Съдържание

Условие Отговори Решение

USAPhO 2000, Четвъртфинал — Задача 4

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

A pendulum consists of a small metal sphere of mass $M$ attached to 0.60-m massless string. It hangs from a support that is located 0.30 m to the right of and 0.40 m above a fixed small rod oriented perpendicular to the plane of swing. Assume that there is no friction between the rod and the string and ignore air resistance. The pendulum is pulled to the right and released from rest when the bob is located 0.15 m higher than the rod (See Fig. 1). As the pendulum swings to the left, the string contacts the rod and the sphere begins to loop around it (See Fig. 2). Ignore the diameter of the rod and take $y$ to be the height of the sphere above the rod. Express your answers to the following in terms of $y,M$ , and $g$ .

Fig.1: Initial pendulum position with support and rod offsets and bob height.

Fig.2: String contacting rod and bob of mass M at height y.

a. Find an expression for the tension $T$ in the string when the string is in contact with the rod. [15 т.]

b. Find the x- and y-components of the force that the rod exerts on the string. [10 т.]

Отговори

Решение

Pendulum contacting rod: tension forces, weight, radius0.10m and geometry.

a. Applying energy conservation between the initial point $h_0=0.15\ m$ and the point $y$

$K_0+U_0=K_y+U_y$ [2]

Initially $K_0=0$ , $U_0=Mgh_0$ [1+1]

At point $y$ $K_y=(\frac12)Mv^2$ , $U_y=Mgy$ [1+1]

$Mgh_0=(\frac12)Mv^2+Mgy$ (4.1) [1]

The sphere is moving in a circle of radius

$R=0.60\ m-\sqrt{(0.40\ m)^2+(0.30\ m)^2}=0.10\ m$ . [1]

The total centripetal force has two contributions, the tension in the string $T$ and the radial component of the sphere’s weight, $Mg\cos\theta$ . Setting the total centripetal force equal to the mass times the centripetal acceleration

$T+Mg\cos\theta=M\frac{v^2}{R}$ (4.2) [4]

Solving (4.1) for $Mv^2$ and substituting into (4.2)

$T+Mg\cos\theta=\frac{2Mg(h_0-y)}{R}$ , (4.3) [1]

From the geometry $\cos\theta=\frac yR$ . [1]

Combining with (4.3) and solving for $T$

T=\frac{2Mg(h_0-y)}R-Mg\frac yR=\frac{Mg(2h_0-3y)}R=\frac{Mg(0.30m-3y)}{0.10m}

$T=3Mg(1-10y)$ where y is in meters. [1]

b. By Newton’s third law, $F$ , the force the rod exerts on the string, is equal and opposite to the force the string exerts on the rod. If there is no friction between the rod and the string, then the tension in both string segments is the same $T$ .

$F_x=-\sum T_{sx}=-(T\sin36.9^\circ-T\sin\theta)$ [2]

and $F_y=-\sum T_{sy}=-(T\cos36.9^\circ+T\cos\theta)$ . [2]

Since $\cos\theta=\frac yR$ , $\sin\theta=\sqrt{1-(\frac yR)^2}$ [1+1]

$F_x=-T\left(0.60-\sqrt{1-(\frac yR)^2}\right)$ [1]

$F_x=-3Mg(1-10y)\left[0.60-\sqrt{1-(10y)^2}\right]$ y in meters. [1]

$F_y=-T\left(0.80+\frac yR\right)$ [1]

$F_y=-3Mg(1-10y)(0.80+10y)$ y in meters. [1]

Теми и трудност

Трудност: неоценена.

Механика · Махала · Движение по окръжност · Работа и енергия

Предпоставки: Разширена олимпиадна подготовка

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