Solution to problem 15.5.6 from the collection of Kepe O.E. 1989

15.5.6. Rotating crank 1 of a hinged parallelogram with a length OA = 0.4 m has an angular velocity of rotation ω1 = 10 rad/s around the axis O. The moments of inertia of cranks 1 and 3 relative to their axes of rotation are 0.1 kg m2. The connecting rod has a mass of 2 m2 = 5 kg. It is necessary to determine the kinetic energy of the mechanism.

To solve this problem, it is necessary to use the formula for the kinetic energy of a system of bodies:

T = Σ(1/2) m v² + Σ(1/2) I ω²,

where Σ(1/2)·m·v² is the kinetic energy of translational motion, Σ(1/2)·I·ω² is the kinetic energy of rotational motion.

Crank mass 1 m2 = 2 m2 = 10 kg. The speed of point A on the circle is equal to v = ω1·OA = 10·0.4 = 4 m/s.

The kinetic energy of translational motion is equal to:

Tpost = (1/2) m v² = (1/2) 10 4² = 80 J.

The moment of inertia of the connecting rod relative to its axis of rotation is equal to:

Ish = (1/12) m2 L² = (1/12) 5 0.4² = 0.0333 kg m².

The moment of inertia of crank 1 relative to its axis of rotation is equal to:

I1 = (1/12)·m1·L² + m1·(L/2)² = (1/12)·10·0.4² + 10·(0.4/2)² = 0.7667 kg·m² .

The kinetic energy of rotational motion is equal to:

Tvracht = (1/2)·(Ish + I1)·ω1² = (1/2)·(0.0333 + 0.7667)·10² = 40 J.

Thus, the kinetic energy of the mechanism is equal to:

T = Tpost + Trot = 80 + 40 = 120 J.

Solution to problem 15.5.6 from the collection of Kepe O.E. 1989

This digital product is a solution to one of the problems from the collection of Kepe O.E. 1989. The solution was completed by a professional specialist in the field of mechanics and represents an accurate and detailed description of the solution to problem 15.5.6.

The solution uses formulas and methods of mechanics, and also provides the calculations necessary to obtain the final result. The solution is made in accordance with the requirements of modern science and represents valuable material for students and specialists in the field of mechanics.

By purchasing this digital product, you get access to a detailed and high-quality solution to problem 15.5.6 from the collection of Kepe O.E. 1989, which can be used as material for studying mechanics and solving similar problems in the future.

Author: Mechanical Professional

Year of manufacture: 1989

Format: PDF

Number of pages: 2

Price: 50 rubles

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Solution to problem 15.5.6 from the collection of Kepe O.E. 1989 is to determine the kinetic energy of the mechanism, which consists of a crank 1 articulated parallelogram with a length OA = 0.4 m. The crank rotates uniformly around the axis O with an angular velocity ω1 = 10 rad/s. The mass of the connecting rod 2 m2 = 5 kg, and the moments of inertia of cranks 1 and 3 relative to their axes of rotation are equal to 0.1 kg m2.

To solve the problem, it is necessary to determine the kinetic energy of each element of the mechanism, and then add them up. For the crank and connecting rod, kinetic energy is determined by the formula:

E = (1/2) * I * ω^2,

where E is the kinetic energy, I is the moment of inertia, ω is the angular velocity of the element.

For crank 1 the kinetic energy will be equal to:

E1 = (1/2) * 0.1 * 10^2 = 5 J.

For the connecting rod, the kinetic energy will be equal to:

E2 = (1/2) * m2 * v^2,

where v is the connecting rod speed. The speed of the connecting rod can be determined from the equation of motion of the connecting rod:

v = r * ω1,

where r is the radius of the crank. Thus,

v = 0.4/2 * 10 = 2 m/s.

Substituting the speed value into the formula for kinetic energy, we get:

E2 = (1/2) * 5 * 2^2 = 10 J.

Thus, the kinetic energy of the mechanism will be equal to:

E = E1 + E2 = 15 J.

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