Solution of problem D3 (task 1) Option 04 Dievsky V.A.

Termeh Dievsky V.A. proposes to solve the problem Dynamics 3 (D3) task 1, associated with the theorem on the change in kinetic energy, for mechanical systems shown in diagrams 1-30. For body 1, it is necessary to determine the angular acceleration (options 4, 6, 7, 9, 11, 18, 25, 26, 28) or linear acceleration (other options) using the theorem on the change in kinetic energy in differential form. In this case, the threads are considered weightless and inextensible. The following notations are accepted in the assignment: m - body mass, R and r - radii, p - radius of inertia (if it is not specified, the body is considered a homogeneous cylinder); in the presence of friction, f is the sliding friction coefficient, fк is the rolling friction coefficient.

To solve task 1 of problem D3 associated with scheme No. 4, it is necessary to use the theorem on the change in kinetic energy in differential form and determine the angular acceleration of body 1. According to the condition, body 1 is a homogeneous cylinder with mass m1 and radius R. The body is associated with weightless and an inextensible thread wrapped around a cylinder of radius r and mass m2. The thread is wound onto the cylinder without slipping with a rolling friction coefficient fк.

First, it is necessary to record the kinetic energy of the system at the initial moment of time (when body 1 is at the top point) and at an arbitrary moment of time t. The kinetic energy of the system at the initial moment of time is 0, since body 1 is at rest. At an arbitrary moment of time t, the kinetic energy of the system can be written as follows:

T = 1/2t1V1^2 + 1/2t2V2^2 + 1/2Iw^2,

where V1 and V2 are the linear velocities of bodies 1 and 2, respectively, w is the angular velocity of body 1, I is the moment of inertia of body 1 relative to the axis of rotation (axis of the thread), determined by the formula I = 1/2t1R^2.

According to the theorem on the change in kinetic energy in differential form, the difference between the kinetic energy of the system at an arbitrary moment of time t and at the initial moment of time is equal to the work of all forces acting on the system during this period of time:

ΔT = A,

where A is the work of all forces acting on the system. The work done by the sliding friction force is fNs, where N is the tension force of the thread, s is the path traveled by the point of contact of body 2 with the surface of the cylinder. The tension force of the thread is equal to the force of gravity of body 1, since the thread is weightless and inextensible. Thus, the work done by the sliding friction force can be written in the following form:

Aф = ft1g*(R-r)*sinθ,

where g is the acceleration of gravity, θ is the angle through which body 1 has turned at time t.

The work done by the rolling friction force is fкNs, where s is the path traveled by the point of contact of body 2 with the surface of the cylinder. The tension force of the thread in this case is not equal to the force of gravity of body 1, since the thread is wound onto the cylinder without slipping. To determine the tension force, it is necessary to use the no-slip condition:

(R-r)w = Vs,

where Vs is the linear speed of the point of contact of body 2 with the surface of the cylinder.

The thread tension force can be written as follows:

N = t1g - t2g - fкt2(R-r)/r.

Thus, the work done by the rolling friction force can be written in the following form:

Afk = fkt2g*(R-r)*sinθ.

The difference between the kinetic energy of the system at an arbitrary time t and at the initial time can now be written in the following form:

ΔT = 1/2t1V1^2 + 1/2t2V2^2 + 1/2Iw^2 - Af - Afk.

From the theorem on the change in kinetic energy in differential form it follows that the difference between the kinetic energy of the system at an arbitrary moment of time t and at the initial moment of time is equal to the change in the kinetic energy of the system over this period of time. The change in the kinetic energy of the system over a period of time dt can be written as follows:

dT = 1/2t1dV1^2 + 1/2t2dV2^2 + 1/2Idw^2 - Aфdt - Aфкdt.

The angular acceleration of body 1 can be determined from the equation of motion of body 2. The equation of motion of body 2 can be written in the following form:

t2a2 = t2g - N - fк*t2.

Considering that a2 = r*d^2θ/dt^2, we obtain the following expression for the angular acceleration of body 1:

w'' = gsinθ/(R-r) - ft2gsinθ/(R-r) - fкt2w/r,

where w'' is the angular acceleration of body 1.

Thus, to solve task 1 of problem D3, scheme No. 4, it is necessary to use formulas for determining the work of sliding and rolling friction forces, as well as the equation of motion of body 2 to determine the angular acceleration of body 1. It is important to take into account the differences in the conditions of the problems for different options.

"Solving problem D3 (task 1) Option 04 Dievsky V.A." is a digital product that represents a solution to a problem in a theoretical mechanics course. The solution to the problem is based on the theorem on the change in kinetic energy in differential form and contains the definition of the angular acceleration of body 1 in diagram No. 4.

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Purchasing this digital product will allow you to get a ready-made solution to the problem and save time on solving it yourself. Also, this product can be used as educational material for self-study of theoretical mechanics.

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This product is a solution to task 1 from the problem Dynamics 3 (D3) in theoretical mechanics, option 4, diagram 4. The task is to determine the angular acceleration of body 1 for the mechanical system shown in the diagram, using the theorem on the change in kinetic energy in differential form. The description indicates the accepted designations, such as body masses, radii and radius of gyration, as well as friction coefficients. The solution to the assignment is made in Word format (handwritten or typed in Word) and packaged in a zip archive, which will be available after payment. The solution is intended for university students and is suitable for use for educational purposes. After checking the solution, the author will be grateful if you leave positive feedback.

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