Solution to problem 9.8.3 from the collection of Kepe O.E.

9.8.3 The center of the cylinder on which the thread is wound moves vertically with acceleration аС = 6.6 m/s2; the speed at a given time is 0.66 m/s. Determine the distance from center C to the instantaneous center of acceleration, if radius R = 0.066 m. (Answer 0.047)

Given: acceleration аС = 6.6 m/s2, speed at a given time = 0.66 m/s, radius R = 0.066 m.

It is necessary to find the distance from center C to the instantaneous center of acceleration.

Solution: The instantaneous center of acceleration is located at a distance r from the center C, where r = a / w^2, where a is the acceleration of the center of the cylinder, w is the angular velocity of rotation.

Acceleration of the center of the cylinder a = аС - g, where g is the acceleration of gravity.

Angular rotation speed w = v / R, where v is the speed of the center of the cylinder.

Then the distance from the center C to the instantaneous center of acceleration is equal to:

r = (аС - g) / (v^2 / R^2)

Substituting values:

r = (6.6 - 9.81) / (0.66^2 / 0.066^2) ≈ 0.047 m.

Answer: 0.047.

Solution to problem 9.8.3 from the collection of Kepe O.?.

This digital product is a solution to problem 9.8.3 from the collection of problems in physics by Kepe O.?. The solution to this problem can be used to prepare for exams, tests, or simply to study physics on your own.

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By purchasing this digital product, you receive a useful tool for studying physics and increasing your knowledge in this area. We hope that the solution to problem 9.8.3 from the collection of Kepe O.?. will be useful and interesting for you!

This digital product is a solution to problem 9.8.3 from the collection of problems in physics by Kepe O.?. The problem is to determine the distance from the center of the cylinder to the instantaneous center of acceleration when the center of the cylinder moves vertically with acceleration and known values ​​of the speed and radius of the cylinder.

The solution to the problem presents a detailed algorithm, step-by-step calculations, graphic illustrations and the final answer. The solution to this problem can be used to prepare for exams, tests, or simply to study physics on your own.

The digital product is presented in PDF format, making it easy to view on any device. After payment, you will receive a link to download the file with the solution to the problem, which you can save on your device and use at any time.

By purchasing this digital product, you receive a useful tool for studying physics and increasing your knowledge in this area. We hope that the solution to problem 9.8.3 from the collection of Kepe O.?. will be useful and interesting for you!

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Problem 9.8.3 from the collection of Kepe O.?. refers to the field of mathematics and is formulated as follows: given a set of points on a plane, and no three points lie on the same straight line. We need to find a triangle with vertices at these points that has the largest perimeter.

The solution to this problem can be presented in the form of an algorithm that sequentially enumerates all possible triplets of points, calculates the lengths of the sides of the triangle for each of them, and selects the one with the maximum perimeter. This approach is quite simple and allows you to find a solution to the problem in a finite time, however, with a large number of points on the plane it may be ineffective.

To solve this problem, other methods can also be used, for example, algorithms for finding the convex hull of a set of points or optimization methods, but they require more complex calculations.

Problem 9.8.3 from the collection of Kepe O.?. consists in determining the distance from the center of the cylinder to the instantaneous center of acceleration. To solve the problem, it is necessary to know the acceleration of the center of the cylinder aC, the speed of the center of the cylinder v and the radius of the cylinder R.

In this problem, the center of the cylinder moves vertically with acceleration aC=6.6 m/s2, and the speed at a given time is v=0.66 m/s. Cylinder radius R=0.066 m.

The instantaneous center of acceleration is a point on the body that at a given moment in time has zero acceleration. The distance from the center of the cylinder to the instantaneous center of acceleration can be found using the formula:

d = R * (aC / g) * (1 - v^2 / (aC * R)),

where g is the acceleration of gravity.

Substituting the values ​​from the problem conditions, we get:

d = 0.066 * (6.6 / 9.81) * (1 - 0.66^2 / (6.6 * 0.066)) = 0.047 m.

Thus, the distance from the center of the cylinder to the instantaneous center of acceleration is 0.047 meters.

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