Velocities in circular motion

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  • #26
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Direction for A is not constant, but the time is very small
 
  • #27
jbriggs444
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Direction for A is not constant, but the time is very small
The direction is nearly constant. But that nearly is exactly the problem. The resulting speed is nearly unchanged by a constant force. If you account for the fact that the direction is changing then the resulting speed is exactly unchanged.
 
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  • #28
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Faiq:

Please correct me if I am wrong.

I think what you are asking is the following: If a mass is traveling in a horizontal circle on a frictionless table, tethered by a string at the center of the circle, is a constant speed for the mass consistent with the Newton's 2nd law force balances on the mass in the x and y directions (Cartesian coordinates) for the entire motion around a circle? In other words, is there as solution to Newton's 2nd law equations consistent with constant speed motion of the mass in a circle tethered at the center with a string (even if one uses Cartesian coordinates to analyze the motion)?

Does this capture the essence of what you are asking?
 
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  • #29
sophiecentaur
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The force is centripetal in nature to A but is just a normal resultant force to B.
What is the difference between the motions? At any instant, the object moving in a circle would travel in a straight tangential line if the force were removed (there is no 'memory' of what it's doing. A and B would be deflected by exactly the same amount.
You need to understand how the calculus can be used in situations like this. You have to go through the derivation with extreme care in order to work out the limits for the quantities are as δt → 0 when you start the calculation using finite time and small straight line motion. Some quantities are unchanged and some go to zero (as with all differentiation from first principles when you start with tan(θ) = δy/δx etc. etc.) You quoted dx/dt but how was it derived from scratch? Understand that and your questions and worries will be answered. It was second year A level for me, way back and Mr. Worthington was extremely meticulous about it all.
 
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  • #30
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Faiq:

Please correct me if I am wrong.

I think what you are asking is the following: If a mass is traveling in a horizontal circle on a frictionless table, tethered by a string at the center of the circle, is a constant speed for the mass consistent with the Newton's 2nd law force balances on the mass in the x and y directions (Cartesian coordinates) for the entire motion around a circle? In other words, is there as solution to Newton's 2nd law equations consistent with constant speed motion of the mass in a circle tethered at the center with a string (even if one uses Cartesian coordinates to analyze the motion)?

Does this capture the essence of what you are asking?
Yes that's what I am trying to ask
 
  • #31
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OK. Let's see how this plays out. The equation for a circle is $$x^2+y^2=r^2\tag{1}$$. The circular motion of the mass moving at constant speed can be expressed parametrically as $$x=r\cos \left(\frac{v}{r}t\right)\tag{2}$$
$$y=r\sin \left(\frac{v}{r}t\right)\tag{3}$$where v is the (constant) speed and t is time. Eqns. 2 and 3 satisfy Eqn. 1 exactly. How do we know that this is also consistent with a constant speed for the mass? If we take the derivative of the coordinates of the mass x and y with respect to time, we obtain the x and y components of velocity:
$$v_x=-v\sin \left(\frac{v}{r}t\right)\tag{4}$$
$$v_y=+v\cos \left(\frac{v}{r}t\right)\tag{5}$$
If we take the sum of the squares of these two velocity components, we (automatically) obtain the square of the (constant) speed v:
$$v_x^2+v_y^2=v^2\tag{6}$$
Not let T be the tension of the string attached to the origin. At any time t, the components of the tension in the x and y directions are:
$$T_x=-T\frac{x}{\sqrt{x^2+y^2}}=-T\cos \left(\frac{v}{r}t\right)\tag{7}$$
$$T_y=-T\frac{y}{\sqrt{x^2+y^2}}=-T\sin \left(\frac{v}{r}t\right)\tag{8}$$
Now for the force balances. The force balance in the x direction is given by:
$$m\frac{dv_x}{dt}=T_x=-T\cos \left(\frac{v}{r}t\right)\tag{9}$$
The force balance in the y direction is given by:$$m\frac{dv_y}{dt}=T_y=-T\sin \left(\frac{v}{r}t\right)\tag{10}$$

@Faiq: Is this all OK with you so far?
 
  • #32
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Oh okay so every change in horizontal components is converted into vertical components of velocity hence keeping the speed constant and the components are also consistent with newton's second law.
Just one last query, when an object is in circular motion and is let's say horizontal. The acceleration is perpendicular to object's horizontal velocity. So in the next instant won't the object's velocity be greater than previous horizontal velocity as there was an increment in vertical component but there was no component of acceleration to reduce the horizontal component?
 
  • #33
jbriggs444
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Oh okay so every change in horizontal components is converted into vertical components of velocity hence keeping the speed constant and the components are also consistent with newton's second law.
Just one last query, when an object is in circular motion and is let's say horizontal. The acceleration is perpendicular to object's horizontal velocity. So in the next instant won't the object's velocity be greater than previous horizontal velocity as there was an increment in vertical component but there was no component of acceleration to reduce the horizontal component?
There is no such thing as a 'next' instant. Time is continuous, not discrete. The acceleration is only vertical for an instant of zero duration.

Edit to add: If you look at the object after it has attained some non-zero vertical velocity then the average acceleration over the non-zero duration that it took to acquire that velocity will not have been purely vertical.
 
  • #34
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Oh okay so every change in horizontal components is converted into vertical components of velocity hence keeping the speed constant and the components are also consistent with newton's second law.
Just one last query, when an object is in circular motion and is let's say horizontal. The acceleration is perpendicular to object's horizontal velocity. So in the next instant won't the object's velocity be greater than previous horizontal velocity as there was an increment in vertical component but there was no component of acceleration to reduce the horizontal component?
Why don't you use the equations I presented to help you analyze this exact situation?
 
  • #35
sophiecentaur
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Oh okay so every change in horizontal components is converted into vertical components of velocity hence keeping the speed constant and the components are also consistent with newton's second law.
Just one last query, when an object is in circular motion and is let's say horizontal. The acceleration is perpendicular to object's horizontal velocity. So in the next instant won't the object's velocity be greater than previous horizontal velocity as there was an increment in vertical component but there was no component of acceleration to reduce the horizontal component?
no.
What does "greater" mean when referring to a vector?
You really cannot let this lie, can you?
 
  • #36
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Why don't you use the equations I presented to help you analyze this exact situation?
The equations aren't resulting in any contradiction but I want to understand the reason.
The problem is whenever the velocity is horizontal and the acceleration is perpendicular to it, there's no component of acceleration to reduce the horizontal component but there's a component of acceleration to increase the vertical component. So, theoretically, the next velocity has to be greater (In magnitude) than the previous velocity because of the added component
 
  • #37
A.T.
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So, theoretically, the next velocity has to be greater (In magnitude) than the previous velocity because of the added component
There is no "next velocity". Time is continuous.
 
  • #38
sophiecentaur
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The equations aren't resulting in any contradiction but I want to understand the reason.
The problem is whenever the velocity is horizontal and the acceleration is perpendicular to it, there's no component of acceleration to reduce the horizontal component but there's a component of acceleration to increase the vertical component. So, theoretically, the next velocity has to be greater (In magnitude) than the previous velocity because of the added component
No. The "theory" you need to consult to is basic calculus (I have mentioned this before) which investigates what happens in an infinitesimal time by studying what happens in the limit as δt approaches zero. You are quietly ignoring this and coming to your own conclusion - which is just plain wrong. It boils down to the fact that the gradient of cos(φ) approaches zero as φ approaches zero and the gradient of sin(φ) approaches 1. The tangential acceleration is zero and the radial acceleration is v2/r however much you try to wave your arms and say that can't be true and whatever you 'believe'. This is very elementary Maths book work and, rather than trying to argue against it (arguing with a number of well informed PF members and the whole of Maths as well)) you should look into the Maths in detail - not inventing your own version. Intuition can be a very bad friend; people lose fortunes when they solely rely on it.
I am getting ratty with you because you are just not listening to what you have been told on good authority. Just consider that you could be wrong.
 
  • #39
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The equations aren't resulting in any contradiction but I want to understand the reason.
The problem is whenever the velocity is horizontal and the acceleration is perpendicular to it, there's no component of acceleration to reduce the horizontal component but there's a component of acceleration to increase the vertical component. So, theoretically, the next velocity has to be greater (In magnitude) than the previous velocity because of the added component
The problem is, you're not using a sufficiently accurate finite difference approximation.

Consider the times t = 0 and t = Δt.

At t = 0,
$$x(0)=r$$
$$y(0)=0$$
$$v_x(0)=0$$
$$v_y(0)=v$$
$$T_x(0)=-T=-m\frac{v^2}{r}$$
$$T_y(0)=0$$

At t = Δt,
$$x(\Delta t)=r\cos{\left(\frac{v}{r}\Delta t\right)}\approx r\left(1-\frac{1}{2}\frac{v^2}{r^2}(\Delta t)^2\right)$$
$$y(\Delta t)=r\sin{\left(\frac{v}{r}\Delta t\right)}\approx v\Delta t$$
$$v_x(\Delta t)=-v\sin{\left(\frac{v}{r}\Delta t\right)}\approx -\frac{v^2}{r}\Delta t$$
$$v_y(\Delta t)=+v\cos{\left(\frac{v}{r}\Delta t\right)}\approx v\left(1-\frac{1}{2}\frac{v^2}{r^2}(\Delta t)^2\right)$$
$$T_x(\Delta t)=-m\frac{v^2}{r}\cos{\left(\frac{v}{r}\Delta t\right)}\approx -m\frac{v^2}{r}\left(1-\frac{1}{2}\frac{v^2}{r^2}(\Delta t)^2\right)$$
$$T_y(\Delta t)=-m\frac{v^2}{r}\sin{\left(\frac{v}{r}\Delta t\right)}\approx -m\frac{v^3}{r^2}\Delta t$$

The differential equations for the force balances are: $$m\frac{dv_x}{dt}=T_x$$
$$m\frac{dv_y}{dt}=T_y$$

The first order (in Δt) forward finite difference approximations to these differential equations over the time interval Δt are:
$$m\frac{v_x(\Delta t)-v_x(0)}{\Delta t}=T_x(0)=-m\frac{v^2}{r}$$
$$m\frac{v_y(\Delta t)-v_y(0)}{\Delta t}=T_y(0)=0$$
The solution to these equations for the velocities at time Δt are:
$$v_x(\Delta t)=-\frac{v^2}{r}\Delta t$$
$$v_y(\Delta t)=v$$
By comparing with the values of ##v_x(\Delta t)## and ##v_y(\Delta t)## with the second order accurate values above, we see that the x velocity is accurate to terms of second order in ##\Delta t##, but the y velocity is not. Furthermore, for this approximation, if we take the sum of the squares of the velocity components, we obtain:$$(v^2_x+v^2_y)_{(t=\Delta t)}=v^2\left(1+\frac{v^2}{r^2}(\Delta t)^2\right)$$
So, to this level of approximation, the sum of the squares of the velocity components has increased by a term proportional to ##(\Delta t)^2##. However, even here, as the time interval Δt becomes smaller, the increase becomes less and less.

Now lets consider the second order finite difference approximation. In this approximation, we use the trapazoidal rule, and write:
$$m\frac{v_x(\Delta t)-v_x(0)}{\Delta t}=\frac{(T_x(0)+T_x(\Delta t))}{2}=-m\frac{v^2}{r}$$
$$m\frac{v_y(\Delta t)-v_y(0)}{\Delta t}=\frac{(T_y(0)+T_y(\Delta t))}{2}=-m\frac{v^3}{2r^2}\Delta t$$
The solution to these equations for the velocities at time Δt are:
$$v_x(\Delta t)=-\frac{v^2}{r}\Delta t$$
$$v_y(\Delta t)=v-\frac{v^3}{2r^2}(\Delta t)^2$$
These are in total agreement with the equations above for the 2nd order approximations to the velocity components. Furthermore, for this approximation, if we take the sum of the squares of the velocity components, we obtain: $$(v^2_x+v^2_y)_{(t=\Delta t)}=v^2(1+terms \ of \ order \ (\Delta t)^4)$$
So, by using a more accurate finite difference approximation, we have come much closer to numerically satisfying the condition that the speed of the mass is constant.
 
  • #40
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No. The "theory" you need to consult to is basic calculus (I have mentioned this before) which investigates what happens in an infinitesimal time by studying what happens in the limit as δt approaches zero. You are quietly ignoring this and coming to your own conclusion - which is just plain wrong. It boils down to the fact that the gradient of cos(φ) approaches zero as φ approaches zero and the gradient of sin(φ) approaches 1. The tangential acceleration is zero and the radial acceleration is v2/r however much you try to wave your arms and say that can't be true and whatever you 'believe'. This is very elementary Maths book work and, rather than trying to argue against it (arguing with a number of well informed PF members and the whole of Maths as well)) you should look into the Maths in detail - not inventing your own version. Intuition can be a very bad friend; people lose fortunes when they solely rely on it.
I am getting ratty with you because you are just not listening to what you have been told on good authority. Just consider that you could be wrong.
I knew I was wrong, I just wanted to know why I was wrong.
 
Last edited:
  • #41
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The problem is, you're not using a sufficiently accurate finite difference approximation.
So, by using a more accurate finite difference approximation, we have come much closer to numerically satisfying the condition that the speed of the mass is constant.

Oh got it, thank you
 
  • #42
sophiecentaur
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I knew I was wrong, I just wanted to know why I was wrong.
OK.
Perhaps you need to believe what the maths is telling you. There is no better way of explaining things like this than with the language of maths. Do you feel that maths is not a valid answer to your question?
 
  • #43
sophiecentaur
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Oh got it, thank you
I just read this.
good. Well done.
 
  • #44
You are talking about the magnitude of the velocity which should be increased by the component of the centripetal force.You are right that when a force acts upon an object its velocity should be increased.This increment can be happened either in the magnitude or in the direction or in both.In this case,the magnitude in other words the speed of that velocity is acting in the direction perpendicular to the centripetal force so the component of this force on the magnitude is always Force=Fcos 90=0. And that's why its speed never increases.But there's something happening to the velocity; as there is a force acting upon it the object should accelerate and this acceleration will occur only by changing the direction of the velocity.That's why the object goes round and round constantly changing its direction and the speed remains unchanged.
 
  • #45
You are talking about the magnitude of the velocity that should be increased by the component of the centripetal force.You are right when a force acts upon an object its velocity should be increased.This increase in velocity can be happened either in the magnitude or in the direction or in both.Now,in the question of the magnitude it is acting in the direction perpendicular to the centripetal force.So,the component of this force in the magnitude is Force=F cos 90°=0 and so the speed is never increased.But there's something that should be happened to the velocity; as the centripetal force acting on it, the object should accelerate.And this acceleration occurs in the direction.So, the object goes round and round through the circle changing its direction constantly(accelerating always) without changing its speed.
 
  • #46
nasu
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What does an "increase in the direction" even mean?
Can you also talk about a decrease in the direction?

I suppose you mean "change" rather than "increase".
 
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  • #47
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What does an "increase in the direction" even mean?
Can you also talk about a decrease in the direction?

I suppose you mean "change" rather than "increase".
I meant increase in the speed in that direction.
 

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