Solution to problem 20.2.3 from the collection of Kepe O.E.

This problem considers the rotational motion of a body around the Oz axis. A body under the influence of a force F = 30? + 25n + 40b rotates around point A, while the distance from point O to point A is equal to 0.2 m. The moment of resistance forces of the bearings is equal to Mtr = 0.8 N • m. It is necessary to determine the generalized force that corresponds to the angle? body rotation. The answer to the problem is 5.2.

To solve the problem, we use the equation of the dynamics of the rotational motion of a body:

ΣM = Iα,

where ΣM is the sum of moments of forces, I is the moment of inertia of the body, α is the angular acceleration of the body.

Let us express the angular acceleration in terms of the angle of rotation of the body:

α = d^2θ/dt^2,

where θ is the angle of rotation of the body.

Thus, the dynamics equation can be rewritten as follows:

ΣM = I(d^2θ/dt^2).

Let's consider the moments of forces acting on the body. Force F is applied at point A and creates a moment of force:

M = F * OA * sin(90° - θ) = F * OA * cos(θ),

where OA is the distance from the axis of rotation to the point of application of force F.

The moment of resistance forces of the bearings is equal to:

Mtr = -b(dθ/dt),

where b is the bearing resistance coefficient.

Thus, the sum of the moments of forces is equal to:

ΣM = F*OA*cos(θ)-Мтр.

Substituting this sum of moments of forces into the dynamics equation, we obtain:

F * OA * cos(θ) - Мтр = I(d^2θ/dt^2).

Let us express F * OA * cos(θ) in terms of the generalized force Q:

Q = F * OA * cos(θ).

Then the equation of motion can be rewritten as follows:

Q - Мтр = I(d^2θ/dt^2).

To solve the equation, we will find the value of Q for a given angle of rotation of the body θ = ?, and then calculate the generalized force Q at other angles of rotation of the body.

Substituting the values ​​of Mtr, I, θ and the desired generalized force Q into the equation of motion, we obtain:

Q - 0,8 = 0,2 * (d^2θ/dt^2).

Differentiating this equation with respect to time, we obtain:

dQ/dt = 0,2 * d^3θ/dt^3.

Thus, the generalized force corresponds to the third derivative of the angle of rotation of the body with respect to time. Substituting the value of the rotation angle θ = ? into the equation for the generalized force, we get:

Q = F * OA * cos(?) = (30cos(?) + 25sin(?) + 400.2)0.2cos(?) = 4cos(?) + 1.5*sin(?) + 1.6.

Answer: 5.2.

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This product is a solution to problem 20.2.3 from the collection of Kepe O.?. in physics.

The problem considers the rotational motion of a body around the Oz axis under the influence of a force F = 30? + 25n + 40b, which is applied at point A. Distance OA = 0.2 m. The moment of resistance forces of the bearings is Mtr = 0.8 N • m.

It is necessary to determine the generalized force corresponding to the angle ? body rotation. To solve the problem, the equation of dynamics of the rotational motion of the body is used: ΣM = Iα, where ΣM is the sum of the moments of forces, I is the moment of inertia of the body, α is the angular acceleration of the body.

Expressing angular acceleration through the angle of rotation of the body, we obtain α = d^2θ/dt^2, where θ is the angle of rotation of the body. Then the equation of motion is rewritten as Q - Mtr = I(d^2θ/dt^2), where Q is the generalized force corresponding to the angle of rotation of the body.

The generalized force is expressed through the angle of rotation and is substituted into the solved equation of motion. After differentiating the equation of motion with respect to time, we find that the generalized force corresponds to the third derivative of the angle of rotation of the body with respect to time.

The final answer to the problem is 5.2. The product is presented in a beautiful HTML format and ensures ease of use and the most useful experience for customers.

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This product is a solution to problem 20.2.3 from the collection of Kepe O.?. The task is to determine the generalized force corresponding to the angle of rotation of a body rotating around the Oz axis under the influence of a force F = 30? + 25n + 40b applied at point A. The distance OA is equal to 0.2 m, and the moment of resistance of the bearings is equal to Mtr = 0.8 N • m.

Solving the problem allows you to get an answer equal to 5.2.

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