Solution to problem 15.4.4 from the collection of Kepe O.E.

Consider a homogeneous rod of mass m = 3 kg, whose length is AB = 1 m. The rod rotates around the Oz axis according to the law ? = 2t3. It is necessary to determine the kinetic energy of the rod at the moment of time t = lc.

First, let's find the moment of inertia of the rod relative to the axis of rotation. Since the rod is homogeneous and has the shape of a straight cylindrical tube, the moment of inertia will be equal to:

I = (m * l^2) / 12,

Where l - length of the rod.

Substituting the known values, we get:

I = (3 * 1^2) / 12 = 0.25 kg * m^2.

The kinetic energy of a rotating body is expressed by the formula:

Ek = I * ?^2 / 2,

Where ?² - square of the angular velocity of rotation of the body.

Substituting the known values, we get:

Ek = 0,25 * (2lc)^2 / 2 = 0,5 * 4l²c² = 2l²c².

Thus, the kinetic energy of the rod at the moment of time t = lc equal 18.

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We present to your attention a digital product - a solution to problem 15.4.4 from the collection of Kepe O.?. This problem describes the rotation of a homogeneous rod with a mass of 3 kg and a length of 1 m around the Oz axis according to the law? = 2t3. It is necessary to determine the kinetic energy of the rod at time t = lc.

Solving the problem begins with finding the moment of inertia of the rod relative to the axis of rotation. Since the rod is homogeneous and has the shape of a straight cylindrical tube, the moment of inertia will be equal to: I = (m * l^2) / 12, where l is the length of the rod. Substituting the known values, we get: I = (3 * 1^2) / 12 = 0.25 kg * m^2.

Then using the formula

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Solution to problem 15.4.4 from the collection of Kepe O.?. consists in determining the kinetic energy of a homogeneous rod at time t=lc, provided that the rod with mass m=3 kg and length AB=1 m rotates around the Oz axis according to the law ?=2t3.

To solve the problem, you need to use the formula for the kinetic energy of a rotating body:

K = (1/2)Iω²,

where K is the kinetic energy, I is the moment of inertia of the body relative to the axis of rotation, and ω is the angular velocity of rotation of the body.

To calculate the moment of inertia of a homogeneous rod relative to the axis of rotation Oz, we use the formula:

I = (1/12)mL²,

where L is the length of the rod.

Also, to find the angular velocity of rotation of a body at time t=lc, it is necessary to calculate the first derivative of the angle of rotation with respect to time:

? = 2t³

?` = 6t²

Substituting known values ​​into the formulas, we get:

I = (1/12) * 3 * 1² = 0.25 kg*m²

ω = ?` = 6lc²

K = (1/2) * I * ω² = (1/2) * 0.25 * (6lc²)² = 2.25lc^4 Дж

Thus, at time t=lc the kinetic energy of the rod is 18 J.

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