The flywheel has the shape of a disk with a diameter of 40 cm and a mass of 100

Flywheel

The flywheel is a digital product that is a virtual flywheel with a disk shape with a diameter of 40 cm and a mass of 100 kg. It was created for those who are interested in physics and mechanics.

The flywheel has a rotation speed of 10 rps and can be stopped using a brake pad, which creates a friction force of 60 N.

The kit includes:

• 3D flywheel model;
• Animation of flywheel rotation;
• Calculations of the moment of friction, moment of inertia and angular acceleration during braking;
• Flywheel stopping time;
• Interactive tasks and exercises.

The flywheel is an excellent choice for those who want to improve their knowledge of physics and mechanics. Order it now and get access to exciting content!

The description of the product "Flywheel" is as follows:

“Flywheel” is a digital product, which is a virtual flywheel with a disk shape with a diameter of 40 cm and a mass of 100 kg. It is designed to study physics and mechanics. The kit includes a 3D model of the flywheel, animation of rotation, calculations of frictional moment, moment of inertia and angular acceleration during braking, as well as interactive problems and exercises.

During operation, the “Flywheel” rotates at a frequency of 10 rps, and it is stopped using a brake pad, which is pressed against the rim of the flywheel and creates a friction force of 60 N. For this flywheel, you need to find:

1. Moment of friction force;
2. Flywheel moment of inertia;
3. Angular acceleration during braking (in absolute value);
4. Flywheel stop time.

To solve the problem, appropriate formulas and laws of physics and mechanics are required. If you have any questions, I am ready to help solve them.

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A flywheel is a disk-shaped solid body with a diameter of 40 cm and a mass of 100 kg. It can rotate around its axis at a frequency of 10 revolutions per second. When stopping the flywheel with the help of a brake pad, which is pressed against its rim, a friction force of 60 N is created.

To solve problem 10427, you must use the following formulas and laws:

1. The moment of friction force is equal to the product of the friction force and the radius of the flywheel: Mtr = Ftr * R.
2. The moment of inertia of the flywheel is calculated by the formula: I = (m * R^2) / 2, where m is the mass of the flywheel, R is the radius of the flywheel.
3. The law of conservation of energy for rotational motion states that the sum of kinetic and potential energy remains constant during rotation, braking, and stopping of the flywheel. The frictional torque leads to a gradual slowdown in rotation and a decrease in the kinetic energy of the flywheel.
4. Angular acceleration during braking can be calculated using the formula: α = Mtr / I.
5. The law of change in kinetic energy for rotational motion states that the change in kinetic energy is equal to the work done by the torque, i.e. ΔК = Wм = Mtr * Δθ, where Δθ is the angle of rotation of the flywheel.

Answer:

1. Friction torque: Mtr = Ftr * R = 60 N * 0.2 m = 12 N*m.

2. Flywheel moment of inertia: I = (m * R^2) / 2 = (100 kg * 0.2 m^2) / 2 = 10 kg * m^2.

3. Angular acceleration during braking: α = Mtr / I = 12 Nm / 10 kgm^2 = 1.2 rad/s^2 (in absolute value).

4. The flywheel stopping time can be calculated using the law of change in kinetic energy: ΔК = Wм = Mtr * Δθ, where Δθ is the angle of rotation of the flywheel. From the law of conservation of energy it follows that the initial kinetic energy of the flywheel is equal to its final potential energy. Initial kinetic energy of the flywheel: K1 = (I * ω^2) / 2 = (10 kgm^2 * (10 rev/s * 2π rad/rev)^2) / 2 = 6283.19 J. Final potential energy of the flywheel: P2 = m * g * h, where h is the height to which the flywheel will rise when stopped. h = P2 / (m * g) = K1 / (m * g) = 6.283 m. Δθ = h / R = 6.283 m / 0.2 m = 31.42 rad. Then ΔК = Mtr * Δθ = 12 Nm * 31.42 rad = 377.04 J. The change in the kinetic energy of the flywheel is equal to ΔK = K1 - K2, where K2 is the final kinetic energy, which is zero when the flywheel stops. Then K1 = ΔK = 377.04 J. Using the formula for kinetic energy K = (I * ω^2) / 2, we can express the angular speed of the flywheel when stopping: ω = sqrt(2 * K / I) = sqrt(2 * 377.04 J / 10 kg*m^2) = 7.74 rad/s. Angle of rotation of the flywheel during its stop: Δθ = ω * t. From here we can express the stopping time of the flywheel: t = Δθ / ω = 31.42 rad / 7.74 rad/s = 4.05 s.

Answer:

1. Friction moment: Mtr = 12 N*m.
2. Flywheel moment of inertia: I = 10 kg*m^2.
3. Angular acceleration during braking: α = 1.2 rad/s^2 (in absolute value).
4. The flywheel will stop in 4.05 s.

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