Calculating Max Tension, Final Angle for Swinging on Rope Over Lake

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The discussion revolves around calculating maximum tension in a rope during a swing over a lake, using concepts from Newton's laws and energy conservation. The original poster seeks clarification on the relationship between velocity and angle at the point of maximum tension, suggesting that maximum velocity occurs at the beginning of the swing. Other participants correct the initial assumptions about the forces involved, emphasizing the need to account for gravitational components in centripetal acceleration calculations. There is a consensus that energy conservation principles can simplify finding the maximum tension, and participants discuss the correct application of kinematic relationships in this context. The conversation highlights the importance of accurately interpreting the problem's parameters and equations.
  • #31
simphys said:
for the derivation, intersting, thanks a lot. One question: so here, you start from the kinematic eq ##a_tds = vdv## I assume, correct?
I start from this equation because the point I want to make in post #28 is that this equation, which is presented by @erobz in post $21 as kinematic, is no different from what one would get using mechanical energy conservation.
 
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  • #32
kuruman said:
I start from this equation because the point I want to make in post #28 is that this equation, which is presented by @erobz in post $21 as kinematic, is no different from what one would get using mechanical energy conservation.
I thought about saying that earlier, but I wanted to let them play around with it, and discover it.
 
  • #33
kuruman said:
It seems to me that this relationship is valid only when v has no radial component, i.e. the object undergoes circular motion. If you use it without derivation, I think you should al teast state its applicability just in case someone might think that it is generally true.

I'm not sure I understand what needs to be done for it to be corrected.
 
  • #34
erobz said:
I'm not sure I understand what needs to be done for it to be corrected.
It's correct in the special case of circular motion when only conservative forces act on the mass undergoing it. So if you choose to use to use and you don't want to be wrong about it, before using it you need to establish that (a) the motion is circular and (b) only conservative forces act on the mass. If you want to use it as part of the solution to a problem that someone else will read, before writing it down, you need to say "because the motion is circular and only conservative forces act on the mass, the following equation can be used." However, because it is a consequence of mechanical energy conservation, you maight as well start from there which is less work as has already been observed.
 
  • #35
If by the index in the ##a_t## you mean tangent to the trajectory, then I think you don't need to correct. By definition, the tangential component of the acceleration is equal to the derivative in respect to time of the magnitude of velocity. There is no velocity perpendicular to the trajectory. "Radial velocity" may reffer to diection along a radius which is not normal to the trajectory in the point of interest. Like for eliptical motion. But here too the change in the magnitude of velocity is given by the component of the force which is tangent to the trajectory and produces a tangential acceleration.
 
  • #36
nasu said:
If by the index in the ##a_t## you mean tangent to the trajectory, then I think you don't need to correct. By definition, the tangential component of the acceleration is equal to the derivative in respect to time of the magnitude of velocity. There is no velocity perpendicular to the trajectory. "Radial velocity" may reffer to diection along a radius which is not normal to the trajectory in the point of interest. Like for eliptical motion. But here too the change in the magnitude of velocity is given by the component of the force which is tangent to the trajectory and produces a tangential acceleration.
Yes.

## a_t ## is the tangential component of the acceleration. We were working in ##n-t## components.
 
  • #37
kuruman said:
It's correct in the special case of circular motion when only conservative forces act on the mass undergoing it.
The definition of the tangential acceleration is not related to a special trajectory or a special type of force.
Nevermind. I did not pay sufficient attention to the formula in discussion. It is not about integrating the tangential acceleration.
Sorry for the confusion.
 
  • #38
erobz said:
I'm not sure I understand what needs to be done for it to be corrected.
In vectors it can be written correctly as ##\vec v.d\vec v=\vec v.\vec a.dt=\vec a.\vec v.dt=\vec a.d\vec s##.
Converting to scalars, the RHS becomes ##a_t.ds##, but I can't find a neat way to write the LHS.

Edit, .. but it could be written ##v.dv##, if ##dv## is defined as the component of the change in velocity in the direction of the existing velocity. I.e. the change in speed, treating speed as a scalar, not a magnitude.
 
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  • #39
erobz said:
Yes.

## a_t ## is the tangential component of the acceleration.
Sorry, my mistake. What I said is true but does not apply to your equation under discussion. I though that by "kinematic equation" is ment ##dv= a_t dt##.
 
  • #40
nasu said:
Sorry, my mistake. What I said is true but does not apply to your equation under discussion. I though that by "kinematic equation" is ment ##dv= a_t dt##.
It's getting worse! Head is spinning! Is it not only the tangential component of the acceleration that contributes to the change in magnitude of the velocity in the ## n-t## coordinates?
 
  • #41
Am I mis-interpreting the author's (R.C. Hibbeler) derivation here?

IMG_1706.jpg


Notice ##(12-19)##
 
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  • #42
erobz said:
It's getting worse! Head is spinning! Is it not only the tangential component of the acceleration that contributes to the change in magnitude of the velocity in the ## n-t## coordinates?
Yes, it is. This is so for any motion. But you have something else in what you call kinematic equation.
 
  • #43
nasu said:
Yes, it is. This is so for any motion. But you have something else in what you call kinematic equation.

I have eq ##(12-19)## in the above derivation. It represents the magnitude of ##\boldsymbol{a_t}##.

nasu said:
Sorry, my mistake. What I said is true but does not apply to your equation under discussion. I thought that by "kinematic equation" is meant ##dv= a_t dt##.

From what you said above it follows from a simple application of the chain rule:

$$ a_t = \frac{dv}{dt} = \frac{dv}{ds} \frac{ds}{dt} = \frac{dv}{ds} v \implies a_t \, ds = v \, dv $$

Why/Where is this supposedly inapplicable, can someone give a simple example?
 
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  • #44
erobz said:
I have eq ##(12-19)## in the above derivation. It represents the magnitude of ##\boldsymbol{a_t}##.
From what you said above it follows from a simple application of the chain rule:

$$ a_t = \frac{dv}{dt} = \frac{dv}{ds} \frac{ds}{dt} = \frac{dv}{ds} v \implies a_t \, ds = v \, dv $$

Why/Where is this supposedly inapplicable, can someone give a simple example?
I think it hinges on how dv is being defined. See edit in post #38.
 

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