Questions about infinitesimal changes in a mechanics problem

In summary: The approximation of cosine to first order is$$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}+...\text{, with $a = 0$}$$.
  • #1
Leo Liu
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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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1. What is an infinitesimal change in a mechanics problem?

An infinitesimal change in a mechanics problem refers to a very small change or variation in a physical quantity, such as position, velocity, or acceleration. It is often denoted by the symbol "d" and represents a change that is so small it can be considered negligible.

2. Why are infinitesimal changes important in mechanics?

Infinitesimal changes are important in mechanics because they allow us to analyze and understand the behavior of complex systems by breaking them down into smaller, more manageable parts. This approach, known as infinitesimal analysis, is a fundamental tool in the study of mechanics and other branches of science.

3. How are infinitesimal changes calculated in mechanics?

Infinitesimal changes are calculated using calculus, specifically the concept of derivatives. By taking the limit as the change becomes infinitely small, we can determine the exact value of the infinitesimal change and its effect on the overall system.

4. Can infinitesimal changes be ignored in mechanics?

No, infinitesimal changes cannot be ignored in mechanics. While they may seem insignificant, they can have a significant impact on the behavior of a system. In fact, many important laws and principles in mechanics, such as Newton's laws of motion and the fundamental theorem of calculus, rely on the concept of infinitesimal changes.

5. Are infinitesimal changes always constant in mechanics?

No, infinitesimal changes are not always constant in mechanics. They can vary depending on the specific system and the variables being analyzed. For example, in a system with changing velocity, the infinitesimal change in velocity will also change over time. However, in some cases, such as a constant acceleration, the infinitesimal change may remain constant.

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