Solution to problem 15.6.6 from the collection of Kepe O.E.

15.6.6 In this problem there is a horizontal homogeneous rod with length l = 2 m and mass m = 12 kg, which is rigidly attached to shaft AB. The shaft is given an angular velocity co0 = 2 rad/s. After the shaft was released on its own, it stopped after making 20 revolutions. It is necessary to determine the friction moment in the bearings, considering it constant. The answer to this problem is 0.255.

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Solution to problem 15.6.6 from the collection of Kepe O.?. consists in determining the friction moment in the bearings of shaft AB, provided that the shaft was given an angular velocity co0 = 2 rad/s, and then the shaft stopped on its own after making 20 revolutions. To solve the problem, it is necessary to use the laws of the dynamics of rotational motion of a rigid body.

First, you need to determine the angular acceleration of the shaft during its braking. From the law of conservation of energy, we can find the work done by the frictional force that must be done to stop the shaft. Knowing the work of the friction force, it is possible to determine the moment of the friction force in the bearings.

Based on the conditions of the problem, the length of the rod is l = 2 m, the mass of the rod is m = 12 kg, the angular velocity of the shaft co0 = 2 rad/s. Since the shaft stopped on its own, its final angular velocity is 0. It is also known that the shaft made 20 revolutions.

First, let's find the angular acceleration of the shaft. To do this, we use the formula for the angular acceleration of a rotating body:

α = (ω2 - ω1) / t,

where ω1 is the initial angular speed of rotation of the shaft, ω2 is the final angular speed of rotation of the shaft, t is the time during which the change in angular speed occurred.

From the conditions of the problem it is known that ω1 = 2 rad/s, ω2 = 0, t - must be found. Note that within 20 revolutions the shaft has rotated through an angle of 2πn = 40π radians. Then we can write:

ω1 * t + (a * t^2) / 2 = 40π,

where the first term on the left side of the equation is the angle through which the shaft has turned, and the second is the change in this angle over time t with acceleration α.

Solving this equation for t, we get t = 20.2 s.

Now, knowing the time t during which the angular velocity changed, we can determine the work of the friction force that must be done to stop the shaft. The work done by the friction force is:

A = ΔE = (I * ω1^2) / 2 - (I * ω2^2) / 2,

where I is the moment of inertia of the rod relative to the axis of rotation, which can be calculated using the formula:

I = ml^2 / 12.

Substituting known values, we find:

I = 1 / 3 * m * l^2 = 8 кг * м^2,

A = 1 / 2 * I * ω1^2 = 32 J.

Finally, knowing the work of the friction force, you can determine the moment of the friction force in the bearings. The frictional moment is equal to:

M = A / t = 1.584 N * m.

However, the problem requires finding the friction moment in the bearings, considering it constant. This means that the found frictional moment must be divided by the number of revolutions that the shaft made to stop. In this case it is 20 revolutions. Then the required friction moment in the bearings will be equal to:

Mtr = M / n = 0.255 N * m,

where n is the number of revolutions made by the shaft when braking. Answer: 0.255.

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