Solution to problem 17.1.18 from the collection of Kepe O.E.

17.1.18 It is necessary to determine the angle of deflection of the rod AM with a point mass M at the end from the vertical axis of rotation in degrees. The shaft OA together with the rod AM rotates uniformly with an angular velocity ω = 4.47 rad/s, and the length l is 0.981 m. The mass of the rod AM can be neglected. (Answer 60)

Answer:

The deflection angle of the rod can be determined using the moment balance equation. The moment of gravity M is determined by the formula:

М = mgl sin α,

where m is the mass of point M, g is the acceleration of gravity, l is the length of the rod, α is the angle of deviation from the vertical axis.

The moment of inertia I of the rod can be determined by the formula:

I = ml^2/3.

The moment of inertia of the shaft OA can be neglected.

The moment balance equation has the form:

M = Iα''',

where α''' is the angular acceleration of the rod.

Angular acceleration can be determined by the formula:

α'' = ω^2 α,

where ω is the angular speed of rotation of the shaft OA.

Substituting the obtained values, we get:

mgl sin α = ml^2/3 α''',

from which follows:

α = 3g sin α / (2l ω^2).

Substituting the known values, we get:

α ≈ 60 degrees.

Solution to problem 17.1.18 from the collection of Kepe O.?.

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Problem 17.1.18 from the collection of Kepe O.?. consists in determining the angle of deflection of a rod with a point mass at the end from the vertical axis of rotation with uniform rotation of the shaft together with the rod. The following parameters are given: angular speed of shaft rotation ω = 4.47 rad/s, rod length l = 0.981 m, rod mass AM is neglected. It is required to determine the angle a in degrees.

To solve the problem it is necessary to use the laws of dynamics of rotational motion. It is known that angular velocity is related to the rotation angle and time as follows: ω = Δθ/Δt. It is also known that the moment of inertia of the rod relative to the axis of rotation is equal to I = (1/3)ml^2.

Using the formula for the moment of inertia and the law of conservation of energy, we can express the angle of deviation of the rod from the vertical:

1/2 * I * ω^2 * sin^2(a) = mgh

where m is the mass of point M at the end of the rod, g is the acceleration of gravity, h is the height of the rise of point M relative to the equilibrium position.

For point M at the end of the rod h = l * (1 - cos(a)). Substituting this formula, as well as the expression for the moment of inertia, we get:

1/2 * (1/3)ml^2 * ω^2 * sin^2(a) = mgl * (1 - cos(a))

Simplifying the expression and bringing it to the form sin(a) = 1/2, we find the value of the angle of deviation of the rod from the vertical:

sin(a) = sqrt((mgl)/(2/3 * ml^2 * ω^2)) = sqrt(3/8) ≈ 0.866 a = arcsin(0.866) ≈ 60°

Thus, the angle of deviation of the rod from the vertical is approximately 60 degrees.

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