3. Oscillating Masses a) See Figure 1. 𝑠𝑖𝑛 ( 𝛼 2 )...

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3. Oscillating Masses a) See Figure 1. 𝑠𝑖𝑛 ( 𝛼 2 ) = 𝑏 3𝑏 = 1 3 β†’ 𝛼 = 38.942441Β° b) i. See Figure 2. Kinetic Energy 𝑇 = 1 2 𝑀1(π‘₯Μ‡1 2 + 𝑦̇1 2 ) + 1 2 𝑀2(𝑦̇2 2 + 𝑦̇2 2 ) Potential Energy π‘ˆ = βˆ’π‘€1𝑔𝑦1 βˆ’ 𝑀2𝑔𝑦2 Coordinate transformation to generalized coordinate πœƒ : π‘₯2 = 3π‘π‘π‘œπ‘ πœƒ , 𝑦2 = βˆ’3π‘π‘ π‘–π‘›πœƒ ; π‘₯1 = βˆ’3π‘π‘π‘œπ‘ (πœƒ + 𝛼) , 𝑦1 = βˆ’3𝑏𝑠𝑖𝑛(πœƒ + 𝛼) β†’ Generalized velocities: π‘₯Μ‡2 = βˆ’3π‘πœƒΜ‡π‘ π‘–π‘›πœƒ , 𝑦̇2 = βˆ’3π‘πœƒΜ‡π‘π‘œπ‘ πœƒ π‘₯Μ‡1 = 3π‘πœƒΜ‡π‘ π‘–π‘›(πœƒ + 𝛼) , 𝑦̇1 = βˆ’3π‘πœƒΜ‡π‘π‘œπ‘ (πœƒ + 𝛼) β†’ 𝑇 = 1 2 𝑀2(9𝑏 2πœƒΜ‡2 𝑠𝑖𝑛2πœƒ + 9𝑏 2πœƒΜ‡2 π‘π‘œπ‘ 2πœƒ) + 1 2 𝑀1[9𝑏 2πœƒΜ‡2 𝑠𝑖𝑛2 (πœƒ + 𝛼) + 9𝑏 2πœƒΜ‡2 π‘π‘œπ‘ 2 (πœƒ + 𝛼)] β†’ 𝑇 = 9 2 𝑀2𝑏 2πœƒΜ‡2 + 9 2 𝑀1𝑏 2πœƒΜ‡2 β†’ 𝑇 = 9 2 𝑏 2πœƒΜ‡2 (𝑀2 + 𝑀1 ) , 𝑀1 = π‘š , 𝑀2 = π›½π‘š β†’ β†’ 𝑇 = 9 2 (1 + 𝛽)π‘šπ‘ 2πœƒΜ‡2 π‘ˆ = βˆ’π‘€1𝑔3𝑏𝑠𝑖𝑛(πœƒ + 𝛼) βˆ’ 𝑀2𝑔3π‘π‘ π‘–π‘›πœƒ β†’ π‘ˆ = βˆ’3π‘šπ‘”π‘π‘ π‘–π‘›(πœƒ + 𝛼) βˆ’ 3π›½π‘šπ‘”π‘ π‘–π‘›πœƒ β†’ π‘ˆ = βˆ’3π‘šπ‘”π‘[𝑠𝑖𝑛(πœƒ + 𝛼) + π›½π‘ π‘–π‘›πœƒ] Lagrangian 𝐿 = 𝑇 βˆ’ π‘ˆ β†’ 𝐿 = 9 2 (1 + 𝛽)π‘šπ‘ 2πœƒΜ‡2 + 3π‘šπ‘”π‘[𝑠𝑖𝑛(πœƒ + 𝛼) + π›½π‘ π‘–π‘›πœƒ] β†’ 𝐿 = 3π‘šπ‘ { 3 2 (1 + 𝛽)π‘πœƒΜ‡2 + 𝑔[𝑠𝑖𝑛(πœƒ + 𝛼) + π›½π‘ π‘–π‘›πœƒ]} β†’ ii. Euler-Lagrange equation: πœ•πΏ πœ•πœƒ βˆ’ 𝑑 𝑑𝑑 ( πœ•πΏ πœ•πœƒΜ‡ ) = 0 β†’ 𝑔[π‘π‘œπ‘ (πœƒ + 𝛼) + π›½π‘π‘œπ‘ πœƒ] βˆ’ 𝑑 𝑑𝑑 [3(1 + 𝛽)π‘πœƒΜ‡] = 0 β†’ 𝑔[π‘π‘œπ‘ (πœƒ + 𝛼) + π›½π‘π‘œπ‘ πœƒ] βˆ’ 3(1 + 𝛽)π‘πœƒΜˆ= 0 iii. Hamiltonian 𝐻 = πœƒΜ‡ πœ•πΏ πœ•πœƒΜ‡ βˆ’ 𝐿 β†’ 𝐻 = 9π‘šπ‘ 2 (1 + 𝛽)πœƒΜ‡2 βˆ’ 3π‘šπ‘ { 3 2 (1 + 𝛽)π‘πœƒΜ‡2 + 𝑔[𝑠𝑖𝑛(πœƒ + 𝛼) + π›½π‘ π‘–π‘›πœƒ]} β†’ ΞΈ Ξ± M M2 1 Figure 2. System in motion x 2 x1 y2 y1 ΞΈ+Ξ± M M2 1 3b Figure 1. Value of Ξ± Ξ±/2 b PΓ‘gina 2 de 4 𝐻 = 9 2 (1 + 𝛽)π‘šπ‘ 2πœƒΜ‡2 βˆ’ 3π‘šπ‘π‘”[𝑠𝑖𝑛(πœƒ + 𝛼) + π›½π‘ π‘–π‘›πœƒ] To express H in term of π‘πœƒ : π‘πœƒ = πœ•πΏ πœ•πœƒΜ‡ β†’ π‘πœƒ = πœ• πœ•πœƒΜ‡ 〈3π‘šπ‘ { 3 2 (1 + 𝛽)π‘πœƒΜ‡2 + 𝑔[𝑠𝑖𝑛(πœƒ + 𝛼) + π›½π‘ π‘–π‘›πœƒ]}βŒͺ β†’ π‘πœƒ = 9(1 + 𝛽)π‘šπ‘ 2πœƒΜ‡ πœƒΜ‡ = π‘πœƒ 9(1+𝛽)π‘šπ‘2 β†’ 𝐻 = 9 2 (1+𝛽)π‘šπ‘ 2π‘πœƒ 2 {9(1+𝛽)π‘šπ‘2}2 βˆ’ 3π‘šπ‘π‘”[𝑠𝑖𝑛(πœƒ + 𝛼) + π›½π‘ π‘–π‘›πœƒ] β†’ 𝐻 = 1 2 π‘πœƒ 2 9(1 + 𝛽)π‘šπ‘ 2 βˆ’ 3π‘šπ‘π‘”[𝑠𝑖𝑛(πœƒ + 𝛼) + π›½π‘ π‘–π‘›πœƒ] Hamilton equations: πœƒΜ‡ = πœ•π» πœ•π‘πœƒ and π‘Μ‡πœƒ = βˆ’ πœ•π» πœ•πœƒ β†’ πœƒΜ‡ = πœ• πœ•π‘πœƒ { 1 2 π‘πœƒ 2 9(1+𝛽)π‘šπ‘2 βˆ’ 3π‘šπ‘π‘”[𝑠𝑖𝑛(πœƒ + 𝛼) + π›½π‘ π‘–π‘›πœƒ]} : expression no simplified β†’ πœƒΜ‡ = π‘πœƒ 9(1+𝛽)π‘šπ‘2 : simplified and π‘Μ‡πœƒ = βˆ’ πœ• πœ•πœƒ { 1 2 π‘πœƒ 2 9(1+𝛽)π‘šπ‘2 βˆ’ 3π‘šπ‘π‘”[𝑠𝑖𝑛(πœƒ + 𝛼) + π›½π‘ π‘–π‘›πœƒ]} : expression no simplified β†’ π‘Μ‡πœƒ = 3π‘šπ‘π‘”[π‘π‘œπ‘ (πœƒ + 𝛼) + π›½π‘π‘œπ‘ πœƒ] : : simplified c) The conditions for stable equilibrium, at the point where it is reached, are two: 1) The Potential Energy function U must be a minimum β†’ βˆ‡βƒ— π‘ˆ = 0 , in this case πœ•π‘ˆ πœ•πœƒ = 0 , and 2) Second derivative of U must be positive, here πœ• 2π‘ˆ πœ•πœƒ2 > 0 Then 1) πœ• πœ•πœƒ {βˆ’3π‘šπ‘”π‘[𝑠𝑖𝑛(πœƒ + 𝛼) + π›½π‘ π‘–π‘›πœƒ]} = 0 β†’ π‘π‘œπ‘ (πœƒ + 𝛼) + π›½π‘π‘œπ‘ πœƒ = 0 , and 2) πœ• 2 πœ•πœƒ2 {βˆ’3π‘šπ‘”π‘[𝑠𝑖𝑛(πœƒ + 𝛼) + π›½π‘ π‘–π‘›πœƒ]} > 0 β†’ [𝑠𝑖𝑛(πœƒ + 𝛼) + π›½π‘ π‘–π‘›πœƒ] > 0 We would have to solve these two equations, to get the angle πœƒ for stable equilibrium. Instead, we will use another approach: The location of the Center of Mass (CM) of the system is shown in Figure 3. PΓ‘gina 3 de 4 π‘₯𝐢𝑀 = 𝑀23𝑏𝑠𝑖𝑛𝛼 2 +𝑀1(βˆ’3𝑏𝑠𝑖𝑛𝛼 2 ) 𝑀1+𝑀2 β†’ π‘₯𝐢𝑀 = 3𝑏𝑠𝑖𝑛 𝛼 2 𝑀2βˆ’π‘€1 𝑀1+𝑀2 , 𝑀1 = π‘š , 𝑀2 = π›½π‘š β†’ π‘₯𝐢𝑀 = 3 π›½βˆ’1 𝛽+1 𝑠𝑖𝑛 𝛼 2 𝑏 The line that passes through the CM is deviated an angle πœ‘ from the y-axis Now, π‘‘π‘Žπ‘›πœ‘ = π‘₯𝐢𝑀 3π‘π‘π‘œπ‘ π›Ό 2 β†’ π‘‘π‘Žπ‘›πœ‘ = 3 π›½βˆ’1 𝛽+1 𝑠𝑖𝑛𝛼 2 𝑏 3π‘π‘π‘œπ‘ π›Ό 2 β†’ π‘‘π‘Žπ‘›πœ‘ = π›½βˆ’1 𝛽+1 π‘‘π‘Žπ‘› 𝛼 2 A mass system will be in stable equilibrium if its Center of Mass is under the supporting point. Looking at Figure 4, we see that it is the case. To find the exact value of πœƒ , let us remember that a mass system under rest, will align vertically its CM on the line passing by the hanging point. In the Figure 4, this situation is shown. There we see that the angle πœƒ for stable equilibrium is πœƒ = πœ‹ 2 βˆ’ ( 𝛼 2 βˆ’ πœ‘) β†’ π‘‘π‘Žπ‘›πœƒ = 1 π‘‘π‘Žπ‘›( 𝛼 2 βˆ’πœ‘) β†’ π‘‘π‘Žπ‘›πœƒ = 1+ π‘‘π‘Žπ‘›( 𝛼 2 )π‘‘π‘Žπ‘›πœ‘ π‘‘π‘Žπ‘›( 𝛼 2 )βˆ’π‘‘π‘Žπ‘›πœ‘+ , π‘‘π‘Žπ‘›πœ‘ = π›½βˆ’1 𝛽+1 π‘‘π‘Žπ‘› 𝛼 2 β†’ π‘‘π‘Žπ‘›πœƒ = 1+ π‘‘π‘Žπ‘›( 𝛼 2 ) π›½βˆ’1 𝛽+1 π‘‘π‘Žπ‘›π›Ό 2 π‘‘π‘Žπ‘›( 𝛼 2 )βˆ’ π›½βˆ’1 𝛽+1 π‘‘π‘Žπ‘›π›Ό 2 β†’ π‘‘π‘Žπ‘›πœƒ = 1+ π›½βˆ’1 𝛽+1 π‘‘π‘Žπ‘›2( 𝛼 2 ) π‘‘π‘Žπ‘›( 𝛼 2 )(1βˆ’ π›½βˆ’1 𝛽+1 ) β†’ π‘‘π‘Žπ‘›πœƒ = 1+ π›½βˆ’1 𝛽+1 π‘‘π‘Žπ‘›2( 𝛼 2 ) 2 𝛽+1 π‘‘π‘Žπ‘›( 𝛼 2 ) β†’ π‘‘π‘Žπ‘›πœƒ = (𝛽+1)+(π›½βˆ’1) π‘‘π‘Žπ‘›2( 𝛼 2 ) 2π‘‘π‘Žπ‘›( 𝛼 2 ) β†’ π‘‘π‘Žπ‘›πœƒ = 𝛽[ 1+π‘‘π‘Žπ‘›2( 𝛼 2 )]+[ 1βˆ’π‘‘π‘Žπ‘›2( 𝛼 2 )] 2π‘‘π‘Žπ‘›( 𝛼 2 ) β†’ π‘‘π‘Žπ‘›πœƒ = 𝛽 𝑠𝑒𝑐2( 𝛼 2 )+[ 2βˆ’π‘ π‘’π‘2( 𝛼 2 )] 2π‘‘π‘Žπ‘›( 𝛼 2 ) β†’ π‘‘π‘Žπ‘›πœƒ = (π›½βˆ’1) 𝑠𝑒𝑐2( 𝛼 2 )+2 2π‘‘π‘Žπ‘›( 𝛼 2 ) β†’ π‘‘π‘Žπ‘›πœƒ = (π›½βˆ’1)+2 π‘π‘œπ‘ 2( 𝛼 2 ) 2𝑠𝑖𝑛( 𝛼 2 )π‘π‘œπ‘ ( 𝛼 2 ) β†’ π‘‘π‘Žπ‘›πœƒ = (π›½βˆ’1)+2 π‘π‘œπ‘ 2( 𝛼 2 ) 𝑠𝑖𝑛𝛼 , 𝛼 = 38.942441Β° β†’ π‘‘π‘Žπ‘›πœƒ = (π›½βˆ’1)+2 π‘π‘œπ‘ 2( 𝛼 2 ) 𝑠𝑖𝑛𝛼 β†’ π‘‘π‘Žπ‘›πœƒ = 𝛽+0.889573 0.628539 β†’ πœƒ = π‘‘π‘Žπ‘›βˆ’1 ( 𝛽+0.889573 0.628539 ) : angle for stable equilibrium of the system. To test the accuracy of this result, let us consider the case when 𝛽 = 1 : two equal masses. In such a case, the stable equilibrium position is achieved, when the valued side of the triangle with magnitude 2b is parallel to x-axis, it is πœƒπ‘ π‘‘π‘Žπ‘βˆ’π‘’π‘ž = πœ‹ 2 βˆ’ 𝛼 2 β†’ πœƒπ‘ π‘‘π‘Žπ‘βˆ’π‘’π‘ž = 70.528780Β° x y Figure 4. Location of the CM at stable equilibrium Ξ±/2 M2 M1 CM Ο• ΞΈ Ξ±/2 M1 M2 x y 3b 2b CM Ο• xCM ΞΈ Figure 3. Location of the CM PΓ‘gina 4 de 4 Using the formula above: π‘‘π‘Žπ‘›πœƒ = (π›½βˆ’1)+2 π‘π‘œπ‘ 2( 𝛼 2 ) 𝑠𝑖𝑛𝛼 , we find: π‘‘π‘Žπ‘›πœƒ = (1βˆ’1)+2 π‘π‘œπ‘ 2( 38.942441 2 ) 𝑠𝑖𝑛(38.942441) β†’ πœƒ = 70.528780 β†’ which is the value obtained before. We use this formula, rather than π‘‘π‘Žπ‘›πœƒ = 𝛽+0.889573 0.628539 , because there are less approxim

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