Pan Pearl River Delta Physics Olympiad 2017 2017 年泛珠三角及中华名校物理奥林匹克邀请赛 Sponsored by Institute for Advanced Study, HKUST
香港科技大学高等研究院赞助
Simplified Chinese Part-1 (Total 7 Problems, 45 Points) 简体版卷-1（共7题，45分）
(9:00 am – 12:00 pm, 3 February, 2017)
1. No-Shadow Day (5 points) 立竿无影（5分）
(a) In the figure, ABCD is a rectangle lying on an inclined plane making an angle ? with the horizontal plane. ABEF is the projection of the rectangle on the horizontal plane. If the measure of the angle DAC is ?, derive an expression for the angle ?. [1]
如图所示，矩形ABCD位于斜面上，斜面与水平面夹角为?。ABEF为该矩形于水平面的投影。设角DAC为?，试推导角?的表达式。[1]
D
C
? F
? A
B
?
E
Let h = AC. Then CE=?sin??.
BC=AD=?cos?.
CE=BCsin?? =?cos??sin??.
Equating the expressions of CE, ?sin??=?cos??sin?? ? ??=arcsin(cos??sin??).
(b) The ecliptic is the plane on which the Earth revolves around the Sun. The axis of rotation of the Earth is inclined at an angle of 23.4o with the normal to the ecliptic. The day of the Summer Solstice (in the Northern Hemisphere) is 21 June. The latitude of Hong Kong is 22.25o, and the no-shadow days are those days on which the Sun does not cast a shadow of a vertical pole at noon in Hong Kong. Using the result of (a) or otherwise, derive the angular displacement of the Earth’s revolution between the Summer Solstice and the no-shadow days in Hong Kong. Give your answer to 3 significant figures. [2]
黄道面是指地球围绕太阳公转的平面。地球的自转轴相对于黄道面的法线倾斜，角度为23.4o。在北半球，夏至的日期为6月21日。香港位于北纬22.25o，而当某日正午的太阳照在一支立于香港的垂直竿子时是没有影子的，那日就是香港的无影日了。试用(a)部结果或其他方法，推导在夏至和香港的无影日之间，地球公转的角位移。答案请给三位有效数字。[2]
In the figure above, consider ABEF to be the equatorial plane of the Earth, and ABCD the ecliptic. Then ? = 23.4o. When the Earth revolves around the Sun, sunlight is incident on the Earth from different directions lying on the plane ABCD. For example, on 21 June, sunlight is incident on the Earth in the direction DA, since this is the northernmost direction of sunlight.
1
Similarly, during Spring Equinox and Autumn Equinox, sunlight is incident on the Earth in the direction AB or BA. Axis of rotation 自转轴 ? Equatorial Plane赤道面 ? Summer Solstice 夏至 0No Shadow Day in Hong Kong香港无影日 Identifying ? = 23.4 and when the Sun does not cast a shadow of the vertical pole at noon in Hong Kong, ? = 22.25o. Hence the angle ? is given by cos??=sin??=sin23.40=0.9534 ? ??=17.560 (c) Write the dates of the no-shadow days in Hong Kong. [2] 试写下香港无影日的日期。[2]
The number of days for the Earth to revolve around the Sun through this angle
17.56
=365(360)=17.8
Hence the days with no shadow in Hong Kong are 18 days before and after the Summer Solstice, that is, 3 June and 9 July.
2. Six Missiles (5 points) 六枚飞弹 （5分）
Six missiles are initially located at the six vertex of a regular hexagon with side length a. The speed of the missiles in the plane is v. Each missile is equipped with an automatic navigation system. The automatic navigation system of each missile guides itself to aim at the current position of its counterclockwise neighbor.
今有飞弹六枚, 分别位于一边长为a 的正六边形的六个角上。 每枚飞弹都装置有自动导航系统。 该系统会指示飞弹永远以速率 v 飞向其逆时针方向之近邻。
(a) Find the radial component of the missile velocities relative to the center of the hexagon. [2] 找出飞弹指向六角形中心的径向速率。[2]
By symmetry, all the six missiles hit at the same time. By symmetry, they must hit at the center of the hexagon.
By symmetry, the missiles are always at the vertex of a rotating hexagon. The radial speed is v cos(?/3) = v/2.
(b) Find the time taken for a missile to hit another. [3] 找出一枚飞弹击中另一枚所需的时间。[3] The time taken is a/(v/2) = 2a/v.
2
sin??sin22.250
3. Falling ladder (10 points) 下跌中的梯子（10分）
A ladder of length 2?? and mass m is standing up against a vertical wall with initial angle ?? relative to the horizontal. There is no friction between the ladder and the wall or the floor. The ladder begins to slide down with zero initial velocity. Denote ??(??) as the angle the ladder makes with the horizontal after it starts to slide and (??(??),??(??)) be the coordinate of the center of mass of the ladder. In this problem, you should take the gravitational potential energy to be zero at y = 0.
一个长度为2??、质量为m 的梯子靠着一道垂直的墙，并与水平形成初始夹角α。梯子与墙身和地面并没有摩擦力。梯子从零初始速度开始下滑。??(??)表示为梯子下滑期间与水平形成的夹角, (??(??),??(??)) 表示为梯子质心的坐标。在本题中，你应将在y = 0处的引力势能取值为零。
y CM = (x(t), y(t) ?(t)
(a) What is the initial total mechanical energy of the ladder in terms of ??? [1] 梯子的初始总机械能是甚么？答案以??表示。[1] Ei?mgy?mglsin?
(b) Write the potential energy of the ladder in terms of ??(??) when it is sliding. [1] 请用??(??)写下梯子下滑时的势能。[1] U?mgy?mglsin?
(c) Write the total kinetic energy of the ladder in terms of ???(??),???(??),??(??) when it is sliding. (Hint: The moment of inertia of a rod of length 2?? and mass m about an axis through the center of mass and perpendicular to its length is ??=????2/3.) [1]
请用???(??),???(??),??(??)写下梯子下滑时的总动能。 (提示: 一条长度为2??、质量为m 的杆子，相对於通过杆子质心并垂直于杆子的转动轴，其??动惯量为??=????2/3。) [1] 11?211?2 ?2?y?2)?I??2?y?2)?ml2?T?m(x?m(x2226(d) As long as the ladder is in contact with the wall, find the relation between ??(??) and ??(??), and similarly the relation between ??(??) and ??(??). [2]
当梯子靠着墙的时候，请找出??(??)和??(??)的关系式，而同样地，找出??(??)和??(??)的关系式。[2]
x?lcos? and y?lsin?
3
x
2017年泛珠三角与中华名校物理奥林匹克邀请赛试题与答案test1_solution
