Theory_1_Solution

Theory_1_Solution - 39th International Physics Olympiad...

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39th International Physics Olympiad - Hanoi - Vietnam - 2008 Theoretical Problem No. 1 /Solution 1 3 == Solution 1. The structure of the mortar 1.1. Calculating the distance TG The volume of water in the bucket is V . The length of the bottom of the bucket is . 33 1000cm 10 m 00 60 0 74 0 12 60 m 0 5322m tan ( . . tan ) . dL h =− = = (as the initial data are given with two significant digits, we shall keep only two significant digits in the final answer, but we keep more digits in the intermediate steps). The height of the water layer in the bucket is calculated from the formula: c 21 / 2 b 0 (2 3 / ) tan60 2 3 cd V d Vb c dbc c + =+ = Inserting numerical values for , and , we find b d 0.01228m c = V . When the lever lies horizontally, the distance, on the horizontal axis, between the rotation axis and the center of mass of water N, is o TH 60 0 4714m 24 tan . dc a ≈+ + = , and (see the figure below). TG ( / )TH 0.01571m mM H T N K R S P Answer: . TG 0.016m = α 1 2 1.2. Calculating the values of and . 1 When the lever tilts with angle , water level is at the edge of the bucket. At that point the water volume is . Assume 10 m PQ d < . From geometry , from which P . The assumption PQ/ 2 Vh b PQ d < Q 0.1111m = is obviously satisfied ( ). 0.5322m d = QS= PQ+ 3 tan / /( ). hh h 1 = 1 To compute the angle , we note that From this we find . o 20.6 1 =

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39th International Physics Olympiad - Hanoi - Vietnam - 2008 Theoretical Problem No. 1 /Solution 2 When the tilt angle is , the bucket is empty : . o 30 o 30 α 2 = G h T R N Q P I S β 1.3. Determining the tilt angle of the lever and the amount of water in the bucket m μ when the total torque on the lever is equal to zero (m) x = Denote PQ . The amount of water in the bucket is water 9( k g ) 2 xhb mx ρ == . =0 when the torque coming from the water in the bucket cancels out the torque coming from the weight of the lever. The cross section of the water in the bucket is the triangle PQR in the figure. The center of mass N of water is located at 2/3 of the meridian RI, therefore NTG lies on a straight line. Then: TN TG mg Mg × or TN TG 30 0.1571 0.4714 mM ×= × = (1) Calculating from x then substitute (1) : TN 2 TN ( 3 ) 0.94 0.08 3 0.8014 32 3 3 x xx La h =+− + = −= which implies (2) 2 TN 9 (0.8014 /3) 3 7.213 x x
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This note was uploaded on 11/08/2011 for the course PHYS 0000 taught by Professor Na during the Spring '11 term at Rensselaer Polytechnic Institute.

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Theory_1_Solution - 39th International Physics Olympiad...

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