2015 年泛珠三角及中华名校物理奥林匹克邀请赛
Pan Pearl River Delta Physics Olympiad 2015
Pan Pearl River Delta Physics Olympiad 2015 2015 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
Part-1 (Total 5 Problems) 卷-1（共5题） (9:00 am – 12:00 pm, 25 February, 2015)
Numerical answers should be given to 3 significant figures.数字答案请给三位有效数字。
1. Retrograde Motion of Mars (9 points) 火星的逆行运动 （9分）
In the history of astronomy, the phenomenon of the retrograde motion played an important role. Suppose we observe the position of Mars at midnight every night for many nights. Using distant stars and constellations as the background, we will find that Mars moves from West to East most of the time. However, there are periods of time that Mars is observed to move in opposite direction, as shown in the figure. The orbital period of Mars is 1.88 y. Assume that the orbits of Earth and Mars are circular, and the tilting of Earth’s axis can be ignored.
在天文史上，行星的逆行运动扮演了重要的角色。假设我们连续多个晚上在午夜观察火星的位置。若以远处的星体和星座为背景，我们会发现大部分时间火星是从西到东运动，但也有些时段是逆向运动，如图所示。火星的轨道周期是1.88年。假设地球和火星的轨道都是圆的，地轴的倾斜可略。
Distant constellation远处的星座 East东 o
t > 0 t = 0 Star T星T t < 0 West西
20
15
o
10
o
5
o
0o
-5
o
-10
o
-15
o
-20
o
(a) What is the orbital radius RM of Mars? Give your answer in AU (Astronomical Units, 1 AU is the average distance between Sun and Earth.) (1 points)
试求火星的轨道半径RM。答案请以AU为单位。(1 AU是太阳与地球的平均距离。)（1分） ?TM??1.88??Using Kepler’s Law, 按开普勒定律，RM?RE??1???1.5233 AU?1.52 AU ?T?1???E?(b) At t = 0, Sun, Earth and Mars lie on a straight line. Sketch a figure indicating the positions of Sun, Earth, Mars, and star T when t > 0. Label them by letters S, E, M, and T respectively. Mark the angular displacements ?E and ?M of Earth and Mars respectively (starting from t =
23231
2015 年泛珠三角及中华名校物理奥林匹克邀请赛
Pan Pearl River Delta Physics Olympiad 2015
0), and the angle ? that gives the angular position of Mars as observed from Earth using distant stars and constellations as the background. (2 points)
在t = 0时，太阳、地球、火星成一直线。试作一草图，显示在t > 0时，太阳、地球、火星和星T的位置，以S、E、M和T标示。在图上标示地球和火星的角位移分别为?E和 ?M（自t = 0开始），和地球观察火星的角位置?（以远处的星体和星座为背景）。（2分） ? M E RE T RM S ?E ?M (c) Derive an expression for the angular position ? of Mars at time t. Express your answer in terms RE, RM, ?E, ?M and t, where ?E and ?M are the orbital angular velocity of Earth and Mars respectively. (4 points)
试推导火星在时间t时的角位置?。答案请以RE, RM, ?E, ?M和t表示，其中?E和?M分别为地球与火星的角速度。（3分） ? ? ?E??Et, ?M??Mt In triangle SEM, we need to find the M exterior angle at E. By constructing a E perpendicular line from M to SE, this T
angle is RE RM 在三角形SEM里，需找出角E的外角。从点M作一线垂直于SE，可见这角为 S RMsin(?E??M)??1???tan??Rcos(???)?R??. ? ?M EME??ME
RMsin(?Et??Mt)?? ??RMcos(?Et??Mt)?RE?(d) Calculate the angular position ? of Mars at t = 0.1 y, 0.2 y and 0.3 y. Give your answer in degrees. (3 points)
试计算火星在t = 0.1年, 0.2年和 0.3年时的角位置?。答案请以度数表示。（3分） ?1.5233sin[2?(0.1)?2?(0.1)/1.88]?ooAt t = 0.1, ??2?(0.1)?tan?1??1.5233cos[2?(0.1)?2?(0.1)/1.88]?1????7.963??8.00 ?????Et?tan?1????1.5233sin[2?(0.2)?2?(0.2)/1.88]?ooAt t = 0.2,??2?(0.2)?tan?1??1.5233cos[2?(0.2)?2?(0.2)/1.88]?1????0.4538??0.454 ??2
2015 年泛珠三角及中华名校物理奥林匹克邀请赛
Pan Pearl River Delta Physics Olympiad 2015
?1.5233sin[2?(0.3)?2?(0.3)/1.88]?ooAt t = 0.3, ??2?(0.3)?tan?1??1.5233cos[2?(0.3)?2?(0.3)/1.88]?1????16.43?16.4 ??
2. Rolling Ball on a Racket (10 points) 球拍滚球（10分）
As shown in the figure, a hollow spherical ball of mass M and radius R is placed on a racket of mass m. The racket has a flat surface with coefficient of static friction ?s and coefficient of kinetic friction ?k and is held horizontally.
如图所示，一个质量为M，半径为R的空心园球被放置在质量为m的球拍上。球拍具有一个平坦的表面，其静摩擦系数为?s，动摩擦系数为?k，并且被保持在水平位置。
M m F
(a) The racket is driven horizontally by a periodic force F(t)?F0cos?0t, with the ball remaining non-slipping. Calculate the maximum velocities of the oscillations of the racket and the ball, denoted as ux and uy respectively. (The moment of inertia of a hollow sphere of mass M and radius R is I = 2MR2/3.) (5 points)
球拍被周期性的力F(t)?F0cos?0t沿水平方向驱动，园球维持在不滑动的状态。试计算球拍与球振动时的最大速度，分别表示为ux和uy。（质量为M，半径为R的空心球体的转动惯量为I = 2MR2/3。） (5分) Let x and y be the displacements of the racket and the ball respectively. Let ? be the angular displacement of the ball (counted in the direction of x at the contact point with the racket). Let f be the frictional force between the racket and the ball. Applying Newton’s law, 令x和y分别为球拍与球的位移。令?为球的角位移（方向按照在与球拍的接触点沿位移x方向计算）。设f是球拍与球之间的摩擦力。运用牛顿定律， ??F?f m?x (1) ??f M?y (2) 2???fR (3) MR2?3x?yCondition for no slipping: 不滑动的条件：?? (4) R2????)?f From Eqs. (3) and (4), 从方程式（3）和（4），M(?(5) xy322? ???? and f?M?Combining with Eq. (2), 结合方程式（2）：? (6) yxx552????F0cos?0t Substituting into Eq. (1), 代入方程式（1）， ?m?M??x5??3F0cos?0t5F0cos?0t2F0cos?0tSolution: 解：x??, , , y?????222(2M?5m)?0(2M?5m)?0(2M?5m)?0R3
R
2015 年泛珠三角及中华名校物理奥林匹克邀请赛
Pan Pearl River Delta Physics Olympiad 2015
2F0sin?0t5F0sin?0t??3F0sin?0t. ??, y, ?(2M?5m)?0(2M?5m)?0(2M?5m)?0R5F02F0Hence 因此 ux?, uy?. (2M?5m)?0(2M?5m)?0(b) At the moment the racket is oscillating at its maximum velocity, its motion is brought to rest abruptly by an external force much stronger than the limiting frictional force between the racket and the ball in a very short duration of time. What is the final velocity of the ball? If the final velocity of the ball is 0, what is the displacement of the ball? (5 points)
在球拍振动至最大速度的一刻，其运动突然被外力煞停，这外力比球拍与球之间的极限摩擦力强得多，作用的时间也很短。问球的最终速度是多少？若球的最终速度为0，其位移是多少？（5分） The impulse acting on the racket is given by the external force multiplied by the time duration of the force, whereas the impulse acting on the ball is given by the limiting frictional force multiplied by the duration. Hence the impulse acting on the ball is negligible. Hence when the racket stops moving, the ball continues to move with the velocity uy and angular 3uy3F0velocity ??. Since uy??R?, the ball will slide until it finally rolls. ?(2M?5m)?0R2RApplying Newton’s law, 作用在球拍的冲量是外力乘以力作用的时间，而作用在球上的冲量是极限摩擦力乘以力作用的时间。因此，作用在球上的冲量可以忽略不计。因此，当球拍停止移动时，3uy3F0?球继续以速度uy和角速度??移动。因uy??R?，球会滑动，直(2M?5m)?0R2R到它最终滚动。运用牛顿定律， ??uy??kgt ????kMg ? yM?y??x2????fR ? R???3u?3?gt MR2?yk322???y??0 ? ? uy??kgt ? yWhen the ball stops sliding, 当球停止滑动时，R?uy1y?uyt??kgt2? 22?kgHence the final velocity of the ball is 0, and its displacement is因此球的最终速度为0，其2uy位移为y?. 2?kg
3. Balloon (10 points) 气球（10分）
2The work done in stretching a spring is converted to its spring energy. Likewise, the work done in stretching a surface of a membrane is converted to its surface energy, given by E = ?S, where ? is called the surface tension of the membrane, and S is its surface area.
拉伸弹簧所做的功被转换成弹簧的内能。同样，拉伸一个薄膜表面所做的功被转换成它的表面能E = ?S，其中?称为薄膜的表面张力，而S是其表面面积。
4
2015 年泛珠三角及中华名校物理奥林匹克邀请赛
Pan Pearl River Delta Physics Olympiad 2015
(a) Consider a balloon of radius R. What is the change in surface energy when the radius changes by dR? Hence derive an expression for the pressure due to surface tension. (2 points) 考虑半径为R的气球。当半径改变为dR时，表面能的变化是多少？由此推导表面张力形成的压力的表达式。（2分） The surface energy of the balloon is E??8?R2 (the balloon has both inner and outer surfaces).气球的表面能是E??8?R2（气球有里外两面）。 Hence dE??16?RdR. Equating this to the work done by pressure 把这等同压强做的功 ?16?RdR4?. dW?pdV?p4?R2dR, ?16?RdR?p4?R2dR ? p??4?R2dRR(b) The surface tension of balloon A is ?. When it is filled with a diatomic ideal gas, its radius becomes R0. The surface tension of balloon B is 2?. When it is filled with the same kind of ideal diatomic gas, its radius becomes R0. The temperature of the environment is T. The two balloons are then connected so that the gases are free to exchange between them until a steady state is reached. The final temperature is the same as that of the environment. What are the final radii of the two balloons respectively? You may neglect the atmospheric pressure in the analysis. (4 points)
气球A的表面张力为?。当它充满了一种双原子的理想气体，其半径是R0。气球B的表面张力为2?。当它被相同的双原子理想气体充满时，其半径是R0。环境的温度为T。然后两个气球被连接，使得气体可以在它们之间自由交流，直至达到稳定状态。最终温度与环境相同。问两个气球最终的半径分别是什么？在分析中你可以忽略大气压力。（4分）
Since the initial pressure in balloon B is higher, the gas will flow from balloon B to A. The radius of balloon B decreases and that of balloon A increases. Hence the pressure in balloon A and B increases and decreases respectively. The pressure difference increases, driving the system further away from equilibrium. This continues until all gases flow into balloon A. Hence RB = 0. 因为气球B的初始压强较高，引致气体从气球B流向A。气球B的半径减少，气球A的半径增加。因此，气球A和B的压强分别增大和减小，使系统进一步远离平衡。这情况持续，直到所有的气体流入气球A。因此，RB = 0。 To find RA, we consider the initial number of moles of gas in balloon A: 要找出RA，我们考虑起初时气球A中气体的摩尔数 3?16??R02pAVA1?4???4?R0????nA???. ????RTRT?R0??3?3RTSimilarly, the initial number of moles of gas in balloon B: 同样，起初时气球B中气体的摩尔数： 3?32??R02pBVB1?8???4?R0????nB???. ???RTRT?R0??3?3RT?Since the number of moles of gas is conserved, 因为气体的摩尔数守恒， 22216??RA16??R032??R0 ? RA?3R0. ??3RT3RT3RT5
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