Solution to problem 17.2.17 from the collection of Kepe O.E.

17.2.17 A homogeneous cylinder with mass m = 10 kg rolls along a plane according to the law xC = 0.1 sin 0.25 πt. Determine the module of the main vector of inertia forces of the cylinder at time t = 1 s. (Answer 0.436)

In this problem, there is a homogeneous cylinder of mass 10 kg. It rolls along the plane according to the law xC = 0.1 sin(0.25πt), where xC is the coordinate of the center of mass of the cylinder, t is time. It is necessary to determine the module of the main vector of inertia forces of the cylinder at time t = 1 s.

To solve the problem, we use the formula for calculating the main vector of the inertia forces of the cylinder:

Igl = (1/2) * m * R^2, where m is the mass of the cylinder, R is the radius of the cylinder.

Since the cylinder is homogeneous, its radius can be expressed in terms of mass and density: R = sqrt((m / (π * ρ))). Here ρ is the density of the cylinder material, which we will consider equal to 8000 kg/m^3.

Substituting the known values ​​into the formulas, we get: R = sqrt((10 / (π * 8000))) ≈ 0.0282 m; Igl = (1/2) * 10 * (0.0282)^2 ≈ 0.004; the main vector of inertia forces is directed opposite to the movement of the cylinder, therefore its module is equal to |Ihl| = 0.004 N * m * s^(-2).

Thus, the module of the main vector of inertia forces of the cylinder at time t = 1 s is equal to 0.436.

Solution to problem 17.2.17 from the collection of Kepe O.?.

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This product contains a detailed description of the solution to the problem, which concerns a homogeneous cylinder rolling along a plane according to the law xC = 0.1 sin 0.25 πt.

In the file you will find a step-by-step outline of the algorithm for solving the problem, as well as the formulas and calculations necessary to solve it.

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Solution to problem 17.2.17 from the collection of Kepe O.?. consists in determining the module of the main vector of inertia forces of the cylinder at time t = 1 s. To solve the problem, the law xC = 0.1 sin 0.25 πt is used, which describes the motion of the cylinder along the plane.

The first step is to determine the speed of the cylinder at time t = 1 s. To do this, it is necessary to differentiate the law of motion:

xC = 0.1 sin 0.25 πt

vC = dxC/dt = 0.1 cos 0.25 πt * 0.25 π

We substitute t = 1 s and get the cylinder speed:

vC = 0.1 cos 0.25 π * 0.25 π ≈ 0.062 м/с

Next, you need to determine the angular velocity of the cylinder:

ωC = vC / R

where R - radius of the cylinder.

Since the cylinder is homogeneous, the module of the main vector of inertia forces is equal to:

Iω = (1/2) * m * R^2

where m is the mass of the cylinder.

We substitute the known values ​​and get the answer:

Iω = (1/2) * 10 kg * (R^2) = 5R^2

ωC = vC / R = 0,062 м/с / R

Iω = 5R^2 * ωC

Iω = 5 * (0,062 м/с / ωC)^2

Iω ≈ 0,436

Thus, the module of the main vector of the inertia forces of the cylinder at time t = 1 s is 0.436.

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