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Direction for A is not constant, but the time is very small

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- Thread starter Faiq
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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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The direction isDirection for A is not constant, but the time is very small

- #28

Chestermiller

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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?

- #29

sophiecentaur

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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.The force is centripetal in nature to A but is just a normal resultant force to B.

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.

- #30

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Yes that's what I am trying to ask

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?

- #31

Chestermiller

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$$y=r\sin \left(\frac{v}{r}t\right)\tag{3}$$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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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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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.

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?

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

Chestermiller

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Why don't you use the equations I presented to help you analyze this exact situation?

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?

- #35

sophiecentaur

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no.

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?

What does "greater" mean when referring to a vector?

You really cannot let this lie, can you?

- #36

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The equations aren't resulting in any contradiction but I want to understand the reason.Why don't you use the equations I presented to help you analyze this exact situation?

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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There is no "next velocity". Time is continuous.So, theoretically, the next velocity has to be greater (In magnitude) than the previous velocity because of the added component

- #38

sophiecentaur

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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 inThe 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

I am getting ratty with you because you are just not listening to what you have been told

- #39

Chestermiller

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The problem is,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

Consider the times t = 0 and t = Δt.

$$x(0)=r$$

$$y(0)=0$$

$$v_x(0)=0$$

$$v_y(0)=v$$

$$T_x(0)=-T=-m\frac{v^2}{r}$$

$$T_y(0)=0$$

$$x(\Delta t)=r\cos{\left(\frac{v}{r}\Delta t\right)}\approx r\left(1-\frac{1}{2}\frac{v^2}{r^2}(\Delta t)^2\right)$$

$$y(\Delta t)=r\sin{\left(\frac{v}{r}\Delta t\right)}\approx v\Delta t$$

$$v_x(\Delta t)=-v\sin{\left(\frac{v}{r}\Delta t\right)}\approx -\frac{v^2}{r}\Delta t$$

$$v_y(\Delta t)=+v\cos{\left(\frac{v}{r}\Delta t\right)}\approx v\left(1-\frac{1}{2}\frac{v^2}{r^2}(\Delta t)^2\right)$$

$$T_x(\Delta t)=-m\frac{v^2}{r}\cos{\left(\frac{v}{r}\Delta t\right)}\approx -m\frac{v^2}{r}\left(1-\frac{1}{2}\frac{v^2}{r^2}(\Delta t)^2\right)$$

$$T_y(\Delta t)=-m\frac{v^2}{r}\sin{\left(\frac{v}{r}\Delta t\right)}\approx -m\frac{v^3}{r^2}\Delta t$$

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

$$m\frac{dv_y}{dt}=T_y$$

The

$$m\frac{v_x(\Delta t)-v_x(0)}{\Delta t}=T_x(0)=-m\frac{v^2}{r}$$

$$m\frac{v_y(\Delta t)-v_y(0)}{\Delta t}=T_y(0)=0$$

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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I knew I was wrong, I just wanted to know why I was wrong.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 inthe 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 v^{2}/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 toldon good authority. Just consider that you could be wrong.

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- #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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OK.I knew I was wrong, I just wanted to know why I was wrong.

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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I just read this.Oh got it, thank you

good. Well done.

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- #46

nasu

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Can you also talk about a decrease in the direction?

I suppose you mean "change" rather than "increase".

- #47

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I meant increase in the speed in that direction.

Can you also talk about a decrease in the direction?

I suppose you mean "change" rather than "increase".

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