Solution to problem 15.7.6 from the collection of Kepe O.E. 1989

In this problem, it is required to determine the speed of movement of a rack at a given distance, if the system began to move from 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 radius of the wheel is 0.1 meters. The total mass of rack 2 and load 3 is 100 kg.

To solve the problem it is necessary to use the laws of conservation of energy and angular momentum. When the system moves, the angular momentum remains constant. Thus, we can write the equation:

Iω = mvr

where I is the moment of inertia of the wheel, ω is its angular velocity, m is the mass of the system, v is the speed of the rack, r is the radius of the wheel.

You can also write the energy conservation equation:

mgh = 1/2Iω2 + 1/2mv2

where h is the lifting height of the rack.

From the equation for conservation of angular momentum we obtain:

ω = mvr / I

Substituting this expression for angular velocity into the energy conservation equation, we obtain:

mgh = 1/2mv2 + 1/2mr2(mv/I)2

Solving the equation for v, we get:

v = √(2gh / (1 + mr2/I))

Substituting the known values, we get:

v = √(2 * 9.81 * 0.2 / (1 + 100 * 0.12 / 0.1)) ≈ 1.25 m/s

Thus, the speed of movement of the rack when it moves a distance s = 0.2 m is approximately 1.25 m/s.

Solution to problem 15.7.6 from the collection of Kepe O.E. 1989

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In this problem, it is necessary to determine the speed of movement of the rack when it moves a distance s = 0.2 m, if the system began to move from rest. To solve the problem, the laws of conservation of energy and angular momentum are used.

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Solution to problem 15.7.6 from the collection of Kepe O.E. 1989 is to determine the speed of movement of the rack when it moves a distance s = 0.2 m, if the system began to move from rest. To solve the problem, the laws of conservation of energy and angular momentum are used.

It is known that the moment of inertia of gear 1 relative to the axis of rotation is 0.1 kg m2, the total mass of rack 2 and load 3 is 100 kg, and the radius of the wheel is r = 0.1 m.

By writing the equation for the conservation of angular momentum, we can obtain an expression for the angular velocity ω = mvr / I. Substituting this expression for the angular velocity into the equation for the conservation of energy, we obtain the equation mgh = 1/2mv^2 + 1/2mr^2(mv/I)^ 2, where h is the height of the rack.

Solving the equation for speed v, we obtain v = √(2gh / (1 + mr^2/I)). Substituting the known values, we get v = √(2 * 9.81 * 0.2 / (1 + 100 * 0.12 / 0.1)) ≈ 1.25 m/s.

Thus, the speed of movement of the rack when it moves a distance s = 0.2 m is approximately 1.25 m/s.

By purchasing the digital product “Solution to problem 15.7.6 from the collection of Kepe O.E. 1989”, you will receive a detailed and understandable solution to the problem, done by hand. This product will help you better understand the material and successfully cope with school assignments.

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This product is a solution to problem 15.7.6 from the collection of Kepe O.E. 1989 in terms of dynamics. In the problem, the moment of inertia of gear 1 relative to the axis of rotation is known, which is equal to 0.1 kg m2, the total mass of rack 2 and load 3 is equal to 100 kg, as well as the radius of the wheel r = 0.1 m. It is necessary to determine the speed of the rack at its moving a distance s = 0.2 m, if at first the system was at rest.

After payment, you will receive the solution to Kepe's problem No. 15.7.6 in the form of a picture in PNG format, handwritten in clear and legible handwriting. The solution is made in accordance with the theorem on the change in kinetic energy of a mechanical system. After purchasing the solution, you can leave a positive review and receive a discount on the next task.

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