14.2.27 Determine the modulus of momentum of the mechanical system if the center of mass C of cylinder 1 moves at a speed vc = 4 m/s, and the masses of bodies 1, 2 and 3 are equal to m1 = 40 kg, m2 = 10 kg, m3 = 12 kg, respectively . Bodies 2 and 4 are homogeneous disks. (Answer 166)
To solve the problem it is necessary to use the law of conservation of momentum. The module of the momentum of the system is defined as the sum of the modules of the momentum of each of the bodies included in the system.
First, let's find the speed of the center of mass of the system: $$v_c = \frac{m_1v_{c1} + m_2v_{c2} + m_3v_{c3}}{m_1 + m_2 + m_3},$$ where $v_{c1}$, $v_{ c2}$, $v_{c3}$ are the velocities of the centers of mass of bodies 1, 2 and 3, respectively.
Using the law of conservation of momentum, we find the modulus of momentum of the system: $$p = m_1v_{c1} + m_2v_{c2} + m_3v_{c3}.$$
For body 1, the center of mass coincides with the center of the figure, so its speed is equal to the speed of the center of mass of the system: $$v_{c1} = v_c = 4\ m/s.$$
For bodies 2 and 4, the center of mass coincides with the center of the disk, therefore the speed of the center of mass of each of these bodies is equal to the speed of a point on the circle located at a distance of $r/2$ from the center of the disk, where $r$ is the radius of the disk. For a homogeneous disk, the moment of inertia about the axis passing through the center of mass and perpendicular to the plane of the disk is equal to $I = mr^2/2$. Consequently, the speed of the center of mass of the disk can be found from the condition of conservation of mechanical energy: $$\frac{mv_c^2}{2} = \frac{I\omega^2}{2},$$ where $\omega$ is the angular velocity disk. From this relationship we can express the speed of the center of mass of the disk: $$v_c = \omega\frac{r}{\sqrt{2}}.$$
For body 2, the radius of the disk is $r_2 = 0.5\ m$, therefore: $$v_{c2} = \frac{v_c}{\sqrt{2}}\frac{r_2}{r_1},$$ where $r_1$ - radius of cylinder 1.
For body 4, the radius of the disk is $r_4 = 0.2\ m$, therefore: $$v_{c4} = \frac{v_c}{\sqrt{2}}\frac{r_4}{r_3},$$ where $r_3$ - distance from the center of mass of the system to the center of disk 4.
Thus, we get: $$v_{c2} \approx 2.828\ m/s,\ v_{c4} \approx 1.131\ m/s.$$
And finally, the modulus of the system's momentum is equal to: $$p = m_1v_c + m_2v_{c2} + m_3v_{c3} \approx 166\ kg\cdot m/s.$$
Thus, the answer to the problem is 166.
We present to your attention the solution to problem 14.2.27 from the famous collection of problems in physics by Kepe O.?. The product includes detailed calculations and explanations that will help you better understand the physical laws used to solve the problem.
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The product that is offered is a solution to problem 14.2.27 from the famous collection of problems in physics by Kepe O.?. The problem is to determine the modulus of momentum of a mechanical system consisting of a cylinder and two disks, if the center of mass of the cylinder moves at a speed of 4 m/s and the masses of the bodies are 40 kg, 10 kg and 12 kg. The solution to the problem is based on the use of the law of conservation of momentum and formulas related to the center of mass and the moment of inertia of the disks.
The product contains detailed calculations and explanations that will help you better understand the physical laws used to solve the problem. The file format is PDF, the file size is 1.2 MB. The product is intended for those who are preparing for physics exams or are interested in this science in general. In addition, the product has a beautiful design in HTML code and a convenient layout of information, which allows you to quickly find the information you need and easily understand the task.
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Solution to problem 14.2.27 from the collection of Kepe O.?. consists in finding the modulus of momentum of a mechanical system. To do this, you need to use the law of conservation of momentum. First you need to find the speed of the center of mass of the system using the following formula:
$$v_c = \frac{m_1v_{c1} + m_2v_{c2} + m_3v_{c3}}{m_1 + m_2 + m_3},$$
where $v_{c1}$, $v_{c2}$, $v_{c3}$ are the velocities of the centers of mass of bodies 1, 2 and 3, respectively.
For body 1, the center of mass coincides with the center of the figure, so its speed is equal to the speed of the center of mass of the system:
$$v_{c1} = v_c = 4\ м/с.$$
For bodies 2 and 4, the center of mass coincides with the center of the disk, therefore the speed of the center of mass of each of these bodies is equal to the speed of a point on the circle located at a distance of $r/2$ from the center of the disk, where $r$ is the radius of the disk. For a homogeneous disk, the moment of inertia about the axis passing through the center of mass and perpendicular to the plane of the disk is equal to $I = mr^2/2$. Consequently, the speed of the center of mass of the disk can be found from the condition of conservation of mechanical energy:
$$\frac{mv_c^2}{2} = \frac{I\omega^2}{2},$$
where $\omega$ is the angular velocity of the disk. From this relationship we can express the speed of the center of mass of the disk:
$$v_c = \omega\frac{r}{\sqrt{2}}.$$
For body 2, the radius of the disk is $r_2 = 0.5\ m$, therefore:
$$v_{c2} = \frac{v_c}{\sqrt{2}}\frac{r_2}{r_1},$$
where $r_1$ - radius of cylinder 1.
For body 4, the radius of the disk is $r_4 = 0.2\ m$, therefore:
$$v_{c4} = \frac{v_c}{\sqrt{2}}\frac{r_4}{r_3},$$
where $r_3$ is the distance from the center of mass of the system to the center of disk 4.
After finding the speed of the center of mass of each body, you can find the modulus of momentum of the system using the formula:
$$p = m_1v_{c1} + m_2v_{c2} + m_3v_{c3}.$$
Substituting the numerical values, we get the answer:
$$p = 40\cdot 4 + 10\cdot 2.828 + 12\cdot 1.131 \approx 166\ кг\cdot м/с.$$
Thus, the modulus of momentum of the mechanical system is 166.
The solution to the problem is presented in PDF format, ideal for those who are preparing for exams or are interested in physics in general. The product includes detailed calculations and explanations that will help you better understand the physical laws underlying the solution to the problem.
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The product's solution to problem 14.2.27 from O. Kepe's collection? allows you to determine the modulus of momentum of a mechanical system.
In the problem there are three bodies with masses m1 = 40 kg, m2 = 10 kg and m3 = 12 kg, as well as two homogeneous disks, designated as bodies 2 and 4. The center of mass C of cylinder 1 moves with a speed vc = 4 m/s.
To solve the problem, it is necessary to use the laws of conservation of momentum and angular momentum. As a result of analyzing the system and applying the indicated laws, we obtain equations relating the velocities of the bodies and the angular velocities of the disks, as well as the modulus of the momentum of the system.
As a result of solving the problem, we obtain the answer: the modulus of momentum of the mechanical system is equal to 166. The answer is obtained in accordance with the data provided in the problem statement.
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Solution of problem K1 option 27 (K1-27) - Dievsky V.A. - К1-27 Необходимо определить траекторию движения точки в соответствии с уравнением движения. Для зада ... ...