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

• simphys
I'm understanding correctly, the tension in the rope would be maximal when the velocity is at it's maximum and the angle between the velocity vector and the tension vector is the smallest? Yes, that is correct.Yes, that is correct.f

#### simphys

Homework Statement
56-kg student runs at grabs a hanging rope,
and swings out over a lake (Fig. 45). He releases the
rope when his velocity
is zero. (a) What is the angle when he releases the rope? (b) What is the tension in the rope just before he releases it? (c) What is the maximum tension in the rope?
Relevant Equations
##a_n = \frac v^2R
Please tell me if I need to post my solution for this.., but I just have a question more or less 'conceptual' question about (c).
so I know that from Newton's 2nd law for centripetal acceleration --> ##F_{rope} - mgcos(\theta) = ma_n## where ##a_n = \frac {v^2}{R}## such that where the normal acceleration is maximal (i.e. also maximum velocity) is where there's max tension in the rope. But.. how do I find where the velocity is going to be maximum?
I would reason about it in terms of energy that in the beginning of the jump he only has kinetic energy after which ##KE## is changed in ##U_{grav}##. From this I know that the velocity is going to be maximum at the beginning of the jump and thus the tension in the rope as well.

[Mentor Note -- Update requested by OP later in the thread:]
vinitial = ##95km/h## and vfinal = ##35km/h##

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picture of the problem

Delta2
so I know that from Newton's 2nd law for centripetal acceleration --> ##F_{rope} - mg = ma_n## where ##a_n = \frac {v^2}{R}##
That is not true, it is the component of the weight mg in the direction towards the centre of the circle that is used in the centripetal acceleration

simphys
That is not true, it is the component of the weight mg in the direction towards the centre of the circle that is used in the centripetal acceleration
apologies you are right, I didn't look in my notes. let me edit.

Homework Statement:: 56-kg student runs at grabs a hanging rope,
and swings out over a lake (Fig. 45). He releases the
rope when his velocity
is zero. (a) What is the angle when he releases the rope? (b) What is the tension in the rope just before he releases it? (c) What is the maximum tension in the rope?
Relevant Equations:: ##a_n = \frac v^2R
Seems to be some information missing, like how fast was he going when he grabbed the rope?

Seems to be some information missing, like how fast was he going when he grabbed the rope?
odd.. my apologies I edited it yesterday.. seems that I didn't save the changes. will do it again.

Seems to be some information missing, like how fast was he going when he grabbed the rope?
do you by chance why it is not editable anymore?

do you by chance why it is not editable anymore?
I think the time limit for editing posts is like 30 min or so. Gold members have 24h edit-span.

Perhaps you can notify a moderator, who can insert the initial speed value.

I think the time limit for editing posts is like 30 min or so. Gold members have 24h edit-span.

Perhaps you can notify a moderator, who can insert the initial speed value.
okay thank you
I will, I presume that would be @berkeman
So @berkeman please insert vinitial = ##95km/h## and vfinal = ##35km/h##

That is not true, it is the component of the weight mg in the direction towards the centre of the circle that is used in the centripetal acceleration
I do not understand what you are saying here. The equation for tension looks correct to me.

Delta2
I do not understand what you are saying here. The equation for tension looks correct to me.
the equation is only correct for the beginning situation, but along the circular path the the Tension force is normal to the path such that the force of gravity makes an angle with the normal to the path which would be ##\theta## as it acts straight downwards.

I jotted it down, but forgot to plug it in.

malawi_glenn
I was assuming we were using it for part (c) and you had intuited the worst case. Indeed the cos needs to be there in general. Thanks for clarification.

Delta2 and simphys
I was assuming we were using it for part (c) and you had intuited the worst case. Indeed the cos needs to be there in general. Thanks for clarification.
Well it is, but what I am not really sure about is.. whether my reasoning about the energies would be the way to go. for part (c)

And so.. yeah I would say before assuming that it's equal we have the general case from which I need to find the max acceleration. Would I go about reasoning about this with kinetic and gravitational energy (mechanical energy), or that I think of it right now, we'll just about immediately know that the biggest force will be obtained when it's equal to mg which is at the beginning, so I guess that the 2nd reasoning would be enough.

okay thank you
I will, I presume that would be @berkeman
So @berkeman please insert vinitial = ##95km/h## and vfinal = ##35km/h##
ok, so using energy conservation write the relationship between the angle and the velocity.

simphys
ok, so using energy conservation write the relationship between the angle and the velocity.
for part (c) right?

for part (c) right?
yes

simphys
okay thank you
I will, I presume that would be @berkeman
So @berkeman please insert vinitial = ##95km/h## and vfinal = ##35km/h##
Are you sure that these value are for the OP problem? For what position is ##v_{final}##? And 95 km/h for a running man?

simphys and Delta2
For what position is ##v_{final}##?
I assumed that is where he releases the rope.

simphys
I assumed that is where he releases the rope.
In the OP it is mentioned that this is when velocity is zero.

simphys
Are you sure that these value are for the OP problem? For what position is ##v_{final}##? And 95 km/h for a running man?
Indeed, something seems amiss. 95 km/h is sufficient that to drive a 10 meter rope swing up past top dead center, thus rendering part a) of the question invalid.

He releases the rope when his velocity is zero.
And 35 km/hour seems decidedly different from zero.

[Found on chegg.com]
"A 56 kg student runs at 5.0m/s, grabs a hanging rope, and swings out over a lake (Fig. 45)"

Google also gives a variety of other versions, not all of which contain the "Fig. 45" text along with various other initial velocities such as 5.2 m/s or 6.0 m/s and various other rider masses. The hits found are generally on cheating sites where students go to harvest answers without actually doing the work.

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simphys and malawi_glenn
Please tell me if I need to post my solution for this.., but I just have a question more or less 'conceptual' question about (c).
so I know that from Newton's 2nd law for centripetal acceleration --> ##F_{rope} - mgcos(\theta) = ma_n## where ##a_n = \frac {v^2}{R}## such that where the normal acceleration is maximal (i.e. also maximum velocity) is where there's max tension in the rope. But.. how do I find where the velocity is going to be maximum?

$$\nearrow^+ \sum F_t = -mg \sin \theta = m a_t \implies a_t = -g \sin \theta$$

Then use the kinematic relationship (apparently on valid for uniform circular motion):

$$\int v \, dv = \int a_t \, ds$$

where ## ds = r d \theta ##

After you do that, you'll end up with ##v^2## ( or ##v##) as a function of ## \theta ##. From that point forward

(1) you can intuit a solution (by judging the size of the terms in the result - where they will make ## v## largest)

or

(2) just set the derivative of that result w.r.t ## \theta = 0## :

$$\frac{dv}{d \theta} = 0$$

and solve for ## \theta ## (applying the necessary tests for extremums)

Either method will get you there.

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simphys and Delta2
But conservation of energy will get you there on a much shorter (and faster) path.

simphys, Delta2, erobz and 1 other person
But conservation of energy will get you there on a much shorter (and faster) path.
They asked for other ways to look at it, so I thought I'd oblige.

simphys, nasu, Delta2 and 1 other person
simphys, Delta2 and jbriggs444
Are you sure that these value are for the OP problem? For what position is ##v_{final}##? And 95 km/h for a running man?
I am sorry guys... total disaster here. Messed up the problem indeed. aiaiai

$$\nearrow^+ \sum F_t = -mg \sin \theta = m a_t \implies a_t = -g \sin \theta$$

Then use the kinematic relationship:

$$\int v dv = \int a_t ds$$

where ## ds = r d \theta ##

After you do that, you'll end up with ##v^2## ( or ##v##) as a function of ## \theta ##. From that point forward

(1) you can intuit a solution (by judging the size of the terms in the result - where they will make ## v## largest)

or

(2) just set the derivative of that result w.r.t ## \theta = 0## :

$$\frac{dv}{d \theta} = 0$$

and solve for ## \theta ## (applying the necessary tests for extremums)

Either method will get you there.
wow thanks a lot for that! That's an interesting way of using n-t components for sure.

Then use the kinematic relationship:$$\int v dv = \int a_t ds$$where ## ds = r d \theta ##
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.

They asked for other ways to look at it, so I thought I'd oblige.

Arguably, \begin{align}\left( \int v dv\right)+\left(- \int a_t ds \right)=0\end{align}is an expression for mechanical energy conservation per unit mass. Here is why.

The first term obviously integrates to the change in kinetic energy per unit mass. Since the work done by the tension is zero, only gravity does work on the mass. This means that the work per unit mass done by the net force is $$\frac{1}{m}W_{\text{net}}=\frac{1}{m}\int \mathbf{F}_{net}\cdot d\mathbf{s}=\int\mathbf{a}\cdot d\mathbf{s}=\int a_tds.$$Since the change in potential energy per unit mass ##\frac{1}{m}\Delta U## is the negative of the work done by gravity per unit mass, it follows that equation (1) is an expression of mechanical energy conservation per unit mass, $$\left(\frac{1}{m}\Delta K\right)+\left(\frac{1}{m}\Delta U\right)=0.$$

simphys
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.

Arguably, \begin{align}\left( \int v dv\right)+\left(- \int a_t ds \right)=0\end{align}is an expression for mechanical energy conservation per unit mass. Here is why.

The first term obviously integrates to the change in kinetic energy per unit mass. Since the work done by the tension is zero, only gravity does work on the mass. This means that the work per unit mass done by the net force is $$\frac{1}{m}W_{\text{net}}=\frac{1}{m}\int \mathbf{F}_{net}\cdot d\mathbf{s}=\int\mathbf{a}\cdot d\mathbf{s}=\int a_tds.$$Since the change in potential energy per unit mass ##\frac{1}{m}\Delta U## is the negative of the work done by gravity per unit mass, it follows that equation (1) is an expression of mechanical energy conservation per unit mass, $$\left(\frac{1}{m}\Delta K\right)+\left(\frac{1}{m}\Delta U\right)=0.$$
nice observation, thank you, I assumed just that (that it's generally true), but why not then?
Because if you look at it, the change in magnitude of the velocity is only determined by the tangential component where the ##a_n## (normal component) doesn't come into play?

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.

Arguably, \begin{align}\left( \int v dv\right)+\left(- \int a_t ds \right)=0\end{align}is an expression for mechanical energy conservation per unit mass. Here is why.

The first term obviously integrates to the change in kinetic energy per unit mass. Since the work done by the tension is zero, only gravity does work on the mass. This means that the work per unit mass done by the net force is $$\frac{1}{m}W_{\text{net}}=\frac{1}{m}\int \mathbf{F}_{net}\cdot d\mathbf{s}=\int\mathbf{a}\cdot d\mathbf{s}=\int a_tds.$$Since the change in potential energy per unit mass ##\frac{1}{m}\Delta U## is the negative of the work done by gravity per unit mass, it follows that equation (1) is an expression of mechanical energy conservation per unit mass, $$\left(\frac{1}{m}\Delta K\right)+\left(\frac{1}{m}\Delta U\right)=0.$$
for the derivation, intersting, thanks a lot. One question: so here, you start from the kinematic eq ##a_tds = vdv## I assume, correct?

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. erobz 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.

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.

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.

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.