3. Oscillating Masses a) See Figure 1. π ππ ( πΌ 2 )...

Transcribed Text

3. Oscillating Masses a) See Figure 1. π ππ ( πΌ 2 ) = π 3π = 1 3 β πΌ = 38.942441Β° b) i. See Figure 2. Kinetic Energy π = 1 2 π1(π₯Μ1 2 + π¦Μ1 2 ) + 1 2 π2(π¦Μ2 2 + π¦Μ2 2 ) Potential Energy π = βπ1ππ¦1 β π2ππ¦2 Coordinate transformation to generalized coordinate π : π₯2 = 3ππππ π , π¦2 = β3ππ πππ ; π₯1 = β3ππππ (π + πΌ) , π¦1 = β3ππ ππ(π + πΌ) β Generalized velocities: π₯Μ2 = β3ππΜπ πππ , π¦Μ2 = β3ππΜπππ π π₯Μ1 = 3ππΜπ ππ(π + πΌ) , π¦Μ1 = β3ππΜπππ (π + πΌ) β π = 1 2 π2(9π 2πΜ2 π ππ2π + 9π 2πΜ2 πππ 2π) + 1 2 π1[9π 2πΜ2 π ππ2 (π + πΌ) + 9π 2πΜ2 πππ 2 (π + πΌ)] β π = 9 2 π2π 2πΜ2 + 9 2 π1π 2πΜ2 β π = 9 2 π 2πΜ2 (π2 + π1 ) , π1 = π , π2 = π½π β β π = 9 2 (1 + π½)ππ 2πΜ2 π = βπ1π3ππ ππ(π + πΌ) β π2π3ππ πππ β π = β3ππππ ππ(π + πΌ) β 3π½πππ πππ β π = β3πππ[π ππ(π + πΌ) + π½π πππ] Lagrangian πΏ = π β π β πΏ = 9 2 (1 + π½)ππ 2πΜ2 + 3πππ[π ππ(π + πΌ) + π½π πππ] β πΏ = 3ππ { 3 2 (1 + π½)ππΜ2 + π[π ππ(π + πΌ) + π½π πππ]} β ii. Euler-Lagrange equation: ππΏ ππ β π ππ‘ ( ππΏ ππΜ ) = 0 β π[πππ (π + πΌ) + π½πππ π] β π ππ‘ [3(1 + π½)ππΜ] = 0 β π[πππ (π + πΌ) + π½πππ π] β 3(1 + π½)ππΜ= 0 iii. Hamiltonian π» = πΜ ππΏ ππΜ β πΏ β π» = 9ππ 2 (1 + π½)πΜ2 β 3ππ { 3 2 (1 + π½)ππΜ2 + π[π ππ(π + πΌ) + π½π πππ]} 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 π» = 9 2 (1 + π½)ππ 2πΜ2 β 3πππ[π ππ(π + πΌ) + π½π πππ] To express H in term of ππ : ππ = ππΏ ππΜ β ππ = π ππΜ β©3ππ { 3 2 (1 + π½)ππΜ2 + π[π ππ(π + πΌ) + π½π πππ]}βͺ β ππ = 9(1 + π½)ππ 2πΜ πΜ = ππ 9(1+π½)ππ2 β π» = 9 2 (1+π½)ππ 2ππ 2 {9(1+π½)ππ2}2 β 3πππ[π ππ(π + πΌ) + π½π πππ] β π» = 1 2 ππ 2 9(1 + π½)ππ 2 β 3πππ[π ππ(π + πΌ) + π½π πππ] Hamilton equations: πΜ = ππ» πππ and πΜπ = β ππ» ππ β πΜ = π πππ { 1 2 ππ 2 9(1+π½)ππ2 β 3πππ[π ππ(π + πΌ) + π½π πππ]} : expression no simplified β πΜ = ππ 9(1+π½)ππ2 : simplified and πΜπ = β π ππ { 1 2 ππ 2 9(1+π½)ππ2 β 3πππ[π ππ(π + πΌ) + π½π πππ]} : expression no simplified β πΜπ = 3πππ[πππ (π + πΌ) + π½πππ π] : : 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 ββ π = 0 , in this case ππ ππ = 0 , and 2) Second derivative of U must be positive, here π 2π ππ2 > 0 Then 1) π ππ {β3πππ[π ππ(π + πΌ) + π½π πππ]} = 0 β πππ (π + πΌ) + π½πππ π = 0 , and 2) π 2 ππ2 {β3πππ[π ππ(π + πΌ) + π½π πππ]} > 0 β [π ππ(π + πΌ) + π½π πππ] > 0 We would have to solve these two equations, to get the angle π 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₯πΆπ = π23ππ πππΌ 2 +π1(β3ππ πππΌ 2 ) π1+π2 β π₯πΆπ = 3ππ ππ πΌ 2 π2βπ1 π1+π2 , π1 = π , π2 = π½π β π₯πΆπ = 3 π½β1 π½+1 π ππ πΌ 2 π The line that passes through the CM is deviated an angle π from the y-axis Now, π‘πππ = π₯πΆπ 3ππππ πΌ 2 β π‘πππ = 3 π½β1 π½+1 π πππΌ 2 π 3ππππ πΌ 2 β π‘πππ = π½β1 π½+1 π‘ππ πΌ 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 , 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 for stable equilibrium is π = π 2 β ( πΌ 2 β π) β π‘πππ = 1 π‘ππ( πΌ 2 βπ) β π‘πππ = 1+ π‘ππ( πΌ 2 )π‘πππ π‘ππ( πΌ 2 )βπ‘πππ+ , π‘πππ = π½β1 π½+1 π‘ππ πΌ 2 β π‘πππ = 1+ π‘ππ( πΌ 2 ) π½β1 π½+1 π‘πππΌ 2 π‘ππ( πΌ 2 )β π½β1 π½+1 π‘πππΌ 2 β π‘πππ = 1+ π½β1 π½+1 π‘ππ2( πΌ 2 ) π‘ππ( πΌ 2 )(1β π½β1 π½+1 ) β π‘πππ = 1+ π½β1 π½+1 π‘ππ2( πΌ 2 ) 2 π½+1 π‘ππ( πΌ 2 ) β π‘πππ = (π½+1)+(π½β1) π‘ππ2( πΌ 2 ) 2π‘ππ( πΌ 2 ) β π‘πππ = π½[ 1+π‘ππ2( πΌ 2 )]+[ 1βπ‘ππ2( πΌ 2 )] 2π‘ππ( πΌ 2 ) β π‘πππ = π½ π ππ2( πΌ 2 )+[ 2βπ ππ2( πΌ 2 )] 2π‘ππ( πΌ 2 ) β π‘πππ = (π½β1) π ππ2( πΌ 2 )+2 2π‘ππ( πΌ 2 ) β π‘πππ = (π½β1)+2 πππ 2( πΌ 2 ) 2π ππ( πΌ 2 )πππ ( πΌ 2 ) β π‘πππ = (π½β1)+2 πππ 2( πΌ 2 ) π πππΌ , πΌ = 38.942441Β° β π‘πππ = (π½β1)+2 πππ 2( πΌ 2 ) π πππΌ β π‘πππ = π½+0.889573 0.628539 β π = π‘ππβ1 ( π½+0.889573 0.628539 ) : angle for stable equilibrium of the system. To test the accuracy of this result, let us consider the case when π½ = 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π π‘ππβππ = π 2 β πΌ 2 β ππ π‘ππβππ = 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ππ = (π½β1)+2 πππ 2( πΌ 2 ) π πππΌ , we find: π‘πππ = (1β1)+2 πππ 2( 38.942441 2 ) π ππ(38.942441) β π = 70.528780 β which is the value obtained before. We use this formula, rather than π‘πππ = π½+0.889573 0.628539 , because there are less approxim

Solution Preview

These solutions may offer step-by-step problem-solving explanations or good writing examples that include modern styles of formatting and construction of bibliographies out of text citations and references. Students may use these solutions for personal skill-building and practice. Unethical use is strictly forbidden.

By purchasing this solution you'll be able to access the following files:
Solution.pdf.

\$8.00
for this solution

PayPal, G Pay, ApplePay, Amazon Pay, and all major credit cards accepted.

Find A Tutor

View available Classical Mechanics Tutors

Get College Homework Help.

Are you sure you don't want to upload any files?

Fast tutor response requires as much info as possible.