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

[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.