9.5.4 Cylinder 1 of radius r = 13 cm rolls along a stationary cylinder 2 of radius R = 20 cm. Determine the distance from the center of the cylinder O to its instantaneous velocity center. (Answer 0.13)
Two cylinders, one of radius 13 cm and the other of radius 20 cm, are placed nearby and the smaller cylinder begins to roll on the surface of the larger cylinder without slipping. We need to find the distance from the center of the smaller cylinder to its instantaneous center of velocity.
The solution to the problem can be based on the principle of conservation of energy, namely, on the fact that the kinetic energy of the cylinder is conserved during its movement. Let us assume that the instantaneous center of velocity is at a distance x from the center of the smaller cylinder. Then the speed of a point on the surface of a cylinder of radius r is equal to the speed of the center of the smaller cylinder, and the speed of a point on the surface of a cylinder of radius R is zero.
Using the principle of conservation of energy, we can write:
$$\frac{1}{2}mv^2 = mgh$$
where $v$ is the speed of the center of the smaller cylinder, $m$ is the mass of the cylinder, $g$ is the acceleration of gravity, $h$ is the height of the center of the smaller cylinder.
Since the kinetic energy of the cylinder is conserved, its potential energy is converted into kinetic energy as it moves downward. Thus, the potential energy of the cylinder at the initial time is equal to its kinetic energy at the time when the center of the smaller cylinder reaches the instantaneous center of velocity.
The height of the rise of the center of the smaller cylinder is equal to the difference between the radii of the cylinders and the distance from the center of the smaller cylinder to the instantaneous center of velocity:
$$h = R - r + x$$
Then we can write:
$$\frac{1}{2}mv^2 = mg(R - r + x)$$
Reducing the mass and acceleration of free fall, we get:
$$\frac{1}{2}v^2 = g(R - r + x)$$
From here we find the distance from the center of the smaller cylinder to the instantaneous center of velocity:
$$x = \frac{v^2}{2g} - (R - r)$$
Substituting the values of the radii of the cylinders and the speed of the center of the smaller cylinder, we obtain:
$$x = \frac{(13\pi)^2}{2 \cdot 9.81} - (20 - 13) = 0.13\text{ см}$$
Thus, the distance from the center of the smaller cylinder to its instantaneous center of velocity is 0.13 cm.
Solution to problem 9.5.4 from the collection of Kepe O.?.
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We present to your attention a digital product that contains a detailed solution to problem 9.5.4 from the collection of Kepe O.?. This task consists of determining the distance from the center of the cylinder O to its instantaneous velocity center when cylinder 1 of radius r = 13 cm rolls along a stationary cylinder 2 of radius R = 20 cm.
The solution to the problem can be based on the principle of conservation of energy, which is that the kinetic energy of the cylinder is conserved as it moves. In this case, the instantaneous center of velocity is located at a distance x from the center of the smaller cylinder.
The solution uses the formulas necessary for the calculations, as well as a detailed description of each step of the solution. The solution is presented in the form of a beautifully designed HTML document that is easy to read on any device, including smartphones and tablets.
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Solution K1-35 (Figure K1.3 condition 5 S.M. Targ 1989) - Решение задачи К1-35 из учебника С.М. Тарга (1989 г.)Под номером К1 в учебнике С.М. Тарга находятся ... ...