A solid cylinder with a mass of 20 kg and a radius of 40 cm rotates

The page contains problem 10315, which describes the rotation of a solid cylinder with a mass of 20 kg and a radius of 40 cm under the influence of friction forces. In this case, the cylinder slows down and stops, and the work done by the friction forces is 1568 J. It is required to find the period of rotation of the cylinder before braking begins.

To solve the problem, you must use the following formulas and laws:

1. The moment of inertia of the cylinder relative to the axis of rotation is equal to I = 0.5 * m * r^2, where m is the mass of the cylinder, r is the radius of the cylinder.
2. The kinetic energy of a rotating cylinder is equal to E = 0.5 * I * w^2, where w is the angular velocity of rotation of the cylinder.
3. The work of friction forces performed when braking the cylinder is equal to A = E1 - E2, where E1 is the kinetic energy of the cylinder before braking, E2 is the kinetic energy of the cylinder after stopping.
4. The period of rotation of the cylinder is determined by the formula T = 2 * pi / w, where pi is the number "pi".

Using these formulas, you can derive a calculation formula for determining the period of rotation of the cylinder before braking begins:

T = 2 * pi * sqrt(I / (2 * A))

Substituting the known values, we get:

T = 2 * pi * sqrt(0.5 * m * r^2 / (2 * 1568))

T = 2 * pi * sqrt(0.5 * 20 * 0.4^2 / (2 * 1568))

T ≈ 3.76 sec

Thus, the period of rotation of the cylinder before braking begins is approximately 3.76 seconds.

Product description

We present to your attention a digital product - an interesting problem about the rotation of a solid cylinder with a mass of 20 kg and a radius of 40 cm!

This problem describes the rotation of a cylinder under the influence of friction forces, its deceleration and stopping, as well as the work of friction forces. The task is intended for those who are interested in physics and mathematics, as well as for those who want to improve their knowledge in this area.

The solution to the problem is based on the use of formulas and laws that are related to the mechanics of rotational motion. As a result, you will be able to calculate the period of rotation of the cylinder before braking begins!

The challenge is a digital product and is available for download in PDF format. Buy a problem right now and improve your knowledge in physics and mathematics!

Product characteristics

• Title: Problem of rotation of a solid cylinder
• Cylinder weight: 20 kg
• Cylinder radius: 40 cm
• Format: PDF

We present to your attention a digital product - an interesting problem about the rotation of a solid cylinder with a mass of 20 kg and a radius of 40 cm! This problem describes the rotation of a cylinder under the influence of friction forces, its deceleration and stopping, as well as the work of friction forces. The task is intended for those who are interested in physics and mathematics, as well as for those who want to improve their knowledge in this area.

To solve the problem, it is necessary to use formulas and laws that are related to the mechanics of rotational motion. The moment of inertia of the cylinder relative to the axis of rotation is equal to I = 0.5 * m * r^2, where m is the mass of the cylinder, r is the radius of the cylinder. The kinetic energy of a rotating cylinder is equal to E = 0.5 * I * w^2, where w is the angular velocity of rotation of the cylinder. The work of friction forces performed when braking the cylinder is equal to A = E1 - E2, where E1 is the kinetic energy of the cylinder before braking, E2 is the kinetic energy of the cylinder after stopping. The period of rotation of the cylinder is determined by the formula T = 2 * pi / w, where pi is the number "pi".

According to the conditions of the problem, a cylinder with a mass of 20 kg and a radius of 40 cm rotated before braking began. The work of friction forces was 1568 J. Using the calculation formula T = 2 * pi * sqrt(I / (2 * A)), where I is the moment of inertia of the cylinder, and A is the work of friction forces, we can calculate the period of rotation of the cylinder before braking begins .

Substituting the known values, we get: T = 2 * pi * sqrt(0.5 * m * r^2 / (2 * 1568)) = 2 * pi * sqrt(0.5 * 20 * 0.4^2 / (2 * 1568)) ≈ 3.76 sec.

Thus, the period of rotation of the cylinder before braking begins is approximately 3.76 seconds. The task is presented in PDF format and is available for download. If you have any questions about the solution, don't hesitate to ask for help.

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Product description:

This product is a solid cylinder with a mass of 20 kg and a radius of 40 cm, which rotates around its axis. Under the influence of friction forces, it moves slowly and stops. The work done by the friction forces that led to the cylinder stopping is 1568 J.

To solve the problem, it is necessary to find the period of rotation of the cylinder before braking begins. To do this, you can use the law of conservation of energy, according to which the sum of the kinetic and potential energy of a body remains constant during movement.

At the initial moment of time, the cylinder rotated with a certain angular velocity corresponding to the kinetic energy. As it slows down, kinetic energy is converted into frictional work, which leads to a decrease in rotation speed. When the speed reaches zero, all kinetic energy will be converted into the work of friction forces, which will lead to the stop of the cylinder.

To find the rotation period, you can use the formula for the kinetic energy of a rotating body:

K = (1/2) I w^2,

where K is kinetic energy, I is the moment of inertia of the cylinder, w is the angular velocity of rotation of the cylinder.

The moment of inertia of the cylinder can be calculated using the formula:

I = (1/2) M R^2,

where M is the mass of the cylinder, R is the radius of the cylinder.

Thus, to find the period of rotation, it is necessary to find the angular velocity of rotation of the cylinder at the initial moment of time and use the formula for the period of oscillation:

T = 2π / w.

Thus, the period of rotation of the cylinder before the start of braking can be found using the following formula:

T = 2π √(I / (2K)) = 2π √(M R^2 / (8K)).

Substituting the known values, we get:

T = 2π √(20 kg * (0.4 m)^2 / (8 * 1568 J)) ≈ 1.43 sec.

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