Questions about infinitesimal changes in a mechanics problem

[Leo Liu]

Homework Statement

This is a statement.

Relevant Equations

$\vec F=\frac {d\vec {\dot P}} {dt}$

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

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.

 

[archaic]

$$\begin{align*}
f=\mu N\Leftrightarrow\Delta T&=2\mu T\sin\left(\frac{\Delta\theta}{2}\right)+\mu\Delta T\sin\left(\frac{\Delta\theta}{2}\right)\\
&\approx\mu T\Delta\theta+\frac{\mu}{2}\Delta T\Delta\theta
\end{align*}$$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
$$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}+...\text{, with $a = 0$}$$

 

[kuruman]

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.

 

[Leo Liu]

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.

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.

What is the approximation of the cosine to first order?

1.

Thanks for your answer.

 

[kuruman]

You got it!