A homogeneous rod with a mass of 0.5 kg and a length of l m can

A homogeneous rod of length l and mass 0.5 kg can rotate freely around a horizontal axis passing through one of its ends. A bullet weighing 10 g, flying horizontally at a speed of 300 m/s, hits the opposite end of the rod and gets stuck in it. It is necessary to determine the amplitude and period of vibration of the rod.

To solve the problem, we will use the law of conservation of angular momentum. Before the bullet collides with the rod, the angular momentum of the system is zero, since the bullet flies horizontally. After the collision, the bullet gets stuck in the rod, and the angular momentum of the system remains constant.

The moment of inertia of the rod relative to its end can be expressed as I = (1/3) * m * l^2, where m is the mass of the rod, l is its length. The angular momentum of the system is L = I * w, where w is the angular velocity of rotation of the rod.

After the bullet collides with the rod, the mass of the system increases to m + M, where M is the mass of the bullet. Consequently, the moment of inertia of the system relative to the end of the rod becomes equal to I' = (1/3) * (m + M) * l^2.

According to the law of conservation of angular momentum, the angular momentum of the system before and after the collision must remain unchanged. It follows from this that I * w = I' * w', where w' is the angular velocity of rotation of the system after the collision.

Let us express the angular velocity of rotation of the system after the collision: w' = I * w / I' = (1/3) * m * l^2 * w / ((1/3) * (m + M) * l^2) = m / (m + M) * w.

The period of oscillation of the rod can be expressed as T = 2 * pi / omega, where omega is the angular frequency of oscillation. The angular frequency of oscillation of the rod is related to its length and moment of inertia relative to the end by the formula omega = sqrt(g * (m + M) * l / (2 * I')), where g is the acceleration of gravity.

Now we can find the amplitude of vibration of the rod. At small deflection angles, the amplitude of oscillations is related to the initial angular velocity of rotation of the rod by the formula A = w * sqrt(I / (m * g * l)). Since before the collision of the bullet with the rod the angular momentum of the system is zero, the initial angular velocity of rotation of the rod is zero. Consequently, the amplitude of oscillations of the rod in this case is zero.

Homogeneous rod

Model: HM-1245

A homogeneous rod with a mass of 0.5 kg and a length of l m can freely rotate around a horizontal axis passing through one of its ends. A bullet weighing 10 g, flying horizontally at a speed of 300 m/s, hits the opposite end of the rod and gets stuck in it. This rod model is made of high-quality materials, ensuring durability and reliability in operation.

• Weight: 0.5 kg
• Length: l m
• Free rotation around a horizontal axis
• Made from high quality materials

Price: 2499 rub.

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A homogeneous rod with a mass of 0.5 kg and a length of l m is model HM-1245. It is made of high quality materials, which guarantees its durability and reliability in operation. The rod can rotate freely around a horizontal axis passing through one of its ends. A bullet weighing 10 g, flying horizontally at a speed of 300 m/s, hits the opposite end of the rod and gets stuck in it.

To determine the amplitude and period of oscillation of the rod, we will use the law of conservation of angular momentum. Before the bullet collides with the rod, the angular momentum of the system is zero, since the bullet flies horizontally. After the collision, the bullet gets stuck in the rod, and the angular momentum of the system remains constant.

The moment of inertia of the rod relative to its end can be expressed as I = (1/3) * m * l^2, where m is the mass of the rod, l is its length. The angular momentum of the system is L = I * w, where w is the angular velocity of rotation of the rod.

After the bullet collides with the rod, the mass of the system increases to m + M, where M is the mass of the bullet. Consequently, the moment of inertia of the system relative to the end of the rod becomes equal to I' = (1/3) * (m + M) * l^2.

According to the law of conservation of angular momentum, the angular momentum of the system before and after the collision must remain unchanged. It follows from this that I * w = I' * w', where w' is the angular velocity of rotation of the system after the collision.

Let us express the angular velocity of rotation of the system after the collision: w' = I * w / I' = (1/3) * m * l^2 * w / ((1/3) * (m + M) * l^2) = m / (m + M) * w.

The period of oscillation of the rod can be expressed as T = 2 * pi / omega, where omega is the angular frequency of oscillation. The angular frequency of oscillation of the rod is related to its length and moment of inertia relative to the end by the formula omega = sqrt(g * (m + M) * l / (2 * I')), where g is the acceleration of gravity.

Now we can find the amplitude of vibration of the rod. At small deflection angles, the amplitude of oscillations is related to the initial angular velocity of rotation of the rod by the formula A = w * sqrt(I / (m * g * l)). Since before the collision of the bullet with the rod the angular momentum of the system is zero, the initial angular velocity of rotation of the rod is zero. Consequently, the amplitude of oscillations of the rod in this case is zero.

The rod of model HM-1245 has a mass of 0.5 kg and a length of l meters. The price of this product is As an artificial intelligence, I do not have access to information about the prices of goods in specific stores or regions. The price of the HM-1245 model may depend on many factors, such as location of sale, taxes, seasonality and market competition. I recommend contacting the sellers or manufacturers of this model to obtain current price information.

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Product description:

A homogeneous rod with a mass of 0.5 kg and a length of l m is proposed, which can freely rotate around a horizontal axis passing through one of its ends. The rod has a cylindrical shape and the same radius along its entire length.

In addition, a bullet weighing 10 g, flying horizontally at a speed of 300 m/s, hits one of the ends of the rod and gets stuck in it.

For this product, it is necessary to solve the problem associated with determining the amplitude and period of vibration of the rod as a result of the impact of a bullet.

To solve the problem, the laws of conservation of energy and angular momentum are used, as well as formulas for calculating the period of oscillation of a mathematical pendulum.

A detailed solution to the problem with a brief record of the conditions, formulas and laws used in the solution, the derivation of the calculation formula and the answer can be provided upon request. If you have any further questions about the solution, don't hesitate to ask for help.

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2. The quality of the material is excellent, the rod is strong and durable.
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Peculiarities:

Excellent uniform rod! The quality of the material is on top, has no defects and corresponds to the declared parameters.

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