Съдържание

Условие Отговори Решение

USAPhO 2000, Четвъртфинал — Задача 2

Оригинал на условието (PDF)Съобщи за грешка

Всички материали от USAPhO 2000

2000 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 / 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.

Possibly Useful Information

Gravitational field at the Earth’s surface $g=9.8\ N/kg$

Newton’s gravitational constant $G=6.67\times10^{-11}\ N\cdot m^2/kg^2$

Binomial expansion $(1+x)^n\approx1+nx$ for $|x|\ll1$

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

(Suggested partial-credit points in square brackets at the right margin)

Условие

An accelerometer consists of a ball of mass $m$ attached to one end of a string. The other end of the string is attached to a cart of mass $M$ . When the cart rolls down an inclined plane of angle $\theta$ with respect to the horizontal, the string makes a constant angle of $2\theta/3$ with respect to the vertical. See diagram at the right. Express your answers to the following in terms of known and given quantities. Assume air resistance is negligible.

Cart M on slope theta carrying hanging ball m; string angle2theta/3.

a. What is the acceleration of the cart? [13 т.]

b. What is the coefficient of friction between the cart and the plane? [12 т.]

Отговори

Решение

a. Resolving the string tension $T$ and the ball weight $mg$ into components parallel and perpendicular to the plane as shown in the diagram at the right.

Forces on the accelerometer ball resolved parallel and perpendicular to the slope.

[1 point for each correct force/component] [6]

we have for the perpendicular component: $\sum F_\perp=0$ [1]

$T\cos(\theta/3)=mg\cos\theta$ (2.1) [1]

And for the parallel component: $\sum F_\parallel=ma$ [1]

$mg\sin\theta-T\sin(\theta/3)=ma$ (2.2) [1]

Solving (2.1) for $T$ $T=mg\frac{\cos\theta}{\cos(\theta/3)}$ [1]

and substituting into (2.2):

$mg\sin\theta-mg\frac{\cos\theta\sin(\theta/3)}{\cos(\theta/3)}=ma$ [1]

$a=g\left(\sin\theta-\cos\theta\frac{\sin(\theta/3)}{\cos(\theta/3)}\right)$ (2.3) [1]

(While it is not required to simplify the above to obtain full credit, it equals $a=2g\sin(\theta/3)$ .)

b. The diagram to the right shows the forces acting on the cart-accelerometer system.

Forces on cart and accelerometer: normal, friction, weight and components.

[1 point for each correct force/component] [5]

For the perpendicular component: $\sum F_\perp=0$ [1]

$F_n-(m+M)g\cos\theta=0$ (2.4) [1]

And for the parallel component:

$\sum F_\parallel=(m+M)a$ [1]

$(m+M)g\sin\theta-\mu F_n=(m+M)a$ (2.5) [1]

Combining equations (2.3), (2.4), and (2.5)

(m+M)g\sin\theta-\mu(m+M)g\cos\theta=(m+M)g\left[\sin\theta-\cos\theta\frac{\sin(\theta/3)}{\cos(\theta/3)}\right]

[1]

Dividing through by $(m+M)g$

$\sin\theta-\mu\cos\theta=\sin\theta-\cos\theta\frac{\sin(\theta/3)}{\cos(\theta/3)}$ [1]

Solving for the coefficient of friction $\mu$ :

$\mu=\frac{\sin(\theta/3)}{\cos(\theta/3)}=\tan(\theta/3)$

(Or any equivalent answer.) [1]

Теми и трудност

Трудност: неоценена.

Механика · Закони на Нютон · Триене

Предпоставки: Разширена олимпиадна подготовка

Оригинал в Архива: Quarterfinal 2000.pdf · официални решения: Quarterfinal 2000 Solutions.pdf