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INPhO 2019, Теория — Задача 3. A chain

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Всички материали от INPhO 2019

Indian National Physics Olympiad – 2019 · 3 февруари 2019 г. · 12 т.

Внимание: Бележка към темата

The complete retained question booklet includes summary answer boxes, eighteen blank detailed-answer pages and three blank graph sheets. All matching archived HBCSE key content is retained, including both full handwritten graph sheets and accepted alternatives/ranges. Printed exponent, sign, factor, notation and unit inconsistencies remain literal with labelled source notes.

Native cover and instructions

Indian National Physics Olympiad – 2019

Date: 03 February 2019. Time: 09:00–12:00 (3 hours). Maximum Marks: 75.

Roll Number: $1\,9\,\square\,\square-\square\,\square\,\square\,\square-\square\,\square\,\square\,\square$ . Extra sheets attached: [blank box]. Centre (e.g. Kota): [blank line].

(Do not write below this line)

This booklet consists of 29 pages (excluding this page) and total of 7 questions.
This booklet is divided in two parts: Questions with Summary Answer Sheet and Detailed Answer Sheet. Write roll number at the top wherever asked.
The final answer to each sub-question should be neatly written in the box provided below each sub-question in the Questions & Summary Answer Sheet.
You are also required to show your detailed work for each question in a reasonably neat and coherent way in the Detailed Answer Sheet. You must write the relevant Question Number(s) on each of these pages.
Marks will be awarded on the basis of what you write on both the Summary Answer Sheet and the Detailed Answer Sheet. Simple short answers and plots may be directly entered in the Summary Answer Sheet. Marks may be deducted for absence of detailed work in questions involving longer calculations. Strike out any rough work that you do not want to be considered for evaluation.
Adequate space has been provided in the answersheet for you to write/calculate your answers. In case you need extra space to write, you may request additional blank sheets from the invigilator. Write your roll number on the extra sheets and get them attached to your answersheet and indicate number of extra sheets attached at the top of this page.
Non-programmable scientific calculators are allowed. Mobile phones cannot be used as calculators.
Use blue or black pen to write answers. Pencil may be used for diagrams/graphs/sketches.
This entire booklet must be returned at the end of the examination.

Table of Constants

Constant	Symbol	Value
Speed of light in vacuum	$c$	$3.00\times10^8\ \mathrm{m\cdot s^{-1}}$
Planck’s constant	$h$	$6.63\times10^{-34}\ \mathrm{J\cdot s}$
	$\hbar$	$h/2\pi$
Universal constant of Gravitation	$G$	$6.67\times10^{-11}\ \mathrm{N\cdot m^2\cdot kg^{-2}}$
Magnitude of electron charge	$e$	$1.60\times10^{-19}\ \mathrm C$
Rest mass of electron	$m_e$	$9.11\times10^{-31}\ \mathrm{kg}$
Value of $1/4\pi\epsilon_0$		$9.00\times10^9\ \mathrm{N\cdot m^2\cdot C^{-2}}$
Avogadro’s number	$N_A$	$6.022\times10^{23}\ \mathrm{mol^{-1}}$
Acceleration due to gravity	$g$	$9.81\ \mathrm{m\cdot s^{-2}}$
Universal Gas Constant	$R$	$8.31\ \mathrm{J\cdot K^{-1}\cdot mol^{-1}}$
	$R$	$0.0821\ \mathrm{l\cdot atm\cdot mol^{-1}\cdot K^{-1}}$
Permeability constant	$\mu_0$	$4\pi\times10^{-7}\ \mathrm{H\cdot m^{-1}}$

Question	Marks	Score
1	9
2	11
3	12
4	7
5	9
6	14
7	13
Total	75

HOMI BHABHA CENTRE FOR SCIENCE EDUCATION

Tata Institute of Fundamental Research

V. N. Purav Marg, Mankhurd, Mumbai, 400 088

Native key cover

Solutions of Indian National Physics Olympiad – 2019

The native key cover repeats the nine instructions, twelve constant rows, seven-question score table totalling 75 marks, roll-number form and HBCSE address from the question cover. It prints “20 pages (excluding this page)”; the retained key contains twenty-one PDF pages, including two appended handwritten graph sheets.

Условие

A chain of length $l$ and linear density $\lambda$ hangs from a horizontal support with both ends A and B fixed to a horizontal support as shown. The two fixed ends are close to each other. At time $t=0$ the end A is released. All vertical distances ( $x$ ) are measured with respect to the horizontal support with the downward direction taken as positive (A and B are initially at $x=0$ ).

Both original chain configurations t=0/t>0, supports, A/B endpoints, l/2 and x arrows.

a) Obtain the momentum $P$ of the center of mass of the system when the end A has fallen by a distance $x$ , in terms of $x$ and speed $\dot x$ .

Summary answer box: $P=$ [blank]. [2 т.]

b) Assume that the end A is falling freely under gravity, i.e., $\ddot x=g$ . Obtain the tension $T$ at the fixed end B just before the chain completes the fall and becomes entirely vertical.

Summary answer box: $T=$ [blank].

Experimentally, the value of tension is found to be different from the above result. We adopt an alternative approach assuming conservation of mechanical energy. [1 т.]

c) Obtain the potential energy $U(x)$ of the chain and plot it versus $x$ . Take the potential energy of a point mass placed at the horizontal support to be zero.

Entire native blank answer/plot frame for the chain potential-energy request.

[2,5 т.]

d) Obtain the speed $\dot x$ when the end A has fallen by a distance $x$ . Assume that all sections of the falling (right side) part of the chain have the same speed.

Summary answer box: $\dot x=$ [blank]. [2,5 т.]

e) Hence obtain $T(x)$ at B as a function of $x$ and related quantities. You are advised to simplify your expression as far as possible.

Summary answer box: $T(x)=$ [blank]. [3 т.]

f) Qualitatively sketch $T(x)$ versus $x$ .

Full original blank T(x) versus x axes and response frame.

Detailed answers can be found on page numbers: [blank]. [1 т.]

Решение

(a)

Let total mass of the chain to be $M=\lambda l$ .

The mass of the left side of the chain $=\lambda\left(\frac{l+x}{2}\right)$ .

CM of the left side of the chain $=\left(\frac{l+x}{4}\right)$ .

The mass of the right side of the chain $=\lambda\left(\frac{l-x}{2}\right)$ .

CM of the right side of the chain $=\left(x+\frac{l-x}{4}\right)$ .

x_{\mathrm{CM}}=\frac{\lambda\left(\frac{l+x}2\right)\left(\frac{l+x}4\right)+\lambda\left(\frac{l-x}2\right)\left(x+\frac{l-x}4\right)}{\lambda l}\qquad (3.1)

\lambda l x_{\mathrm{CM}}=\frac{\lambda l^2}4+\frac{\lambda lx}2-\frac{\lambda x^2}4\qquad (3.2)

P=\lambda l\dot x_{\mathrm{CM}}=\lambda\left(\frac{l-x}2\right)\dot x\qquad (3.3)

(b)

\ddot x=g\qquad (3.4)

\dot x=gt=\sqrt{2gx}\qquad (3.5)

\dot P=Mg-T\qquad (3.6)

From Eq. (3.3)

\dot P=\frac\lambda2\left[\ddot x(l-x)-\dot x^2\right]\qquad (3.7)

\frac\lambda2[gl-3gx]=Mg-T\qquad (3.8)

T(x)=\frac{Mg}2\left(1+\frac{3x}l\right)=\frac{\lambda lg}2\left(1+\frac{3x}l\right)\qquad (3.9)

T(x=l)=2\lambda lg\qquad (3.10)

Experimentally, the value of tension is found to be different from the above result. We adopt an alternative approach assuming conservation of mechanical energy.

(c)

In general, $U=-\lambda lgx_{\mathrm{CM}}$ . Using Eq. (3.2)

U=\frac{-\lambda g}4(l^2+2lx-x^2)\qquad (3.11)

Entire native potential-energy graph, all axes, 0,L endpoint and both negative lambda g l squared levels.

(d)

From Eq. (3.3), kinetic energy of the chain

K(x)=\frac12\lambda\left(\frac{l-x}2\right)\dot x^2\qquad (3.12)

As the total energy is conserved,

U(x)+K(x)=U(t=0)\qquad (3.13)

\dot x^2=\left[\frac{g(2lx-x^2)}{(l-x)}\right]^{1/2}\qquad (3.14)

\dot x=\left[\frac{g(2lx-x^2)}{(l-x)}\right]^{1/2}\qquad (3.15)

Source note: Printed Eq. (3.14) has $\dot x^2$ on the left and a square-root exponent on the right; the next Eq. (3.15) repeats that right-hand side with $\dot x$ on the left. Both original equations are retained.

(e)

\frac d{dt}\dot x^2=2\dot x\ddot x

Substituing Eq. (3.15) in the above equation yields

\ddot x=g+\frac{g(2lx-x^2)}{2(l-x)^2}\qquad (3.16)

Combining Eqs. (3.6), (3.7) and (3.16)

T=\lambda lg-\frac\lambda2\left[\ddot x(l-x)-\dot x^2\right]\qquad (3.17)

=\frac{\lambda g}{4(l-x)}[2l^2+2lx-3x^2]\qquad (3.18)

(f)

Entire original T versus x curve with lambda g l/2 intercept, 0,L, asymptote and all labels.

Source note: Both original key plots mark the chain endpoint with uppercase $L$ ; the question and key formulas define the chain length as lowercase $l$ . The actual native plots are retained.

Оригинал в Архива: INPhO2019-Question.pdf · решение от архива: INPhO2019-Solution-20190219.pdf