Solution to problem 20.6.12 from the collection of Kepe O.E.

In a conservative system, kinetic energy T and potential energy P are related by the equation P = T - Pot, where Pot is potential energy. For a given system, kinetic energy T is equal to x1^2 + x2^2 + 2x1x2, and the potential energy P is equal to 0.5*x1^2 + x2.

The differential equation of motion of the system for the generalized coordinate x2 has the form d/dt(dT/dx2) - dП/dx2 = 0. Substituting the values ​​of T and П, we obtain d/dt(2x2 + 2x1) - 1 = 0.

The derivative d/dt(x2) is equal to the acceleration x2. Substituting x1 = 0.25 and solving the equation, we get the acceleration x2 at the time when x1 = 0.25 m. Answer: -0.25.

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This digital product is a solution to problem 20.6.12 from the collection of problems by Kepe O.?. according to conservative systems. The problem requires determining the acceleration of the generalized coordinate x2 at the moment of time when x1 = 0.25 m.

The solution to the problem was carried out in accordance with high standards of quality and accuracy. It uses the equation of connection between the kinetic T and potential P energies of a conservative system, as well as the differential equation of motion of the system for the generalized coordinate x2.

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Solution to problem 20.6.12 from the collection of Kepe O.?. consists in determining the acceleration of the generalized coordinate x2 of the conservative system at the moment of time when the generalized coordinate x1 is equal to 0.25 m.

To solve the problem, it is necessary to use the differential equation of motion of the system corresponding to the generalized coordinate x2, and the formula for calculating the acceleration of the second generalized coordinate:

Lagrange method: d/dt(dL/dq') - dL/dq = Q

where L is the Lagrangian of the system, q are generalized coordinates, Q are generalized forces.

In this problem, potential energy P is a function only of coordinate x1, and kinetic energy T is a function of coordinates x1 and x2. Therefore, the Lagrangian of the system can be written as follows:

L = T - П = x1^2 + x2^2 + 2x1x2 - 0.5*x1^2 - x2

Differentiating the Lagrangian with respect to the velocity of the generalized coordinate x2, we obtain:

dL/dx2' = 2x1 + 2x2

Differentiating the Lagrangian with respect to the generalized coordinate x2, we obtain the equation of motion:

d/dt(dL/dx2') - dL/dx2 = 0

d/dt(2x1 + 2x2) - (-1) = 0

2dx2/dt + 2dx1/dt = 1

dx2/dt = (1 - 2*dx1/dt)/2

At the moment of time when the generalized coordinate x1 is equal to 0.25 m, it is necessary to substitute x1 = 0.25 into the resulting expression for acceleration and calculate the value:

dx2/dt = (1 - 2*0)/2 = 0.5 м/c^2

Answer: the acceleration of the generalized coordinate x2 at the moment of time when the generalized coordinate x1 is equal to 0.25 m is equal to -0.25 m/s^2.

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