Solution to problem 16.1.13 from the collection of Kepe O.E.

It is necessary to determine the angular acceleration of a homogeneous rod with length l = 1 m and mass m = 4 kg, which rotates around the Oz axis. It is known that a torque Mz = 3N•m is applied to the rod.

Answer:

We use the formula to calculate angular acceleration:

α = Mz / I,

where α is the angular acceleration, Mz is the torque, and I is the moment of inertia of the rod.

The moment of inertia of a rod rotating around its axis is equal to:

I = m * l^2 / 12.

Substituting the values ​​of m and l, we get:

I = 1 / 3 * m * (l / 2)^2 = 1/3 * 4 * (1/2)^2 = 1/3 * 4 * 1/4 = 1/3 кг * м^2.

Substituting the values ​​of Mz and I, we get:

α = Mz / I = 3 / (1/3) = 9 (rad/s^2).

Answer: the angular acceleration of the rod is 9 rad/s^2.

Solution to problem 16.1.13 from the collection of Kepe O..

This digital product is a solution to problem 16.1.13 from the collection of Kepe O.. in physics. The solution is presented in the form of a detailed description with a step-by-step solution algorithm and an answer to the question at the end.

The problem is to determine the angular acceleration of a homogeneous rod with length l = 1 m and mass m = 4 kg, which rotates around the Oz axis. It is known that a torque Mz = 3N•m is applied to the rod. The solution is based on the use of a formula to calculate the angular acceleration and moment of inertia of the rod.

By purchasing this digital product, you receive a detailed solution to problem 16.1.13 from the collection of Kepe O.. and confidence in your knowledge in the field of physics.

This product is a digital solution to problem 16.1.13 from the collection of Kepe O.?. in physics. The problem is to determine the angular acceleration of a homogeneous rod with a length of 1 m and a mass of 4 kg, which rotates around the Oz axis, with a known torque Mz = 3 N m. The solution to the problem is based on the use of a formula for calculating the angular acceleration and moment of inertia of the rod.

By purchasing this digital product, you receive a detailed description, step by step, of the algorithm for solving the problem, as well as an answer to the question. As a result, you will gain more confidence in your knowledge of physics.

This digital product is a solution to problem 16.1.13 from the collection of Kepe O.?. in physics. The problem is to determine the angular acceleration of a homogeneous rod with length l = 1 m and mass m = 4 kg, which rotates around the Oz axis. It is known that a torque Mz = 3N•m is applied to the rod.

The solution to the problem is based on the use of a formula for calculating the angular acceleration and moment of inertia of the rod. The moment of inertia of a rod rotating around its axis is equal to I = m * l^2 / 12. Substituting the values ​​of the mass and length of the rod, we get I = 1 / 3 kg * m^2. Then, using the formula for calculating the angular acceleration α = Mz / I, we find the angular acceleration of the rod: α = 9 rad/s^2.

By purchasing this digital product, you receive a detailed description of the solution to problem 16.1.13 from the collection of Kepe O.?., which includes a step-by-step solution algorithm and an answer to the question at the end. Also, this product will help you strengthen your knowledge in the field of physics.

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Solution to problem 16.1.13 from the collection of Kepe O.?. consists in determining the angular acceleration of a homogeneous rod, which has a mass m = 4 kg and a length l = 1 m, and rotates around the Oz axis in the presence of a torque Mz = 3 N•m.

To solve the problem, it is necessary to use the law of conservation of angular momentum, which states that the angular momentum of a system remains constant if the system is not acted upon by external moments. In our case, the angular momentum of the system consists of the angular momentum of the rod and the angular momentum of the rotation. The angular momentum of the rod can be expressed as Iω, where I is the moment of inertia of the rod and ω is its angular velocity. The moment of inertia of the rod is equal to (1/12)mL², where m is the mass of the rod and L is its length. The angular momentum of rotation is equal to Lω, where L is the moment of the force creating the rotation of the rod, and ω is its angular velocity.

Using this information, we can write the conservation equation for angular momentum:

Iω + Lω = const

Knowing that the moment of inertia of the rod is equal to (1/12)mL², and the moment of force Mz = 3 N•m, we can write the equation for finding the angular acceleration α:

(1/12)mL²α = Mz

α = 12Mz/(mL²)

Substituting the known values, we get the answer:

α = 12 * 3 N•m / (4 kg * (1 m)²) = 9 rad/s²

Thus, the angular acceleration of a homogeneous rod with a mass of 4 kg and a length of 1 m, rotating around the Oz axis in the presence of a torque Mz = 3 N•m, is equal to 9 rad/s².

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