Equilibrium Position of a Hanging Candy Cane

[TensorCalculus]

Homework Statement

A thin ribbon is tied to the curved end P of a thin, uniform candy cane.

The Candy Cane is made up of 2 sections:

- A semicircular section of radius r,
- A straight section, which is twice as long as the length of the semicircular section

The ribbon and candy cane are hung up together as a decoration. The candy cane is hinged freely about P and pivots until it lies in equilibrium in a vertical plane.

Calculate the angle θ that the straight part of the candy cane makes with the vertical. Give your answer in degrees to 3 significant figures.

You may use the result that the centre of mass of a thin uniform semicircular arc lies at a distance of $\frac{2r}{\pi}$ from its centre.

Relevant Equations

Condition for equilibrium -> $F_{res} = 0, \; \tau_{res} = 0$
Position of Center of Mass -> $x_{cm} = \frac{\sum_i m_i r_i}{\sum_i m_i}$

Hello!

This is not a homework problem but an IsaacPhysics problem that I stumbled across and have been trying (and getting incorrect quite a few times by now). I'm not really sure where I went wrong, so if someone could help me out that would be much appreciated...

(Diagram of the problem)

Since the resultant moment is 0, that means that the center of mass of the whole candy cane will be directly below point P... so what I did is set up an expression for the x position of center of mass of the whole system (taking point P as the origin) and then set that to zero. I've drawn a diagram to illustrate how I attained this expression - hopefully it is clear enough, if not please tell me and I will try and redraw it...

Horizontal distance $x_1$ of the COM of the curved segment from point P:
$$x_1=\sin(\arctan(\frac{2}{\pi})+90-\theta)(\sqrt{(\frac{2R}{\pi})^2 + R^2})$$

Horizontal distance $x_2$ of the COM of the straight segment from point P:
$$x_2 = \sin(\arctan(\frac{\pi}{2}) - 90 + \theta)(\sqrt{(2R)^2 + (\pi R)^2})$$

Then, using the formula for the x position of the center of mass:
$$x_{com} = \frac{x_1 + 2x_2}{3} = 0$$
(Multiply x_2 by 2 because it is double the mass of x_1)

I plugged everything in, and then spent around 1.5 hours trying to solve it... to not much avail. The step-by-step would make this a rather long request so I won't put it in for now (unless you would like it, in which case tell me and I'll write it out) but essentially what I tried to do was first simplify, by multiplying by 3, simplifying what was inside the sine for each of the terms (by plugging in the value of the arctan, adding/subtracting 90 degrees etc etc) and also simplifying the square root. Then I played around with various trig identities such as the identity for sin(a-b) but I just kept getting lead to dead ends - either I would go in a huge circle and end up with the same expression that I started with or I would make the expression so much longer/more complicated than it originally was, with no elegant cancellations in sight (apart from eliminating the R). I'm not sure how you would go about solving this... I do not know if I took the problem from the wrong angle and ended up with an overcomplicated expression or I'm just bad at maths (probably a mix of both). If you know - I'd like to do a bit more playing around and try to figure it out myself for now...

Eventually I gave up and (admittedly, and annoyedly) fed WolframAlpha the expression I came up with (substituting the 90 for a pi/2 since Wolfram works in radians) and then converting the answer(s) it spit out into degrees. Said answers were all wrong - which leads me to believe the original expression isn't correct. After restarting the problem 6 times to ensure I definitely did all of the geometry right, and getting the same answer every time (well, once it was different, because of a silly mistake), I come to PF asking for help. Could anyone give me a hint on where I went wrong?

Thank you for reading all of this, it's quite long, but hopefully thorough enough that someone can provide help...

(for some reason the TeX is not rendering on my device, so if it is wrong please do tell me)

 

[kuruman]

1. Draw the candy cane to scale with the straight piece along the vertical.
2. You know where the CM of the semicircle is, say at P1. The straight piece has its CM at its midpoint, say at P2. Add the two points to scale.
3. The CM of the whole candy cane is at point P3 on the line joining P1 and P2. Draw it in to scale noting that the straight piece has twice the mass of the curved piece.
4. Draw a dotted line connecting P3 with the point of suspension.
5. The angle of that dotted line w.r.t. the vertical is what you want.

I am suggesting a strategy - you have to put in the trigonometry and analytic geometry. Personally, I think that Wolfram Alpha is overkill.

 

[haruspex]

Homework Statement: A thin ribbon is tied to the curved end P of a thin, uniform candy cane.

The Candy Cane is made up of 2 sections:

- A semicircular section of radius r,
- A straight section, which is twice as long as the length of the semicircular section

The ribbon and candy cane are hung up together as a decoration. The candy cane is hinged freely about P and pivots until it lies in equilibrium in a vertical plane.

Calculate the angle θ that the straight part of the candy cane makes with the vertical. Give your answer in degrees to 3 significant figures.

You may use the result that the centre of mass of a thin uniform semicircular arc lies at a distance of $\frac{2r}{\pi}$ from its centre.
Relevant Equations: Condition for equilibrium -> $F_{res} = 0, \; \tau_{res} = 0$
Position of Center of Mass -> $x_{cm} = \frac{\sum_i m_i r_i}{\sum_i m_i}$

Hello!

This is not a homework problem but an IsaacPhysics problem that I stumbled across and have been trying (and getting incorrect quite a few times by now). I'm not really sure where I went wrong, so if someone could help me out that would be much appreciated...

(Diagram of the problem)
View attachment 362880

Since the resultant moment is 0, that means that the center of mass of the whole candy cane will be directly below point P... so what I did is set up an expression for the x position of center of mass of the whole system (taking point P as the origin) and then set that to zero. I've drawn a diagram to illustrate how I attained this expression - hopefully it is clear enough, if not please tell me and I will try and redraw it... View attachment 362879

Horizontal distance $x_1$ of the COM of the curved segment from point P:
$$x_1=\sin(\arctan(\frac{2}{\pi})+90-\theta)(\sqrt{(\frac{2R}{\pi})^2 + R^2})$$

Horizontal distance $x_2$ of the COM of the straight segment from point P:
$$x_2 = \sin(\arctan(\frac{\pi}{2}) - 90 + \theta)(\sqrt{(2R)^2 + (\pi R)^2})$$

Then, using the formula for the x position of the center of mass:
$$x_{com} = \frac{x_1 + 2x_2}{3} = 0$$
(Multiply x_2 by 2 because it is double the mass of x_1)

I plugged everything in, and then spent around 1.5 hours trying to solve it... to not much avail. The step-by-step would make this a rather long request so I won't put it in for now (unless you would like it, in which case tell me and I'll write it out) but essentially what I tried to do was first simplify, by multiplying by 3, simplifying what was inside the sine for each of the terms (by plugging in the value of the arctan, adding/subtracting 90 degrees etc etc) and also simplifying the square root. Then I played around with various trig identities such as the identity for sin(a-b) but I just kept getting lead to dead ends - either I would go in a huge circle and end up with the same expression that I started with or I would make the expression so much longer/more complicated than it originally was, with no elegant cancellations in sight (apart from eliminating the R). I'm not sure how you would go about solving this... I do not know if I took the problem from the wrong angle and ended up with an overcomplicated expression or I'm just bad at maths (probably a mix of both). If you know - I'd like to do a bit more playing around and try to figure it out myself for now...

Eventually I gave up and (admittedly, and annoyedly) fed WolframAlpha the expression I came up with (substituting the 90 for a pi/2 since Wolfram works in radians) and then converting the answer(s) it spit out into degrees. Said answers were all wrong - which leads me to believe the original expression isn't correct. After restarting the problem 6 times to ensure I definitely did all of the geometry right, and getting the same answer every time (well, once it was different, because of a silly mistake), I come to PF asking for help. Could anyone give me a hint on where I went wrong?

Thank you for reading all of this, it's quite long, but hopefully thorough enough that someone can provide help...

(for some reason the TeX is not rendering on my device, so if it is wrong please do tell me)

It is always possible that the given answer is wrong, but you do not say what answer you are getting, so we cannot tell.
Wrt solving your equations, I get an equation of the form $A\cos(\beta)+B\sin(\beta)=0$.

 

[Steve4Physics]

Horizontal distance $x_1$ of the COM of the curved segment from point P:
$$x_1=\sin(\arctan(\frac{2}{\pi})+90-\theta)(\sqrt{(\frac{2R}{\pi})^2 + R^2})$$

Horizontal distance $x_2$ of the COM of the straight segment from point P:
$$x_2 = \sin(\arctan(\frac{\pi}{2}) - 90 + \theta)(\sqrt{(2R)^2 + (\pi R)^2})$$

You can check your expressions for $x_1$ and $x_2$. Put $\theta = 0$ into the expressions – do you get the expected values?

(BTW, you can factor-out $R$ in your right-hand sides.)

 

[TensorCalculus]

1. Draw the candy cane to scale with the straight piece along the vertical.
2. You know where the CM of the semicircle is, say at P1. The straight piece has its CM at its midpoint, say at P2. Add the two points to scale.
3. The CM of the whole candy cane is at point P3 on the line joining P1 and P2. Draw it in to scale noting that the straight piece has twice the mass of the curved piece.
4. Draw a dotted line connecting P3 with the point of suspension.
5. The angle of that dotted line w.r.t. the vertical is what you want.

I am suggesting a strategy - you have to put in the trigonometry and analytic geometry. Personally, I think that Wolfram Alpha is overkill.

This is a very smart strategy! it took me a second to realise why this works, but it does, and simplified the problem drastically. Since they wanted the answer to 3 sig figs I couldn't just measure the angle no matter how accurate my diagram was (sadly), but it turned the problem into a much more doable geometry puzzle :D. This strategy landed me with the right answer, thank you so much! I have never thought about tackling geometry problems by just sort of... thinking about them from a different angle I guess - and it works!

I have a horrible habit of learning the physics first, then the maths needed to do that physics afterwards. As a result, I have ended up using WolframAlpha to help with my physics problems far more than I should...

It is always possible that the given answer is wrong, but you do not say what answer you are getting, so we cannot tell.
Wrt solving your equations, I get an equation of the form $A\cos(\beta)+B\sin(\beta)=0$.

suddenly realises that she could have used the identity sin(x) = cos(90-x) the whole time
By some miracle of the gods I never saw that way of rearranging the equation, despite spending so long on it. I guess once you've tried one incorrect method you end up with tunnel vision - because I have not tried it yet but I see how an equation of that form will arise (and then, you make use of the fact that tan = sin/cos to solve for Beta, correct?). The given answer wasn't wrong in the end I guess, but this is what wolfram gave me (I plugged in n=0 and n=1 for both, unless it was going to return a negative number, and then converted from radians to degrees)
$$\theta \approx 2\pi n-2.7325$$
and
$$\theta \approx 2(\pi n + 0.20454)$$

The value of $\theta$ that I get from this (this is the only one in the range of $0<\theta<180$) is 23.44 degrees which isn't right (the correct answer was 41.5, which I got using the method @kuruman suggested)

You can check your expressions for $x_1$ and $x_2$. Put $\theta = 0$ into the expressions – do you get the expected values?

(BTW, you can factor-out $R$ in your right-hand sides.)

That's also smart... and the answer is no, for both of them. Slightly concerningly. (Unless I did the substitutions wrong which is perfectly possible). I'm probably going to go back and check it now that it's a day later... maybe with a fresh mind I will be able to spot some dumb mistake I made yesterday. Despite already getting the correct answer it's always interesting to look at nice problems from various angles, not to mention I spent so long checking those expressions I feel somewhat obliged to correct them haha

Thank you all for the help :)

 

[kuruman]

For the record, here is my solution. In the figure on the right the two black dots represent the centers of mass of the semicircle (top) and the straight piece of the candy cane (bottom). The magenta dot between the two represents the CM of the candy cane. It is located at $\frac{2}{3}$ of the way from the top black dot to the bottom black dot.

The CM of the semicircle is at distance $y=\dfrac{2R}{\pi}$ above the horizontal diameter.
The position of the semicircle CM relative to the point of suspension is represented by green vector $\mathbf v_1=R~\mathbf {\hat x}+\dfrac{2R}{\pi}~\mathbf {\hat y}.$
The position of the candy cane's CM relative to the semicircle's CM is represented by black vector
$\mathbf v_2=R~\mathbf {\hat x}-\left(\dfrac{2R}{\pi}+\pi R~\right)\mathbf {\hat y}.$
Finally, the position of the point of suspension relative to the candy cane's CM is represented by magenta vector $\mathbf v_3.$

From the drawing, $$\mathbf v_1+\frac{2}{3}\mathbf v_2+\mathbf v_3=0\implies \mathbf v_3=-\left(\mathbf v_1+\frac{2}{3}\mathbf v_2\right)=-\frac{5}{3}R~\mathbf {\hat x}+\left(\frac{2}{3}\pi R-\frac{2R}{3\pi}\right)~\mathbf {\hat y}.$$The cosine of the desired angle is given by $$\cos\!\beta=\frac{\mathbf v_3 \cdot \mathbf {\hat y}}{|\mathbf v_3|}.$$ Doing this on a spreadsheet gave me $\beta=41.5^{\circ}.$

On edit
My thanks to @Steve4Physics who detected an error in my original expression for the semicircle center of mass.. It has now been fixed and my answer is in line with OP's answer.

 

[Lnewqban]

Scale drawing using R=1:

 

[haruspex]

you make use of the fact that tan = sin/cos to solve for Beta, correct?

Yes. In the more tricky case of $A\sin(x)+B\cos(x)=C$, set $\tan(y)=A/B$, producing $\sin(y)\sin(x)+\cos(y)\cos(x)=C\cos(y)/B=\sin(x+y)$.

 

[TensorCalculus]

Scale drawing using R=1:

View attachment 362922

Very cool! How did you draw this - with CAD/vector drawing software?

Yes. In the more tricky case of $A\sin(x)+B\cos(x)=C$, set $\tan(y)=A/B$, producing $\sin(y)\sin(x)+\cos(y)\cos(x)=C\cos(y)/B=\sin(x+y)$.

Oh I didn't know that - thanks! That's useful to know :)
Hopefully next time I stumble across an equation like the one I had, I remember that cos(x) = sin(90-x)

 

[TensorCalculus]

@kuruman - I really liked the vector algebra method you gave to solve the problem, I rarely ever use vectors to solve geometry problems.
Do you think that it's a good strategy to solve physics olympiad/maths Olympiad geometry? Or just in general, using vectors expressed in terms of $\hat \imath$ and $\hat\jmath$ (or $\hat x$ and $\hat y$ ) to express forces rather than how most (British) Olympiad do it which is express them in terms of magnitude and direction? In this case the former was more useful, but are there cases where the latter is more useful? How do you know when to use one over the other?

 

[Lnewqban]

Very cool! How did you draw this - with CAD/vector drawing software?

AutoCAD.
Please, note that the distances among the pivot and both individual centers of mass (the semicircular section and the straight section) are proportional to the difference in the weight of each part: 2 to 1.
That is the condition of static equilibrium that determines the unique calculated hanging angle.

 

[kuruman]

How do you know when to use one over the other?

I can think of no specific guidelines for when you use which one. As long as you put both approaches in your mathematical toolbox and know they're there, their circumstances of use will be apparent to you as you try one, then the other and see which works better on a case by case basis. Eventually, you will gain experience and know which one might work better than the other.

 

[kuruman]

Yes. In the more tricky case of $A\sin(x)+B\cos(x)=C$, set $\tan(y)=A/B$, producing $\sin(y)\sin(x)+\cos(y)\cos(x)=C\cos(y)/B=\sin(x+y)$.

Did you mean to say
$\sin(y)\sin(x)+\cos(y)\cos(x)=C\cos(y)/B=\cos(x-y)?$

 

[TensorCalculus]

I can think of no specific guidelines for when you use which one. As long as you put both approaches in your mathematical toolbox and know they're there, their circumstances of use will be apparent to you as you try one, then the other and see which works better on a case by case basis. Eventually, you will gain experience and know which one might work better than the other.

Oh okay! I'll try using vectors expressed in terms of unit vectors more often then, and see where they come in useful. Thank you for the advice.

Did you mean to say
$\sin(y)\sin(x)+\cos(y)\cos(x)=C\cos(y)/B=\cos(x-y)?$

Aren't they the same?

EDIT: Never mind, I didn't see the last part of the equation since I'm on mobile. Yes, it should be cos(x-y) right?

 

[haruspex]

Did you mean to say
$\sin(y)\sin(x)+\cos(y)\cos(x)=C\cos(y)/B=\cos(x-y)?$

Doh. I had written $\tan(y)=B/A$ but mistranscribed it, then only half adjusted what followed.

 

[Steve4Physics]

FWIW here's a (quite short) solution using the 2D rotation matrix – or rather just its first row for transforming the x-coordinate.

Please excuse the non-use of LaTeX as it's not working for me.

Rotating a point with coordinates (x, y) anticlockwise about the origin through angle θ gives the new x-coordinate x' = x cos θ - y sin θ. For convenience in this solution we use a clockwise rotation which means:
x’ = x cos θ + y sin θ (equation 1)

Start with the cane 'upright':

The weights of the curved and straight sections are W and 2W respectively.

If the cane is rotated by angle θ clockwise about P we can immediately write the new x coordinates of the CMs using equation 1:

When the cane is in equilibrium the moments about P balance:
W x₁ + 2Wx₂ = 0

Substituting x₁ and x₂ and cancelling W and R:

cos θ + (2/π) sin θ + 4cos θ - 2π sin θ = 0

5cos θ = (2π - 2/π) sin θ

tan θ = 5/(2π - 2/π)

θ = arctan (5/(2π - 2/π))

θ = 41.5° or (41.5° + 180°) = 221.5°

(The angle 221.5° corresponds to the unstable equilibrium position with the cane's CM vertically above P.)

Edited to (hopefully) make the explanation easier to follow (see Post #17).

 

[kuruman]

Your solution is hard to follow. Which angle is α in the drawing?

 

[Steve4Physics]

Your solution is hard to follow. Which angle is α in the drawing?

α was 2π - θ, i.e. the angle needed to achieve the required orientation by rotating the cane anticlockwise about P (to allow the standard 2D rotation matrix to be used).

However I've changed the working to avoid using α, so hopefully it's now clearer.