Questions about infinitesimal changes in a mechanics problem

  • #1
Leo Liu
353
156
Homework Statement
This is a statement.
Relevant Equations
##\vec F=\frac {d\vec {\dot P}} {dt}##
1595637817647.png

While reading Kleppner's book, I came across the question above whose solution given by an answer book, is shown below.
1595637876843.png

I wrote out an equation for inward force and another equation for horizontal forces:
$$\begin{cases}
f_{\Delta \theta}=\mu N=\mu \frac{\Delta\theta} 2 (T+T'),\text{ where T' is the force exerted by the next segment of the rope}\\
\\
T=T'+f=T'+\mu \frac{\Delta\theta} 2 (T+T')
\end{cases}$$

As you can see, I used ##T'## instead of ##T+\Delta T## to denote the force on the right side of the diagram. My first question is why the changes of the force for each tiny segment of the rope are the same (##\Delta T##).

The author omitted ##\Delta T## when he was approximating the inward force; whereas, he kept it when finding an approximation for the horizontal forces. I would like to know why he has treated the same term differently.

I understand that ##\sin(\Delta\theta /2)\approx\Delta\theta /2## when ##\Delta\theta## is small because the first order polynomial is a good approximation of ##\sin(x)## around the origin. However, I wonder what I should do to approximate ##\cos(\Delta\theta /2)##.

Could you please answer the three questions I asked above? Thank you.
 
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  • #2
$$\begin{align*}
f=\mu N\Leftrightarrow\Delta T&=2\mu T\sin\left(\frac{\Delta\theta}{2}\right)+\mu\Delta T\sin\left(\frac{\Delta\theta}{2}\right)\\
&\approx\mu T\Delta\theta+\frac{\mu}{2}\Delta T\Delta\theta
\end{align*}$$As ##\Delta\theta\to0##, ##\Delta T\to0##, and so you have ##\mu T\Delta\theta>>\frac{\mu}{2}\Delta T\Delta\theta##.

When you want to approximate a function near some input value ##a##, use Taylor's formula (##a## is ##0## in this problem):
$$f(x)=f(a)+(x-a)f'(a)+\frac{(x-a)^2}{2!}f''(a)+\frac{(x-a)^3}{3!}f'''(a)+...$$
Jump to "Trigonometric functions" https://en.wikipedia.org/wiki/Taylor_series
$$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}+...\text{, with $a = 0$}$$
 
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  • #3
1. What do you mean "the same"? The same as what? ##\Delta T## here denotes the difference between the tensions at each end for a sample arc. The equation ##\Delta T=\mu T \Delta \theta## tells you how it is related to ##T## and ##\Delta \theta##.

2. For the inward direction if you don't omit ##\Delta T##, you get for the inward force$$F_{in}=T\sin\left( \frac{\Delta \theta}{2}\right)+(T+\Delta T)\sin\left( \frac{\Delta \theta}{2}\right)\approx 2T\left( \frac{\Delta \theta}{2}\right)+\Delta T\left( \frac{\Delta \theta}{2}\right).$$ Which of the two terms is the leading one to first order?

3. The series expansions are
##\sin\left( \dfrac{\Delta \theta}{2}\right)\approx \dfrac{\Delta \theta}{2}-\dots##
##\cos\left( \dfrac{\Delta \theta}{2}\right)\approx 1- \dfrac{1}{2}\left(\dfrac{\Delta \theta}{2}\right)^2+\dots##
What is the approximation of the cosine to first order?

I see that @archaic beat me to the punch.
 
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  • #4
kuruman said:
1. What do you mean "the same"? The same as what? ΔT here denotes the difference between the tensions at each end for a sample arc.
I see. I assumed this because I thought the corresponding tiny angular displacements are the same.
kuruman said:
Which of the two terms is the leading one to first order?
Probably the first term since the second term is a very small number times a very small number.
kuruman said:
What is the approximation of the cosine to first order?
1.

Thanks for your answer.
 
  • #5
You got it!
 
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