Съдържание

Условие Expt. 1:Expt. 2:Решение

USAPhO 2006, Четвъртфинал — Задача 3

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Всички материали от USAPhO 2006

Preliminary Exam — Open Response Questions · 25 т.

Указания

DO NOT OPEN THIS TEST UNTIL YOU ARE TOLD TO BEGIN

Show all your work. Partial credit will be given.
Start each question on a new sheet of paper. Be sure to put your name in the upper right-hand corner of each page, along with the question number and the page number/total pages for this problem. For example,

Doe, Jamie

#1 – P. 1/3

A hand-held calculator may be used. Its memory must be cleared of data and programs. You may use only the basic functions found on a simple scientific calculator. Calculators may not be shared. Cell phones may not be used during the exam or while the exam papers are present. You may not use any tables, books, or collections of formulas.
Each of the four questions is worth 25 points. The questions are not necessarily of the same difficulty. Good luck!
In order to maintain exam security, do not communicate any information about the problems or the solutions on this exam (or on the multiple-choice exam) until after February 13th.

Possibly Useful Information

Quantity	Value
Gravitational field at the Earth's surface	$g=9.8$ N/kg
Newton's gravitational constant	$G=6.67\times10^{-11}$ N·m $^2$ /kg $^2$
Average Earth-Sun Distance	1 Astronomical Unit (A.U.) $=1.5\times10^{11}$ m
Binomial expansion	$(1+x)^n\simeq1+nx$ for $\lvert x\rvert\ll1$

Moment of Inertia about Center of Mass - Uniform Object

(will not be provided on the second screening exam)

Object	Moment of Inertia
Disk	$\frac12MR^2$
Sphere	$\frac25MR^2$
Rod	$\frac1{12}ML^2$

Solutions to Problems

Each question is worth 25 points. Any correct solution should be awarded equivalent points. Suggested partial-credit points are presented in square brackets in the right margin after the equation number. You may further break down the listed points into one point increments. If it is clear they have done an intermediate step, they should get credit for it even if they have not presented it. Students should not be penalized in a subsequent part for using the wrong answer to a previous part. (No double jeopardy.)

Условие

A uniform thin rod is suspended from the ceiling by two identical ideal vertical springs attached to the ends of the rod. Two experiments are performed:

Expt. 1:

From equilibrium, both ends of the rod are pulled down by an equal, small amount. When released, the rod oscillates with period $T_1$ .

Expt. 2:

From equilibrium, one end of the rod is pulled down by a small amount, and the other end is pushed up by an equal amount. When released, the rod oscillates with a period $T_2$ .

Find the numerical value of the ratio $T_1/T_2$ .

Решение

There are at least two ways to approach this problem. Other solution methods may exist as well.

Method 1

This approach focuses on the differential equations of oscillatory motion.

In general, any system satisfying an equation of the form

\ddot z=-\omega^2z\qquad\text{(3-1)}

will oscillate with angular frequency $\omega$ . $\ddot z$ is shorthand for the second time derivative of the variable $z$ , or

\ddot z=\frac{d^2z}{dt^2}\qquad\text{(3-2)}

Students don't need to write this down, but they will need to use it if they take this approach.

Let the rod have mass $M$ and length $2r$ . Let the springs each have constant $k$ .

In experiment 1, let $x$ be the vertical displacement of the rod from equilibrium. Then the equation of motion is

Ma=F_{\mathrm{net}}\qquad\text{(3-3)}

M\ddot x=-2kx\qquad\text{(3-4)}

\ddot x=-\frac{2k}Mx\qquad\text{(3-5)}

\omega_1=\sqrt{\frac{2k}M}\qquad\text{(3-6)}

[6 pt]

where the 2 on the right hand side of Eq. 3-4 arises from the presence of two springs. In the last line we take advantage of Eq. 3-1; it does not need to be derived. Note the absence of gravity from this expression. $x$ is the change in length from the equilibrium position, and then net force at the equilibrium position is zero.

In experiment 2, let $\theta$ be the angular displacement of the rod from equilibrium. Note that for small $\theta$ , the extension of each spring from equilibrium is given by

x=r\theta\qquad\text{(3-7)}

[2 pt]

The force of the spring on each end of the rod is

F=-kx=-kr\theta\qquad\text{(3-8)}

[2 pt]

so the net torque on the rod about the center of rotation is

\tau_{\mathrm{net}}=-2(kr\theta)r\qquad\text{(3-9)}

[2 pt]

The rotational inertia of the rod of length $2r$ about the center is

I=\frac1{12}M(2r)^2=\frac13Mr^2\qquad\text{(3-10)}

[3 pt]

Then, combining Eq. 3-7 through Eq. 3-10,

I\alpha=\tau\qquad\text{(3-11)}

\frac13Mr^2\ddot\theta=-2kr^2\theta\qquad\text{(3-12)}

\ddot\theta=-\frac{6k}M\theta\qquad\text{(3-13)}

\omega_2=\sqrt{\frac{6k}M}\qquad\text{(3-14)}

[6 pt]

Finally, since

T=\frac{2\pi}\omega\qquad\text{(3-15)}

[2 pt]

we have from Eq. 3-6 and Eq. 3-14

\frac{T_1}{T_2}=\frac{\omega_2}{\omega_1}\qquad\text{(3-16)}

=\sqrt3\qquad\text{(3-17)}

[2 pt]

Method 2

This method focuses on the fundamental expression for the period of an oscillating system

The period of oscillation of a mass spring system is given by

T=2\pi\sqrt{\frac mk}\qquad\text{(3-18)}

[3 pt]

where $k$ is the spring constant and $m$ is the mass.

In the first experiment there are two springs, each effectively moving half of the total mass $M$ of the thin rod, so

T_1=2\pi\sqrt{\frac{M/2}k}=2\pi\sqrt{\frac M{2k}}\qquad\text{(3-19)}

[2 pt]

The period of oscillation of a rotational system is given by

T_2=2\pi\sqrt{\frac I\kappa}\qquad\text{(3-20)}

[3 pt]

where $\kappa$ is the rotational spring constant and $I$ the rotational inertia. In this case,

\kappa\theta=\tau\qquad\text{(3-21)}

[2 pt]

where $\tau$ is the torque about the center of rotation. That torque is provided for by the force of the spring,

\tau=rF=\frac L2kx\qquad\text{(3-22)}

[4 pt]

where $x$ is the extension of the spring and $L$ the length of the thin rod.

The angle $\theta$ is given by

\theta\simeq\frac x{L/2}\qquad\text{(3-23)}

[2 pt]

for small $x$ , so

\kappa=\tau/\theta\qquad\text{(3-24)}

=k\frac{L^2}4\qquad\text{(3-25)}

[2 pt]

for one spring.

The rotational inertia for a thin rod about the center is

I_{\mathrm{rod}}=\frac1{12}ML^2\qquad\text{(3-26)}

[1 pt]

but we are interested in only half, so

I=\frac1{24}ML^2\qquad\text{(3-27)}

[2 pt]

Combine Eq. 3-20, Eq. 3-25, and Eq. 3-27. The period of oscillation is then

T_2=2\pi\sqrt{\frac I\kappa}\qquad\text{(3-28)}

=2\pi\sqrt{\frac{ML^2/24}{kL^2/4}}\qquad\text{(3-29)}

=2\pi\sqrt{\frac M{6k}}\qquad\text{(3-30)}

[2 pt]

The ratio is found by dividing Eq. 3-19 by Eq. 3-30:

\frac{T_1}{T_2}=\sqrt3\qquad\text{(3-31)}

[2 pt]

Оригинал в Архива: Quarterfinal 2006.pdf · официални решения: Quarterfinal 2006 Solutions.pdf