Съдържание

Условие Отговори Решение

USAPhO 2003, Четвъртфинал — Задача 3

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Всички материали от USAPhO 2003

AAPT AIP United States Physics Team 2003 — Preliminary Exam, Open Response Questions · 25 т.

Указания

4 Questions, 60 minutes

INSTRUCTIONS

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

Q1 – 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. 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!

Possibly Useful Information

Gravitational field at the Earth’s surface	$g = 9.8\ \mathrm{N/kg}$
Newton’s gravitational constant	$G = 6.67 \times 10^{-11}\ \mathrm{N \cdot m^2/kg^2}$
Binomial expansion	$(1 + x)^n \approx 1 + nx$ for $\lvert x\rvert \ll 1$

Moment of Inertia about Center of Mass – Uniform Object

(will not be provided on the second screening exam)

Disk	$\tfrac{1}{2} MR^2$
Sphere	$\tfrac{2}{5} MR^2$
Rod	$\tfrac{1}{12} ML^2$

Solutions to Problems

Unless explicitly told to use a specific method, any correct solution should be awarded equivalent points. Suggested partial-credit points are presented in square brackets at the right margin. 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. For example, in 2.a., if a student wrote down $2v_o t - \frac{1}{2}gt^2 = H - v_o t - \frac{1}{2}gt^2$ , they should get 8 points credit. Students should not be penalized in a subsequent part for using the wrong answer to a previous part. (No double jeopardy.)

Условие

A disk of radius $R$ is located in the $x$ - $y$ plane with its center at the origin. The disk has four disk-shaped holes of radius $R/4$ which are centered at $(R/2, 0)$ , $(0, R/2)$ , $(-R/2, 0)$ , and $(0, -R/2)$ , as shown in the accompanying diagram. The disk has mass $M$ and uniform density. The moment of inertia (rotational inertia) of this disk about the $z$ axis is $CMR^2$ where $C$ is a constant.

a. The disk starts from rest at the top of an incline of height $H$ and rolls without slipping to the bottom. What is the speed of its center of mass at the bottom? Express your answer in terms of $H$ , $C$ , and fundamental constants. [8 т.]

A large circle of radius R (labelled M) centred at the origin of grey x and y axes, with four smaller circular holes centred on the axes at distance R/2 from the centre; grey arrows mark the radius R, the hole radius R/4 and the distance R/2.

b. A nail of negligible mass is glued to the top of the disk at $(0, R)$ . The nail points in the $z$ direction and is mounted in a frictionless pivot so that the disk hangs in the vertical plane. The disk is displaced by a small angle from its equilibrium position and released from rest. What is the period of oscillation of the disk in terms of $R$ , $C$ , and fundamental constants? [7 т.]

c. Find the numerical value of the constant $C$ . [10 т.]

Отговори

Решение

3. a. Using energy conservation

\mathrm{PE_i + KE_i = PE_f + KE_f}

[1]

“i” refers to the starting point at the top of the plane and “f” refers to the bottom of the plane.

Taking the bottom of the planes as our zero of potential energy $\mathrm{PE_f} = 0.$ [1]

At the top of the plane $\mathrm{PE_i} = MgH.$ $\mathrm{KE_i} = 0.$ [1]

At the bottom of the plane, the kinetic energy consists of translational kinetic energy plus rotational kinetic energy

\mathrm{KE_f} = \tfrac{1}{2}Mv^2 + \tfrac{1}{2}I\omega^2.

[2]

Substituting into the energy conservation equation

MgH = \tfrac{1}{2}Mv^2 + \tfrac{1}{2}I\omega^2. \qquad (3-1)

Since it rolls without slipping

\omega = \frac{v}{R}.

[1]

Combining terms in Eq (3-1)

MgH = \tfrac{1}{2}Mv^2 + \tfrac{1}{2}CMR^2\left(\frac{v}{R}\right)^2 = \tfrac{1}{2}Mv^2(1 + C).

[1]

Solving for $v$

v = \sqrt{\frac{2gH}{1 + C}}.

[1]

b. By the parallel axis theorem, the moment of inertia about a point on the rim is

I = I_{cm} + MR^2 = CMR^2 + MR^2 = (1 + C)MR^2.

[1]

Either remembering or deriving the equation for the period of a physical pendulum

T = 2\pi\sqrt{\frac{I}{Mgd}}

[4]

where I is the moment of inertia about the pivot point and d is the distance between the pivot point and the center of mass $d = R.$ [1]

Combining these equations

T = 2\pi\sqrt{\frac{(1 + C)MR^2}{MgR}} = 2\pi\sqrt{\frac{(1 + C)R}{g}}.

[1]

c. Let $\sigma$ = the mass per unit area in the $x$ - $y$ plane. Consider a solid disk of radius $R$ and density $\sigma$ and four solid disks with radius $R/4$ and density $-\sigma$ .

The mass of a solid disk is

M_{SD} = \sigma\pi R^2.

[1]

The moment of inertia of the solid disk is

I_{SD} = \tfrac{1}{2}M_{SD}R^2 = \tfrac{1}{2}(\sigma\pi R^2)R^2 = \tfrac{1}{2}\sigma\pi R^4.

[1]

The mass of a hole is

M_H = -\sigma\pi\left(\frac{R}{4}\right)^2.

[1]

The moment of inertia a hole about its center

I_{H,cm} = \tfrac{1}{2}M_H\left(\frac{R}{4}\right)^2.

[1]

Using the parallel axis theorem to find the moment of inertia of a hole about the center of the disk, a distance $d = R/2$ from the center of the hole,

I_H = I_{H,cm} + M_H d^2 = \tfrac{1}{2}M_H\left(\frac{R}{4}\right)^2 + M_H\left(\frac{R}{2}\right)^2 = \frac{1}{32}M_H R^2(1 + 8) = \frac{9}{32}M_H R^2.

[1]

I_H = \frac{9}{32}\left(-\sigma\pi\left(\frac{R}{4}\right)^2\right)R^2 = -\frac{9}{512}\sigma\pi R^4.

[1]

The total mass of the disk

M = M_{SD} + 4M_H = \sigma\pi R^2 + 4\left(-\sigma\pi\left(\frac{R}{4}\right)^2\right) = \sigma\pi R^2\left(1 - 4\left(\frac{1}{16}\right)\right) = \frac{3}{4}\sigma\pi R^2.

[2]

The total moment of inertia of the disk is

I = I_{SD} + 4I_H = \frac{1}{2}\sigma\pi R^4 + 4\left(-\frac{9}{512}\sigma\pi R^4\right) = \frac{1}{128}\sigma\pi R^4(64 - 9) = \frac{55}{128}\sigma\pi R^4.

Rewriting it in terms of $M$

I = \frac{55}{128}\sigma\pi R^4\left(\frac{M}{\frac{3}{4}\sigma\pi R^2}\right) = \frac{55}{128}\,\frac{4}{3}MR^2 = \frac{55}{96}MR^2.

[1]

To find $C$

C = \frac{I}{MR^2} = \frac{55}{96}

[1]

Оригинал в Архива: Quarterfinal 2003.pdf · официални решения: Quarterfinal 2003 Solutions.pdf