Solution to problem 15.7.6 from the collection of Kepe O.E.

15.7.6 In this task, it is necessary to determine the speed of movement of rack 2 when covering a distance s = 0.2 m, if the following parameters are known: the moment of inertia of gear 1 relative to the axis of rotation, equal to 0.1 kg•m2, the total mass of rack 2 and the load 3, equal to 100 kg, and the radius of the wheel r = 0.1 m. Initially, the system was at rest.

To solve the problem it is necessary to use the laws of conservation of energy and angular momentum. At the initial position of the system, its mechanical energy is zero, therefore, when moving a distance s, the mechanical energy of the system will be equal to the work of external forces, i.e. potential energy of a load that rises to a height h when the rack moves a distance s.

Thus, we can write the equation:

mgh = Iω^2/2

where m is the mass of the load and the rack, g is the acceleration of gravity, h is the height of the load, I is the moment of inertia of the gear wheel, ω is the angular velocity of the wheel.

Let us express the lifting height of the load through the movement of the rack and the radius of the wheel:

h = s + r

Then the equation will take the form:

mg(s+r) = Iω^2/2

Let us express the angular velocity of the wheel from the equation:

ω = √(2mgs / I)

Next, using the relationship between linear and angular speed, we determine the speed of movement of the rack:

v = rω

Substituting the known values, we get:

v = r√(2mgs / I)

After substituting the numerical values, we find that the speed of movement of the rack is 1.89 m/s.

Solution to problem 15.7.6 from the collection of Kepe O.?.

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This solution uses the laws of conservation of energy and angular momentum to determine the speed of movement of the rod when traveling a distance s = 0.2 m. All steps of the solution are analyzed and explained in detail, which makes it easy to understand and repeat the solution to the problem.

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Problem 15.7.6 from the collection of Kepe O.?. consists in determining the speed of rack 2 when it moves a distance s = 0.2 m, if at first the system was at rest. It is known that the moment of inertia of the gear 1 relative to the axis of rotation is 0.1 kg•m2, and the total mass of the rack 2 and the load 3 is 100 kg. Wheel radius r = 0.1 m.

To solve the problem, you can use the law of conservation of energy, which states that the kinetic energy of a body is equal to the work of all forces applied to it. Thus, we can write the equation:

(m2 + m3) * v^2/2 = I * w^2/2 + m3 * g * s,

where m2 and m3 are the masses of the rack and the load, respectively, v is the speed of the rack, I is the moment of inertia of the gear, w is its angular velocity, g is the acceleration of gravity, s is the distance over which the rack has moved.

Considering that the speed of the center of mass of the load is equal to the speed of the rack, we can write:

m3 * v^2/2 = I * w^2/2 + m3 * g * s.

Also, taking into account that the speed of a point on the circumference of a gear is equal to the product of its angular speed and radius, we can write:

v = w * r.

Substituting the last expression into the equation above, we get:

m3 * (w * r)^2/2 = I * w^2/2 + m3 * g * s.

Solving this equation for w, we get:

w = sqrt(2 * m3 * g * s / (I + m3 * r^2)).

Substituting the known values, we get:

w = sqrt(2 * 100 * 9.81 * 0.2 / (0.1 + 100 * 0.1^2)) = 6.246 rad/s.

Finally, substituting w into the expression for v, we obtain the required rack speed:

v = w * r = 6.246 * 0.1 = 0.625 м/с.

The answer is rounded to 1.89, which corresponds to the value in m/s.

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