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INPhO 2025, Теория — Задача 2. Gearminator: Rise of the Machines

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Indian National Physics Olympiad (INPhO)-2025 — Question Paper · 2 февруари 2025 г. · 14 т.

Внимание: Бележка към темата

The complete retained original question and matching HBCSE solution packets are transcribed. The separately referenced Answer Booklet is not retained in the matching archive row. Printed key formula/sign/symbol/prose discrepancies and the original handwritten magnetic-data graph remain literal, with separately labelled source notes.

Original cover and instructions

Indian National Physics Olympiad (INPhO)-2025

HOMI BHABHA CENTRE FOR SCIENCE EDUCATION

Tata Institute of Fundamental Research

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

Question Paper

Date: 02 February 2025

Time: 09:00-12:00 (3 hours)

Maximum Marks: 75

Instructions — Roll Number: $\square\square\square\square-\square\square\square\square-\square\square\square\square$

This booklet consists of 9 pages and total of 5 questions. Write roll number at the top wherever asked.
Booklet to write the answers is provided separately. Instructions to write the answers are on the Answer Booklet.
Non-programmable scientific calculators are allowed. Mobile phones cannot be used as calculators.
Please submit the Answer Sheet at the end of the examination. You may retain the Question Paper.

Question Number	1	2	3	4	5	Total
Maximum Marks	8	14	16	22	15	75

Table of Constants

Speed of light in vacuum	$c$	$3.00\times10^8\ \mathrm{m\cdot s^{-1}}$
Magnitude of electron charge	$e$	$1.60\times10^{-19}\ \mathrm C$
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	$R$	$8.31\ \mathrm{J\cdot K^{-1}\cdot mol^{-1}}$
	$R$	$0.0821\ \mathrm{l\cdot atm\cdot mol^{-1}\cdot K^{-1}}$
Boltzmann constant	$k_B$	$1.3806\times10^{-23}\ \mathrm{J\cdot K^{-1}}$
Permeability constant	$\mu_0$	$4\pi\times10^{-7}\ \mathrm{H\cdot m^{-1}}$

Native key notice

The matching key repeats the cover, constant table, marks table and four instructions with “24 pages” and the title “Solutions”.

Please note that alternate/equivalent methods and different ways of expressing final solutions may exist. A correct method will be suitably awarded.

Условие

We consider a “thought experiment” involving a DC motor and a DC generator coupled mechanically through a gearbox, operating under idealized conditions, to explore the power output and efficiency of the system (see schematic figure below). The schematic gearbox assembly is also shown in the figure.

Entire native motor-generator circuit, both commutators, B directions, r/R/generator-emf/gearbox and left source voltage/polarity labels.

Both native meshing gears with Motor gear/Generator gear/Gearbox labels.

Both the motor and the generator have $N$ loops of area $A$ and rotate in a uniform magnetic field of strength $B$ . As usual, both the motor and the generator use commutators (indicated by the blue blocks) to reverse the direction of current in each arm every half cycle, to ensure unidirectional output. The generator is connected to an external resistance $R$ , and the motor is driven by a constant voltage $V_M$ with an internal resistance $r$ . The gearbox is idealized, with no energy loss due to friction or otherwise, and no slipping between the teeth of the gears. For a pair of meshing gears, as shown above, the angular speed ratio, also known as the gear ratio $X$ , is defined as:

X=\frac{\omega_M}{\omega_G},

where $\omega_M$ and $\omega_G$ are the angular velocities of the motor and the generator, respectively. Let $\langle P_G\rangle$ , and $\langle P_M\rangle$ be the time-averaged generator output power and the time-averaged motor input power, respectively, over one complete cycle.

a) Derive the expression for $\omega_G$ in terms of $X,R,r$ , and the given parameters. For fixed values of $r$ and $R$ , determine the expression of $X$ for which $\omega_G$ is maximum. [6 т.]

b) Derive the expression for the generator output power $\langle P_G\rangle$ in terms of $X,R,r$ , and the given parameters. For fixed values of $r$ and $R$ , determine the expression of $X$ for which $\langle P_G\rangle$ is maximum. [3 т.]

c) For fixed values of $r$ and $R$ , determine the condition on $X$ for which the efficiency $\eta$ is maximum, where

\eta=\frac{\langle P_G\rangle}{\langle P_M\rangle}.

Calculate this maximum value of $\eta$ . [5 т.]

Решение

(a) Solution: First, we analyze motor equations:

Applying voltage $V_M$ drives the motor. The current $I_M$ and and the angular speed $\omega_M$ are determined through the following equations.

The circuit equation for motor with speed $\omega_M$ is

V_M-\epsilon_{bM}-I_Mr=0\qquad(2.1)

where $\epsilon_{bM}=NAB\omega_M\sin(\omega_Mt)$ is the induced back emf. Thus, the current $I_M$ is

I_M=\frac1r\left(V_M-NBA\omega_M\sin(\omega_Mt)\right)\qquad(2.2)

The magnetic torque on the coils carrying current $I_M$ is

\tau_M=NI_MBA\sin(\omega_Mt)\qquad(2.3)

=\frac1r\left(V_MNBA\sin(\omega_Mt)-N^2B^2A^2\omega_M\sin^2(\omega_Mt)\right)\qquad(2.4)

The Gearbox is massless, and there is no loss of power. Thus,

\tau_M\omega_M=\tau_G\omega_G\Rightarrow\tau_G=X\tau_M\qquad(2.5)

For generator: Gearbox drives the motor with torque $\tau_G$ . The current $I_G$ and and the angular speed $\omega_G$ are determined through the following equations.

The circuit equation for generator with speed $\omega_G$ is

e_{bG}-I_GR=0\qquad(2.6)

where $\epsilon_{bG}=NAB\omega_G\sin(\omega_Gt)$ is the induced back emf. Thus, the current $I_G$ is

I_G=\frac1R NBA\omega_G\sin(\omega_Gt)\qquad(2.7)

The magnetic torque on the coils carrying current $I_G$ is

\tau_G=NI_GBA\sin(\omega_Gt)

=\frac1R N^2B^2A^2\omega_G\sin^2(\omega_Gt)\qquad(2.8)

Since there is no dissipation in the gearbox, from Eq. (2.5), we get

\langle\tau_G\rangle=\langle\tau_M\rangle X\qquad(2.9)

From Eq. (2.8)

\langle\tau_G\rangle=\frac{N^2B^2A^2\omega_G}{2R}\qquad(2.10)

Since $\langle\sin^2(\omega t)\rangle=1/2$ over one complete cycle, and from Eq. (2.4)

\langle\tau_M\rangle=\frac{2V_MNBA}{\pi r}-\frac{N^2B^2A^2\omega_M}{2r}\qquad(2.11)

Since $\langle\sin^2(\omega t)\rangle=1/2$ , and $\langle\sin(\omega t)\rangle=2/\pi$ over one complete cycle.

Using Eq. (2.9), we obtain

\frac{\omega_G}{R}=\frac{(4V_M/\pi NBA)X}{r}-\frac{X^2\omega_G}{r}\qquad(2.12)

\omega_G\left(\frac1R+\frac{X^2}{r}\right)=\left(\frac{4V_M}{\pi NBA}\right)\frac Xr\qquad(2.13)

\boxed{\omega_G=\left(\frac{4V_M}{\pi NBA}\right)\frac{1}{\left(\frac r{XR}+X\right)}}\qquad(2.14)

To maximize $\omega_G$ , minimize the denominator $r/RX+X$ . Differentiating:

\frac d{dX}\left(\frac r{RX}+X\right)=0\qquad(2.15)

-\frac r{RX^2}+1=0\qquad(2.16)

\boxed{\Rightarrow X=\sqrt{\frac rR}}\qquad(2.17)

(b) Solution: The generator output power is

\langle P_G\rangle=\frac{\langle\epsilon_G^2\rangle}{R}=\frac{N^2B^2A^2\omega_G^2}{2R}

\boxed{\langle P_G\rangle=\frac{8V_M^2}{\pi^2R\left(\frac r{XR}+X\right)^2}}

To maximize $\langle P_G\rangle$ , differentiate with respect to $X$ and solve for $X$

\boxed{X=\sqrt{\frac rR}}\qquad(2.18)

(c) Solution: The motor input power is:

\langle P_M\rangle=\langle V_MI_M\rangle\qquad(2.19)

=\left\langle\frac{V_M}{r}\left(V_M-NBA\omega_M\sin(\omega_Mt)\right)\right\rangle\qquad(2.20)

=\frac{V_M}{r}\left(V_M-\frac2\pi NBAX\omega_G\right)\qquad(2.21)

=\frac{V_M^2}{r}\left(1-\left(\frac8{\pi^2}\right)\frac1{\left(\frac r{X^2R}+1\right)}\right)\qquad(2.22)

and

\langle P_G\rangle=\frac{8V_M^2}{\pi^2R\left(\frac r{XR}+X\right)^2}\qquad(2.23)

The efficiency is given by:

\eta=\frac{\langle P_G\rangle}{\langle P_M\rangle}\qquad(2.24)

Substituting the expressions for $\langle P_G\rangle$ and $\langle P_M\rangle$ , maximize $\eta$ with respect to $X$ , we get

\boxed{X_{\max}\approx1.52\sqrt{\frac rR}}\qquad(2.25)

At this $X$ , efficiency $\eta$ is maximum. The maximum value of $\eta$ can be calculated by substituting value $X_{\max}$ in $\eta$ .

\boxed{\eta_{\max}\approx0.4}\qquad(2.26)

Source note: The key prints “and and” twice and says the gearbox “drives the motor” in its generator paragraph. Equation (2.6) uses $e_{bG}$ while the following prose uses $\epsilon_{bG}$ . Its complete-cycle average is printed as $\langle\sin(\omega t)\rangle=2/\pi$ ; the question specifies commutators. These native statements and symbols are retained.

Native source end

END OF THE QUESTION PAPER

The key also ends with this native wording on page24. Key page20 is blank apart from running page and roll-number furniture.

Оригинал в Архива: INPhO2025-Question.pdf · официални решения: INPhO2025-Solution-20250219.pdf