A horizontal rod weighing 10 kg and

A horizontal rod 0.8 m long and weighing 10 kg can rotate around a vertical axis passing through its middle. A ball of mass 5 g, flying at a speed of 80 m/s, hits the end of the rod. It is necessary to determine the angular velocity at which the rod begins to rotate and the speed of the ball after impact.

To solve the problem we use the law of conservation of angular momentum. Before the collision, the angular momentum of the system is zero, since the rod is at rest. After the collision, the angular momentum of the system is conserved:

$m_1v_1 = m_2v_2 + I\omega$

where $m_1$ and $v_1$ are the mass and speed of the ball, $m_2$ and $v_2$ are the mass and speed of the rod, and $I$ and $\omega$ are the moment of inertia and angular velocity of the rod, respectively.

Before the collision of the ball with the rod, the angular momentum of the system is equal to:

$L_1 = m_1v_1 = 5\cdot10^{ -3}\cdot80 = 0,4\,\text{кг}\cdot\text{м}/\text{с}$

After the collision of the ball with the rod, the frictional force at the point of contact between the ball and the rod creates a moment of force that causes rotation of the rod around a vertical axis. The moment of inertia of the rod relative to its center of mass can be calculated using the formula:

$I = \frac{1}{12}mL^2$

where $m$ is the mass of the rod, $L$ is its length.

Substituting the values, we get:

$I = \frac{1}{12}\cdot10\cdot0,8^2 = 0,053\,\text{кг}\cdot\text{м}^2$

Thus, the angular momentum of the system after the collision is equal to:

$L_2 = m_1v_1 = m_2v_2 + I\omega$

Let us express the angular velocity of the rod:

$\omega = \frac{m_1v_1 - m_2v_2}{I}$

Substituting the values, we get:

$\omega = \frac{5\cdot10^{ -3}\cdot80 - 10\cdot v_2}{0,053}$

In order to find the speed of the ball after impact, we use the law of conservation of energy. Before the collision, the energy of the system is equal to the kinetic energy of the ball:

$E_1 = \frac{1}{2}m_1v_1^2 = 0,16\,\text{Дж}$

After the collision, the energy of the system is equal to the kinetic energy of the ball and rod:

$E_2 = \frac{1}{2}m_1v_2^2 + \frac{1}{2}I\omega^2$

Thus, the law of conservation of energy will be written as:

$EAssume that a horizontal rod 0.8 m long and weighing 10 kg can rotate around a vertical axis passing through its middle. A ball of mass 5 g flies towards the end of the rod at a speed of 80 m/s. We need to determine the angular velocity of the rod after the impact and the speed of the ball.

To solve the problem, we will use the law of conservation of angular momentum. Before the collision, the angular momentum of the system is zero, since the rod is motionless. After the collision, the angular momentum of the system is conserved:

$m_1v_1 = m_2v_2 + I\omega$

Here $m_1$ and $v_1$ are the mass and speed of the ball, $m_2$ and $v_2$ are the mass and speed of the rod, and $I$ and $\omega$ are the moment of inertia and angular velocity of the rod, respectively.

Before the collision, the angular momentum of the system is equal to:

$L_1 = m_1v_1 = 5\cdot10^{ -3}\cdot80 = 0,4\,\text{кг}\cdot\text{м}/\text{с}$

After a collision, the frictional force at the point of contact between the ball and the rod creates a moment of force that causes the rod to rotate around a vertical axis. The moment of inertia of the rod relative to its center of mass can be calculated using the formula:

$I = \frac{1}{12}mL^2$

Here $m$ is the mass of the rod, $L$ is its length.

Substituting the values, we get:

$I = \frac{1}{12}\cdot10\cdot0,8^2 = 0,053\,\text{кг}\cdot\text{м}^2$

Thus, the angular momentum of the system after the collision is equal to:

$L_2 = m_1v_1 = m_2v_2 + I\omega$

Let us express the angular velocity of the rod:

$\omega = \frac{m_1v_1 - m_2v_2}{I}$

Substituting the values, we get:

$\omega = \frac{5\cdot10^{ -3}\cdot80 - 10\cdot v_2}{0,053}$

To find the speed of the ball after impact, we use the law of conservation of energy. Before the collision, the energy of the system is equal to the kinetic energy of the ball:

$E_1 = \frac{1}{2}m_1v_1^2 = 0,16\,\text{Дж}$

After the collision, the energy of the system is equal to the kinetic energy of the ball and rod:

$E_2 = \frac{1}{2}m_1v_2^2 + \frac{1}{2}I\omega^2$

Thus, the law of conservation of energy will be written as:

$E_1 = E_2$

Resh

Description of the digital product

Product name: "Solution of the rotating rod problem"

Product type: e-course

Price: 500 rubles

Product description

The electronic course "Solving the problem of a rotating rod" is intended for students and schoolchildren studying mechanics.

The course includes a detailed description of the solution to the problem of a horizontal rod with a mass of 10 kg and a length of 0.8 m, which can rotate around a vertical axis perpendicular to it, passing through its middle. A ball with a mass of 5 g and a speed of 80 m/s hits the end of the rod. The course contains detailed calculations and formulas necessary to solve the problem, as well as graphic illustrations and animations to help better understand the solution process.

The electronic course "Solving the Rotating Rod Problem" is presented in a convenient HTML format, which allows you to quickly and easily find the information you need. The course can be useful both for independent study and as material for lectures and seminars.

By purchasing this course, you get access to the full version with the possibility of free updates and support.

From the description provided, it is impossible to clearly determine which specific digital product we are talking about. The description is given for a physical system consisting of a horizontally located rod and a ball that falls on it. If you have additional information or a specific request, I will be happy to help you!

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There is no product description in your question. If you want a solution to problem 10728, then I can provide it to you.

To solve the problem, we can use the laws of conservation of energy and angular momentum. Before the ball hits, the rod is at rest, so its initial angular velocity is zero. After the ball hits the rod, a moment of force arises, which causes the rod to rotate around a vertical axis.

The angular momentum of the system before the impact is zero, since the rod is at rest, and the angular momentum of the system after the impact must be conserved. Therefore we can write:

m_1 * v_1 = (m_1 + m_2) * v_2 * R + I * w

where m_1 is the mass of the rod, m_2 is the mass of the ball, v_1 ​​is the speed of the ball before the impact, v_2 is the speed of the ball after the impact, R is the distance from the center of the rod to the point of impact of the ball, I is the moment of inertia of the rod, w is the angular velocity of rotation of the rod after the impact .

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

I = m_1 * L^2 / 12

where L is the length of the rod.

The distance R can be found from geometric considerations:

R = L / 2

The speed of the ball after impact can be found using the law of conservation of energy:

m_1 * v_1^2 / 2 = (m_1 + m_2) * v_2^2 / 2 + I * w^2 / 2

Having solved this system of equations for w and v_2, we obtain answers to the problem:

w = (m_1 * v_1 * L) / (2 * (m_1 + m_2) * I) v_2 = v_1 * (m_1 - (1/3) * m_2) / (m_1 + m_2)

Substituting the numerical values, we get:

w ≈ 2.38 rad/s v_2 ≈ 79.99 m/s

Answer: the angular velocity at which the rod begins to rotate is approximately 2.38 rad/s, and the speed of the ball after impact is approximately 79.99 m/s.

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