Solution to problem 17.2.4 from the collection of Kepe O.E.

17.2.4 It is necessary to determine the main moment of inertia of the wheel relative to its center of mass O. This calculation is necessary, provided that the wheel rotates around its center of mass at a speed described by the law φ = 2t2. The mass of the wheel is 2 kg and is uniformly distributed over the rim of radius r = 20 cm.

Answer:

The main moment of inertia of the wheel is calculated by the formula I = Σmr², where Σmr² is the sum of the products of the particle mass by the square of the distance to the axis of rotation.

First, let's find the mass of an elementary particle of the wheel. To do this, divide the mass of the wheel by the number of elementary particles into which the wheel can be divided.

The mass of an elementary particle can be found using the formula m = M/N, where M is the mass of the wheel, and N is the number of elementary particles.

Thus, the mass of an elementary particle is m = 2 kg / (2πr/N) = 2N/π grams.

Now we can calculate the main moment of inertia of the wheel. Note that the distance from the center of mass O to the axis of rotation is equal to the radius of the wheel r. Also note that the angular acceleration of the wheel is equal to the second derivative of the angle φ with respect to time: ω = d²φ/dt² = 4 rad/s².

I = Σmr² = ∫(0→2πr) (2N/π)r²dφ = 4Nr²

Thus, the main moment of inertia of the wheel is equal to -0.32 kg * m² (-4Nr² * ω).

Solution to problem 17.2.4 from the collection of Kepe O.?.

This solution is a digital product available in our digital product store. Solution to problem 17.2.4 from the collection of Kepe O.?. intended for students and teachers studying mechanics.

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Digital product "Solution to problem 17.2.4 from the collection of Kepe O.?" is a detailed solution to problem 17.2.4 from the collection of Kepe O.?, where it is required to determine the main moment of inertia of the wheel relative to its center of mass O, provided that the wheel rotates around its center of mass with a speed described by the law φ = 2t2. The mass of the wheel is 2 kg and is uniformly distributed over the rim of radius r = 20 cm.

The solution is made using the appropriate formulas and is given step by step with a description of all the necessary steps. First, the mass of the elementary particle of the wheel was found, and then the main moment of inertia of the wheel was calculated using the formula I = Σmr². The calculation result is -0.32 kg*m².

The design of the presented digital product is made in a beautiful html format, which ensures convenient and understandable viewing on any device. By purchasing this solution to the problem, you receive not only a high-quality product, but also confidence in completing the task correctly and successfully passing the exam.

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Problem 17.2.4 from the collection of Kepe O.?. consists in determining the main moment of inertia of the wheel relative to the center of mass O under given conditions.

A wheel with a mass of 2 kg rotates around the center of mass O according to the law φ = 2t2, where φ is the angle of rotation of the wheel, and t is time. The radius of the wheel is 20 cm, and the mass is evenly distributed along the rim. It is necessary to find the main moment of inertia of the wheel relative to the center of mass O.

The solution to this problem involves calculating the moment of inertia of the wheel relative to the center of mass O and applying the formula for calculating the main moment of inertia forces. The moment of inertia of the wheel can be calculated using the formula:

I = mr^2/2,

where m is the mass of the wheel, and r is the radius of the wheel.

Substituting the given values, we get:

I = 2 * (0.2)^2 / 2 = 0.02 kg * m^2.

Next, to calculate the main moment of inertia, it is necessary to multiply the moment of inertia of the wheel by the square of the wheel speed:

J = I * ω^2,

where ω is the angular speed of rotation of the wheel.

The angular speed of rotation of the wheel can be found as the derivative of the angle φ with respect to time t:

ω = dφ/dt = 4t.

Substituting the given values, we get:

ω = 4t,

I = 0.02 kg * m^2,

J = I * ω^2 = 0.02 * (4t)^2 = 0.32t^2.

Thus, the main moment of inertia of the wheel relative to the center of mass O is equal to 0.32t^2. At t = -0.4 seconds this moment is -0.32 kg * m^2.

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