How Do You Mathematically Determine the Highest Point of a Pendulum?

[jackkk_gatz]

Homework Statement

The sphere of mass 𝑚 of a pendulum of length 𝐿 is initially held vertically. When a wind blows with a constant force 𝐹 non-conservative, show that if the sphere starts moving from rest, the maximum height it reaches is
H=2L/(1+(mg/F)^2)

Relevant Equations

L(1-cos(&))=Hmax

I have tried to use the intial velocity v=(2gH)^(1/2) and tried to use conservation of energy, using potential energy to find the maximum height but still can't arrive to the answer provided. Don't know what concepts are relevant here, apparently I can't use velocity neither cosine or sine. I can understand that a bigger force will rise higher the pendulum but, How do I write that mathematically? The way I was approaching the problem, always used velocity, but can't jump from velocity to acceleration or force without having time

 

[berkeman]

Welcome to PF.

I have tried to use the intial velocity v=(2gH)^2

But the problem says that the mass is released from rest, so $v_i$ would be zero, no?

Can you show us your Free Body Diagram (FBD) for the mass when it is as some angle $\theta$ while being blown sideways by the wind? Use the "Attach files" link below the Edit window to upload a PDF or JPEG copy of your FBD. Thanks.

 

[jackkk_gatz]

Yes! The initial velocity is zero, what I meant by v=(2gH)^(1/2), is that I supposed once the wind stopped blowing the pendulum would start to swing, since it swings to its maximum height on every end. It would fall from the right and have a velocity at its lowest point v=(2gH)^(1/2) and raise again but on the other side, the left side. So I took the new velocity as my initial velocity and I tried to calculate how far it would rise on the left side, because it would be equal to the same height it raise on the right side, when the intial velocity was zero. By free body diagram do you mean the one that I made? Because the problem doesn't gives you one.

 

[berkeman]

By free body diagram do you mean the one that I made? Because the problem doesn't gives you one.

Yes, one that you draw. Most problems like this will not do the FBD for you; that is often the first step in your solution of the problem, and usually will give you insights into possible ways to set up the equations to solve the problem.

 

[jackkk_gatz]

I made this two

 

[topsquark]

I made this two

Where is the force F in your first diagram? What is the sum of the forces on the pendulum bob when it gets to its highest point?

-Dan

 

[berkeman]

I made this two

That's a start. You can define your x and y axes however you want and you should still get the right answer, but it's more traditional (and more intuitive for me) to define the y-axis pointing up and the x-axis pointing to the right. Could you try re-drawing the left diagram using that definition of the axes, and then write the sum of forces in the x-direction and also in the y-direction? When the wind is blowing continuously and holding the mass out steady at an angle $\theta$ with respect to the vertical y-axis, what do those two sum-of-forces equations have to equal?

 

[jackkk_gatz]

Is this correct? If so, I used some algebra and now I have cos(&)/sin(&)=mg/F. But I'm not sure what to do with that, I could find the angle only if I could use the inverse of the trigonometric functions.

 

[berkeman]

Good, I got the same equations. Now we look back at the original question, and see that we need to solve for H, the height along the y-axis that the mass rises to. So add in the lines on your FBD to show what H is, and think about how to write the trig equation for H and how to use your equations to isolate H for the solution to the problem...

 

[berkeman]

So add in the lines on your FBD to show what H is,

BTW, you diagram is not quite to scale to show H, so maybe just redraw it to show the mass at the distance L for $\theta = 0$ initially and the final $\theta$ position/value.

 

[haruspex]

What is the sum of the forces on the pendulum bob when it gets to its highest point?

Good, I got the same equations.

Are you both treating this as a statics problem? That's not how I read it.
@jackkk_gatz , you need to consider the general situation, i.e. with the sphere still being accelerated. What equations do you get for that?

 

[jackkk_gatz]

Good, I got the same equations. Now we look back at the original question, and see that we need to solve for H, the height along the y-axis that the mass rises to. So add in the lines on your FBD to show what H is, and think about how to write the trig equation for H and how to use your equations to isolate H for the solution to the problem...

Do you mean something like this? lengthen horizontal lines to see the height on the y-axis?

 

[berkeman]

Are you both treating this as a statics problem? That's not how I read it.

Oh, interesting point! I was treating it like the wind blew the mass up to some steady-state height, but I see your point now that it may be a more dynamic problem where the mass initially makes it a bit higher than the steady-state position and then oscillates/settles into the slightly lower final position.

Back to the drawing board...

 

[jackkk_gatz]

Are you both treating this as a statics problem? That's not how I read it.
@jackkk_gatz , you need to consider the general situation, i.e. with the sphere still being accelerated. What equations do you get for that?

hmm still being accelerated on what direction? Do you mean by the force applied or gravitational force? I understood, from the help I am getting, that the sphere would float due to the constant force applied or something like that, now I'm getting a little confused about the acceleration you mention

 

[TSny]

There are different ways to approach this problem. Here's one way that you could consider. The gravitational force mg and the wind force F add as vectors to produce a total constant force that we could denote F'. This force F' acts like an "effective gravitational force" in a "tilted" direction.

The pendulum is "released" from position A. Can you describe the motion of the pendulum assuming F' remains constant in magnitude and direction?

 

[erobz]

You can also get there with Work/Energy.

$$ \int \boldsymbol F \cdot d \boldsymbol s = mgH$$

 

[jackkk_gatz]

You can also get there with Work/Energy.

$$ \int \boldsymbol F \cdot d \boldsymbol s = mgH$$

That was one of the things I thought of at the beginning, but I thought I should describe the trajectory with parametric equations and I'm not that familiar with using calculus or line integrals on physics problems. Also thought it could be instead of parametric equations, substitute F and dS with something else to make the integral easier the thing is, don't know how to do that and I have never done something like that, can't think of a way to start

 

[jackkk_gatz]

There are different ways to approach this problem. Here's one way that you could consider. The gravitational force mg and the wind force F add as vectors to produce a total constant force that we could denote F'. This force F' acts like an "effective gravitational force" in a "tilted" direction.

View attachment 314035The pendulum is "released" from position A. Can you describe the motion of the pendulum assuming F' remains constant in magnitude and direction?

Isn't that a similar approach to the first ones mentioned? I suppose that F' would make the pendulum stop at a certain point, where the angle of F' and the angle of rope to respect the vertical, would have the same magnitude.
Wait I think I get it, the angle that F' forms is just half of the angle that the pendulum would form at its highest point? If i treat it like an effective gravity I think that's how it should behave, and if so, why does it behave like that? Because of momentum?

 

[TSny]

Wait I think I get it, the angle that F' forms is just half of the angle that the pendulum would form at its highest point?

Yes

If i treat it like an effective gravity I think that's how it should behave, and if so, why does it behave like that? Because of momentum?

Yes. It behaves just like a regular pendulum, except that it oscillates about the configuration where the string and F' are colinear. The inertia of the bob keeps the bob moving past this configuration.

 

[jackkk_gatz]

Thanks! I found the result with your approach @TSny but I found it REALLY complicated, is there a simpler way? I had to use trigonometric identities and algebra, which was really hard especially not knowing what I was aiming for

 

[erobz]

That was one of the things I thought of at the beginning, but I thought I should describe the trajectory with parametric equations and I'm not that familiar with using calculus or line integrals on physics problems. Also thought it could be instead of parametric equations, substitute F and dS with something else to make the integral easier the thing is, don't know how to do that and I have never done something like that, can't think of a way to start

This path $s$ is circular arc of radius $L$ and angle $\theta$. In the dot product we are getting the component of the force $\boldsymbol F$ tangent to the path $\boldsymbol s$ :

$$ \int \boldsymbol F \cdot d \boldsymbol s = F L \int \cos \theta \, d \theta = mgH $$

That is a straightforward integral to evaluate. After evaluation you put the result in terms of $H$ and $L$ using some light trigonometry, and then solve the resulting equation for $H$.

 

[jackkk_gatz]

This path $s$ is circular arc of radius $L$ and angle $\theta$. In the dot product we are getting the component of the force $\boldsymbol F$ tangent to the path $\boldsymbol s$ :

$$ \int \boldsymbol F \cdot d \boldsymbol s = F L \int \cos \theta \, d \theta = mgH $$

That is a straightforward integral to evaluate. After evaluation you put the result in terms of $H$ and $L$ using some light trigonometry, and then solve the resulting equation for $H$.

Thanks! I'll try this

 

[TSny]

Thanks! I found the result with your approach @TSny but I found it REALLY complicated, is there a simpler way? I had to use trigonometric identities and algebra, which was really hard especially not knowing what I was aiming for

The trig isn't too bad. The angle $\beta$ that F' makes to the vertical is easily found to satisfy $\tan \beta = \large \frac{F}{mg}$, where $F$ is the wind force. Then the max height is $$h_{\rm max} = L(1-\cos 2\beta) = 2L \sin^2 \beta = \frac{2L}{\csc^2 \beta} = \frac{2L}{1+\cot^2 \beta} = \frac{2L}{1+(\frac {mg}{F})^2}$$

 

[jackkk_gatz]

The trig isn't too bad. The angle $\beta$ that F' makes to the vertical is easily found to satisfy $\tan \beta = \large \frac{F}{mg}$, where $F$ is the wind force. Then the max height is $$h_{\rm max} = L(1-\cos 2\beta) = 2L \sin^2 \beta = \frac{2L}{\csc^2 \beta} = \frac{2L}{1+\cot^2 \beta} = \frac{2L}{1+(\frac {mg}{F})^2}$$

well uuhhh that is indeed simpler than what I did, I made a lot of things just to get tan(&)^2 and a lot more other things to reach the final expression

 

[jackkk_gatz]

This path $s$ is circular arc of radius $L$ and angle $\theta$. In the dot product we are getting the component of the force $\boldsymbol F$ tangent to the path $\boldsymbol s$ :

$$ \int \boldsymbol F \cdot d \boldsymbol s = F L \int \cos \theta \, d \theta = mgH $$

That is a straightforward integral to evaluate. After evaluation you put the result in terms of $H$ and $L$ using some light trigonometry, and then solve the resulting equation for $H$.

is the integral on polar coordinates?

 

[erobz]

is the integral on polar coordinates?

Yeah, $r,\theta$ coordinates. Are you stuck on how the integral was initially transformed, or stuck evaluating the integral?

 

[jackkk_gatz]

Yeah, $r,\theta$ coordinates. Are you stuck on how the integral was initially transformed, or stuck evaluating the integral?

yeah I am stuck at how the integral was initially transformed, I'm not familiar with those kinds of things, i tried to understand it by reading my calculus book but I'm still not quite sure how to do it

 

[Steve4Physics]

@jackkk_gatz, the work done by $F$ (acting horizontally) is simply $F \times \text {(horizontal displacement)}$. If $\alpha$ is the pendulum’s maximum angle to the vertical, then (conservation of energy):
$mgH = FL \sin \alpha$

It’s convenient to rearrange and square the above equation:
$(\frac {mg}{F})^2 = (\frac LH)^2 \sin^2 \alpha$

(The reason being that the working is less messy when you find the expression for $\sin^2 \alpha$ in term of $L$ and $H$ and substitute the expression into the above equation.)

A bit of algebra, then gives the required expression for H. (No calculus needed!)

 

[erobz]

yeah I am stuck at how the integral was initially transformed, I'm not familiar with those kinds of things, i tried to understand it by reading my calculus book but I'm still not quite sure how to do it

I was afraid that was the part you were going to ask about. Yeah, you won't run into an explanation of this in a Calculus textbook.

Are you familiar with vectors and dot products?

 

[jackkk_gatz]

I was afraid that was the part you were going to ask about. Yeah, you won't run into an explanation of this in a Calculus textbook.

Are you familiar with vectors and dot products?

Yes! I'm familiar with that and the basics of vector analysis

 

[jackkk_gatz]

Why does

@jackkk_gatz, the work done by $F$ (acting horizontally) is simply $F \times \text {(horizontal displacement)}$. If $\alpha$ is the pendulum’s maximum angle to the vertical, then (conservation of energy):
$mgH = FL \sin \alpha$

It’s convenient to rearrange and square the above equation:
$(\frac {mg}{F})^2 = (\frac LH)^2 \sin^2 \alpha$

(The reason being that the working is less messy when you find the expression for $\sin^2 \alpha$ in term of $L$ and $H$ and substitute the expression into the above equation.)

A bit of algebra, then gives the required expression for H. (No calculus needed!)

Why does work equal to that? I'd expect It to be F x (diagonal displacement)

 

[erobz]

Draw the mass at some angle $\theta$ and then draw a line tangent to the circular arc at the mass. That is the instantaneous direction of the infinitesimal "arc" $d \vec{ \boldsymbol s}$ at $\theta$. Call it $\vec{ \boldsymbol u}$ (for unit vector). Next, determine the angle between $\vec{ \boldsymbol F}$ and $d \vec{ \boldsymbol s}$ ( or $\vec{ \boldsymbol u}$ ) and take the dot product inside the integral. What is the result of that part?

 

[Steve4Physics]

Why does work equal to that? I'd expect It to be F x (diagonal displacement)

If F acted in the same direction as the diagonal displacement, then you would be correct.

But it is implied (i.e. we assume!) that the wind is blowing horizontally, so F acts horizontally and has no vertical component.

Imagine the path of the mass, from start to finish, as a stairway of little steps each of size dx (horizontally) and dy (vertically).

The work done by F when the mass rises up a dy-step is zero because the movement is perpendicular to F. Total work done moving in y-direction is zero

The work done by F when the mass moves along a dx-step is $Fdx$ because the movement is in the same direction as F. Total work done moving in the x-direction = $\int Fdx = F\Delta x$ where $\Delta x$ is the total horizontal displacement.

Overall work done by F = $F\Delta x$

Of course using notation for line integral of a dot product, the above working could be presented more elegantly.

 

[erobz]

Of course using notation for line integral of a dot product, the above working could be presented more elegantly.

But not more insightfully.

 

[jbriggs444]

Why does

Why does work equal to that? I'd expect It to be F x (diagonal displacement)

The work done by the horizontal component of the force is the horizontal force component times the horizontal displacement component.

The work done by the vertical component of the force is the vertical force component times the vertical displacement component.

If you want, you could calculate the work done by the component of the [unchanging] net force in the direction of the total displacement. Or, for that matter as the component of the total displacement in the direction of the net force. The result will be the same regardless.

If your instructor prefers to use "projection" instead of "component", that works too.

Or you can always take the product of the magnitudes of the net force and total distance and multiply that by the cosine of the angle between the two directions. Lots and lots of ways to compute dot products.

 

[jackkk_gatz]

Draw the mass at some angle $\theta$ and then draw a line tangent to the circular arc at the mass. That is the instantaneous direction of the infinitesimal "arc" $d \vec{ \boldsymbol s}$ at $\theta$. Call it $\vec{ \boldsymbol u}$ (for unit vector). Next, determine the angle between $\vec{ \boldsymbol F}$ and $d \vec{ \boldsymbol s}$ ( or $\vec{ \boldsymbol u}$ ) and take the dot product inside the integral. What is the result of that part?

I got that the angle between u and the force applied is theta, and the dot product is equal to Fcos(&), since u is a unit vector his magnitude is one. So that's what I got, don't know if I'm correct. Or using the vector dS it yields (F)(ds)cos(&)
Also, one quick doubt, what's the reason we introduce a unit vector?

 

[jackkk_gatz]

If F acted in the same direction as the diagonal displacement, then you would be correct.

But it is implied (i.e. we assume!) that the wind is blowing horizontally, so F acts horizontally and has no vertical component.

Imagine the path of the mass, from start to finish, as a stairway of little steps each of size dx (horizontally) and dy (vertically).

The work done by F when the mass rises up a dy-step is zero because the movement is perpendicular to F. Total work done moving in y-direction is zero

The work done by F when the mass moves along a dx-step is $Fdx$ because the movement is in the same direction as F. Total work done moving in the x-direction = $\int Fdx = F\Delta x$ where $\Delta x$ is the total horizontal displacement.

Overall work done by F = $F\Delta x$

Of course using notation for line integral of a dot product, the above working could be presented more elegantly.

Ohh okay, that's pretty comprehensible. I looked up the definition of work on one of my books and it makes sense, if the force isn't in the same direction it won't make any work. But there is still something that isn't clear to me, then why the object goes up on dy? If the force being applied doesn't have a component on the y-axis, why does it go up? In order to change its direction it must have work being made on dy, then what force is making that work and why we aren't taking it as Wtotal=W1+W2? Or is it mgh+Fdx=Wtotal?

Sorry for asking too many questions

 

[haruspex]

If the force being applied doesn't have a component on the y-axis, why does it go up?

It goes up because of the tension, but since there is never a movement in the current direction of the tension, the tension does no work. Rather, it serves to translate the work done by the wind into a partially vertical movement.

 

[erobz]

I got that the angle between u and the force applied is theta, and the dot product is equal to Fcos(&), since u is a unit vector his magnitude is one. So that's what I got, don't know if I'm correct. Or using the vector dS it yields (F)(ds)cos(&)

$$ \int \vec{ \boldsymbol F } \cdot d \vec{ \boldsymbol s} = \int F \vec{ \boldsymbol u_F } \cdot \vec{ \boldsymbol u_s } ds = \int \left( \vec{ \boldsymbol u_F } \cdot \vec{ \boldsymbol u_s } \right) F ds $$
By the definition of the Dot Product:

$$ \vec{ \boldsymbol u_F } \cdot \vec{ \boldsymbol u_s } = \left\| \vec{ \boldsymbol u_F }\right\| \left\|\vec{ \boldsymbol u_s }\right\| \cos \theta = 1\cdot 1 \cos \theta $$

And,

$$ \int \left( \vec{ \boldsymbol u_F } \cdot \vec{ \boldsymbol u_s } \right) F ds = \int \cos \theta F \, ds $$

The only thing remaining is to change the variable of integration $s$ to the variable in the integrand $\theta$.

Spoiler

differentiate the relationship for the length of a circular arc:

$s = r \theta$

Also, one quick doubt, what's the reason we introduce a unit vector?

Usually when we want to represent the vector as having some scalar magnitude.

$$ \vec{ \boldsymbol s } = s \vec{ \boldsymbol u_s } $$

 

[haruspex]

Usually when we want to represent the vector as having some scalar magnitude.

$$ \vec{ \boldsymbol s } = s \vec{ \boldsymbol u_s } $$

or using the hat notation, $\vec s=s\hat s$.

 

[Gezstarski]

Much of this discussion is correct but not very revealing. The key thing to remember is that the work done by a force is the product of that force and the movement of its point of application IN THE DIRECTION OF THAT FORCE. Using cartesian coordinates with origin at the initial position the ball moves from (0,0) to (x,h) and you can just equate the work done by the wind with that done against gravity
F x = m g h
because the wind acts only horizontally and gravity only vertically. Kinetic energy doesn't come in because the ball is staionary at start and end. The tension in the string doesn't do any work because its length doesn't change. It just constrains the ball to move in a circle so that
x^2 + (L-h)^2 = L^2
Substitute x from the first equation into the 2nd and you get the answer after only a little rearrangement.
This is (in my view) an equivalent but simpler way of looking at what is written above by Steve4Physics and others,

But Bob Dylan had a still better way of looking at it
"The answer, my friend, is ..."

 

[haruspex]

Much of this discussion is correct but not very revealing

?
working in terms of x and y is perhaps slightly simpler than #28 etc., but #15, #23 offer a very instructive insight. And don't overlook that subsequent discussion was needed to address the OP's misunderstanding of how force x distance works.