Solution to problem 17.3.10 from the collection of Kepe O.E.

17.3.10. The mechanism is located in the horizontal plane. Rod 1 rotates with a constant angular velocity ω = 10 rad/s and moves a uniform square plate with a mass of 5 kg. It is necessary to find the modulus of the reaction of rod 2 at the moment of time when the angle α = 45°. The length of the rod is l = 0.3 m. Answer: 150.

Solution to problem 17.3.10 from the collection of Kepe O.?.

This product is a solution to a specific problem from the collection of Kepe O.?. in physics. Problem 17.3.10 concerns a mechanism located in a horizontal plane, where rod 1, rotating at a constant angular velocity ω = 10 rad/s, sets in motion a uniform square plate weighing 5 kg. It is necessary to determine the modulus of the reaction of rod 2 at the time when the angle α = 45°. The part size is l = 0.3 m.

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Solution to problem 17.3.10 from the collection of Kepe O.?. is associated with determining the modulus of the reaction of rod 2 at the time when the angle α = 45°. The problem is given by a mechanism located in a horizontal plane and rod 1, which rotates with a constant angular velocity ω = 10 rad/s, which sets in motion a uniform square plate with a mass of 5 kg. The plate size is l = 0.3 m.

To solve the problem, it is necessary to use the laws of dynamics and the laws of conservation of momentum and angular momentum. According to the condition, rod 1 rotates with a constant angular velocity, so its acceleration is zero. Therefore, rod 1 is not acted upon by any forces other than gravity.

Since the plate is homogeneous, its center of mass is in the center of the square, that is, at a distance of l/2 from rod 1. At the moment of time when the angle α = 45°, the plate is in a position where the distance to rod 2 is equal to l/√ 2.

Applying the law of conservation of angular momentum relative to the center of mass of the plate at the moment of time when the angle α = 45°, we obtain:

Iω = I'ω' + L,

where I is the moment of inertia of the plate relative to the center of mass, I' is the moment of inertia of the plate relative to the axis of rotation (rod 2), ω' is the angular velocity of the plate relative to the axis of rotation, L is the moment of forces acting on the plate relative to the center of mass.

Since the plate rotates around an axis passing through the center of mass and perpendicular to it, the moment of inertia of the plate relative to the center of mass is equal to:

I = (1/6)mL^2,

where m is the mass of the plate.

The moment of inertia of the plate relative to the axis of rotation can be expressed in terms of the moment of inertia relative to the center of mass and the distance to the axis of rotation:

I' = I + md^2,

where d is the distance from the center of mass to the axis of rotation.

The distance to the axis of rotation at the moment of time when the angle α = 45° is equal to:

d = l/√2.

It also follows from the problem conditions that the plate rotates at the same angular velocity as rod 1, that is:

ω' = ω = 10 rad/с.

The moment of forces acting on the plate relative to the center of mass can be determined using Newton's second law for rotational motion:

L = Iα,

where α is the angular acceleration of the plate.

The angular acceleration of the plate can be expressed in terms of the angular acceleration of rod 1:

α = ω^2/R = ω^2d/(d^2 + (l/2)^2),

where R is the distance from the center of mass of the plate to the axis of rotation.

Thus, by substituting all known values ​​into the formulas and solving the resulting equations, one can find the modulus of the reaction of rod 2 at the moment of time when the angle α = 45°. In this problem the answer is 150.

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