Solution to problem 9.5.1 from the collection of Kepe O.E.

Solution to problem 9.5.1 from the collection of Kepe O.?.

This digital product is a solution to problem 9.5.1 from the collection of problems in physics by Kepe O.?. The solution was written by a professional teacher and describes in detail the process of solving the problem.

The problem is to determine the distance from the geometric center of the disk to the instantaneous center of velocities when a disk of radius 50 cm moves along a plane. Solving this problem will help students and schoolchildren better understand the concept of an instantaneous center of velocities and correctly apply it in solving similar problems.

By purchasing this product, you will have access to a complete and detailed solution to the problem, which is presented in an easy-to-read format. You can use it as a sample for performing similar tasks or as additional material for preparing for exams and Olympiads in physics.

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This digital product is a solution to problem 9.5.1 from the collection of problems in physics by Kepe O.?. The problem is to determine the distance from the geometric center of the disk to the instantaneous center of velocities when a disk of radius 50 cm moves along a plane. The solution was completed by a professional teacher and presented in an easy-to-read format. Solving this problem will help students and schoolchildren better understand the concept of an instantaneous center of velocities and correctly apply it in solving similar problems.

To solve the problem it is necessary to use the concept of an instantaneous velocity center. To determine the distance from the geometric center of the disk to the instantaneous center of velocities, it is necessary to draw a perpendicular to the axis of rotation of the disk from the geometric center of the disk and determine the point of intersection of this perpendicular with the straight line passing through the point of contact of the disk with the plane and the instantaneous center of velocities. The distance from the geometric center of the disk to the instantaneous velocity center is equal to half the radius of the disk, that is, 0.5 cm.

By purchasing this product, you will get access to a complete and detailed solution to the problem, which can be used as a sample for performing similar tasks or as additional material for preparing for exams and Olympiads in physics.

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Solution to problem 9.5.1 from the collection of Kepe O.?. is associated with determining the distance from the geometric center of a disk of radius R = 50 cm to the instantaneous center of velocities when the disk rolls along a plane.

The instantaneous center of velocities is the point on the disk at which the velocity of motion is zero. It is the center of curvature of the disk's trajectory.

To solve the problem, you can use the formula for the radius of curvature of the motion trajectory:

R = v^2 / a,

where v is the speed of movement of the disk, and is the acceleration caused by friction.

Since the velocity at the instantaneous velocity center is zero, the acceleration a can be found from the formula for velocity:

v = ωR,

where ω is the angular speed of rotation of the disk.

Substituting the expression for speed into the expression for the radius of curvature, we get:

R = (ωR)^2 / a,

a = ω^2R.

Now you can find the distance from the geometric center of the disk to the instantaneous center of velocities, which is equal to R - r, where r is the distance from the center of the disk to the geometric center.

For a disk of radius R = 50 cm, the answer is 0.5 cm.

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