have to predict a final result to the accuracy one desires, but there are also times when it is fine to make assumptions that make the method easier and change the final answer by an insignificant amount. You are correct though in that one should at least know how to if needed be more and more accurate in developing an equation to predict an outcome for a result that has yet to occur, as we as scientists are supposed to do this. Mathematicians get to have fun in the world of theory and playing with equations; but we're supposed to make something useful or fix something we've already made.

So I do see your point and I appreciate it, and also the way in which you replied to my deliberate insults was also dignified, I do apologize for stating some of the things I did write, and lastly, thank you for welcoming to this forum

Many Smiles,
Craig

Last edited: Apr 24, 2010
20. Apr 24, 2010