ipho2014-第45届国际物理奥林匹克竞赛理论试题与解答

上传人:第*** 文档编号:60797002 上传时间:2018-11-18 格式:PDF 页数:16 大小:2.40MB
返回 下载 相关 举报
ipho2014-第45届国际物理奥林匹克竞赛理论试题与解答_第1页
第1页 / 共16页
ipho2014-第45届国际物理奥林匹克竞赛理论试题与解答_第2页
第2页 / 共16页
ipho2014-第45届国际物理奥林匹克竞赛理论试题与解答_第3页
第3页 / 共16页
ipho2014-第45届国际物理奥林匹克竞赛理论试题与解答_第4页
第4页 / 共16页
ipho2014-第45届国际物理奥林匹克竞赛理论试题与解答_第5页
第5页 / 共16页
点击查看更多>>
资源描述

《ipho2014-第45届国际物理奥林匹克竞赛理论试题与解答》由会员分享,可在线阅读,更多相关《ipho2014-第45届国际物理奥林匹克竞赛理论试题与解答(16页珍藏版)》请在金锄头文库上搜索。

1、 Theoretical competition. Tuesday, 15 July 2014 1/1 Problem 1 (9 points) This problem consists of three independent parts. Part A (3 points) A small puck of mass is carefully placed onto the inner surface of the thin hollow thin cylinder of mass and of radius . Initially, the cylinder rests on the h

2、orizontal plane and the puck is located at the height above the plane as shown in the figure on the left. Find the interaction force between the puck and the cylinder at the moment when the puck passes the lowest point of its trajectory. Assume that the friction between the puck and the inner surfac

3、e of the cylinder is absent, and the cylinder moves on the plane without slipping. The free fall acceleration is . Part B (3 points) A bubble of radius = 5.00 cm, containing a diatomic ideal gas, has the soap film of thickness = 10.0 m and is placed in vacuum. The soap film has the surface tension =

4、 4.00 102 N m and the density = 1.10 g cm3. 1) Find formula for the molar heat capacity of the gas in the bubble for such a process when the gas is heated so slowly that the bubble remains in a mechanical equilibrium and evaluate it; 2) Find formula for the frequency of the small radial oscillations

5、 of the bubble and evaluate it under the assumption that the heat capacity of the soap film is much greater than the heat capacity of the gas in the bubble. Assume that the thermal equilibrium inside the bubble is reached much faster than the period of oscillations. Hint: Laplace showed that there i

6、s pressure difference between inside and outside of a curved surface, caused by surface tension of the interface between liquid and gas, so that = 2 . Part C (3 points) Initially, a switch is unshorted in the circuit shown in the figure on the right, a capacitor of capacitance 2 carries the electric

7、 charge 0, a capacitor of capacitance is uncharged, and there are no electric currents in both coils of inductance and 2, respectively. The capacitor starts to discharge and at the moment when the current in the coils reaches its maximum value, the switch is instantly shorted. Find the maximum curre

8、nt max through the switch thereafter. Theoretical competition. Tuesday, 15 July 2014 1/4 Problem 1 Solution Part A Consider the forces acting on the puck and the cylinder and depicted in the figure on the right. The puck is subject to the gravity force and the reaction force from the cylinder . The

9、cylinder is subject to the gravity force , the reaction force from the plane 1, the friction force and the pressure force from the puck = . The idea is to write the horizontal projections of the equations of motion. It is written for the puck as follows = sin, (A.1) where is the horizontal projectio

10、n of the puck acceleration. For the cylinder the equation of motion with the acceleration is found as = sin . (A.2) Since the cylinder moves along the plane without sliding its angular acceleration is obtained as = / (A.3) Then the equation of rotational motion around the center of mass of the cylin

11、der takes the form = , (A.4) where the inertia moment of the hollow cylinder is given by = 2. (A.5) Solving (A.2)-(A.5) yields 2 = sin. (A.6) From equations (A.1) and (A.6) it is easily concluded that = 2. (A.7) Since the initial velocities of the puck and of the cylinder are both equal to zero, the

12、n, it follows from (A.7) after integrating that = 2. (A.8) It is obvious that the conservation law for the system is written as = 2 2 + 2 2 + 2 2 , (A.9) where the angular velocity of the cylinder is found to be = , (A.10) since it does not slide over the plane. Solving (A.8)-(A.10) results in veloc

13、ities at the lowest point of the puck trajectory written as = 2 (2+), (A.12) = (2+). (A.13) In the reference frame sliding progressively along with the cylinder axis, the puck moves in a circle of radius and, at the lowest point of its trajectory, have the velocity = + (A.14) and the acceleration re

14、l= rel 2 . (A.15) At the lowest point of the puck trajectory the acceleration of the cylinder axis is equal to zero, therefore, the puck acceleration in the laboratory reference frame is also given by (A.15). = 2 . (A.16) then the interaction force between the puck and the cylinder is finally found

15、as = 3 1 + 3 . (A.17) Theoretical competition. Tuesday, 15 July 2014 2/4 Part B 1) According to the first law of thermodynamics, the amount of heat transmitted to the gas in the bubble is found as = + , (B.1) where the molar heat capacity at arbitrary process is as follows = 1 = + . (B.2) Here stands for the molar heat capacity of the gas at constant volu

展开阅读全文
相关资源
正为您匹配相似的精品文档
相关搜索

最新文档


当前位置:首页 > 中学教育 > 教学课件 > 高中课件

电脑版 |金锄头文库版权所有
经营许可证:蜀ICP备13022795号 | 川公网安备 51140202000112号