When solving the equations of rotational or oscillatory (oscillating) motion, it is necessary to know the moment of inertia of the system under consideration. This article is devoted to the study of various kinds of pendulums and the moment of inertia with which they are characterized.
The concept of the pendulum. Kinds
Before giving a determination of the moment of inertia of the pendulum, it is necessary to consider what this device is. In physics, it means absolutely any system that can oscillate or rotate around a point or axis under the influence of a gravitational field, that is, gravity. This definition implies that the pendulum must necessarily have a finite mass, while the center of mass of the system should not be at the point through which the axis of rotation passes.
There are various types of pendulums. In this article, we will consider only 3 of them:
- math, or simple;
- physical (for example, a homogeneous rod);
- Oberbek's pendulum.
The first two are oscillatory-type pendulums, the third - rotational.
Rotation and moment of inertia
When a body with a certain mass begins to rotate around its axis, then its motion is usually described by the following equation:
M = I * α.
Here M is the total, or resulting, moment of all external forces that act on the system, I is its moment of inertia, and α is the angular acceleration.
The moment of force M, by definition, is a value equal to the product of the acting force on the shoulder, which is equal to the distance from the point of applied force to the axis of rotation.
The moment of inertia is a quantity characterizing the inertial properties of the system, that is, how quickly it can be untwisted by applying a certain moment M. Also I characterizes the kinetic energy stored by the rotating system. The moment of inertia I for a material point (an imaginary object whose mass is concentrated in an infinitely small volume of space) that performs circular motion at a distance from the r axis can be calculated by the following formula:
I = m * r 2 .
In the general case, when determining I for an arbitrary shape body, one should use the following expressions:
1) I = ∑m i * r i 2 .
2) I = ∫dm * r i 2 = ρ * ∫dV * r i 2 .
The first equality is used for a discrete arrangement of masses in the system, the second - for continuous.
It can be seen from these expressions that I is a function of the distance to the axis of rotation and the mass distribution in the system relative to this axis and does not depend on the applied moments of forces M, nor on the rotation speed ω.
Mathematical (simple) pendulum
Since this type of oscillatory system is the simplest, we will consider it in more detail. The mathematical pendulum is a material point that is suspended on a weightless and inextensible thread. If this point is slightly deviated from the equilibrium position and then released, then it will begin to oscillate. It is also assumed that there is no friction force at the point of attachment of the thread, and air resistance is neglected.
As is clear from the description above, a mathematical pendulum is an ideal case that is not realized in practice. Nevertheless, its study allows us to obtain some important conclusions for the type of motion under consideration.
The figure below shows this pendulum, and also the forces acting in the system when it oscillates.
Applying the equation of motion to it, we obtain the following equality:
M = -m * g * sin (θ) * L; I = m * L 2 ; α = d 2 θ / dt 2 =>
=> -m * g * sin (θ) * L = m * L 2 * d 2 θ / dt 2 , whence:
L * d 2 θ / dt 2 + g * sin (θ) = 0.
Let us explain some points: the moment of force from the thread tension T (see. Fig.) Is equal to zero, since it acts directly on the axis; the moment of gravity is taken with a minus sign, since it is directed clockwise; L is the length of the thread; angular acceleration α, by definition, is the second derivative of the angle of rotation in time or the first derivative in time of the angular velocity ω; the formula for the moment of inertia of a pendulum of this type coincides with that for a material point with mass m located at a distance L from the axis of rotation.
The expression obtained above can be simplified if we take the approximation: sin (θ) ≈θ. It is true when the angles of oscillation are small (up to θ = 10 o the error does not exceed 0.5%). In this case we get:
L * d 2 θ / dt 2 + g * θ = 0.
We have obtained the second-order classical differential equation (differential ur.). Its solution is the sine function:
θ = A * sin (ω * t + θ 0 ).
Here A and θ 0 are the oscillation amplitude and the initial angle of deviation from equilibrium, respectively. If this decision is substituted in the differential. ur higher, then you can get the angular velocity and the oscillation period:
ω = √ (g / L) and T = 2 * pi / ω = 2 * pi * √ (L / g).
We obtained an amazing result: the oscillation period of the mathematical pendulum does not depend on the initial conditions (A and θ 0 ), as well as on the mass m.
The behavior of a mathematical pendulum was first studied by Galileo. Subsequently, Huygens showed the possibility of using the obtained formula to determine the acceleration of gravity of the Earth.
General physical pendulum
This device is a solid body of arbitrary shape (its mass can be unevenly distributed over its volume), which oscillates about a horizontal axis that does not pass through the center of mass of the body.
When solving the equation of motion of this device, an ideal object is considered, the mass of which is concentrated in its center of gravity. Such an assumption leads to the following formula for the period of its oscillation:
T = 2 * pi * √ (I o / (m * g * h)).
Here h is the distance from the center of gravity to the axis of rotation O, I o is the moment of inertia of the physical pendulum. Note that if, for calculating the moment of gravity, you can use the additivity property of this quantity and reduce the sum of all moments to one applied to the center of gravity, then you cannot do this to calculate the moment of inertia I o , it should be calculated using the general formulas that were given earlier.
Oscillating rod and its moment of inertia
Imagine that there is a solid rod of mass m and length L, which is suspended vertically from one of the ends. This design is capable of oscillating under the influence of gravity.
If we apply integration with respect to the axis to such a rod, we can obtain that the moment of inertia of the pendulum of the physical structure indicated is equal to:
I o = m * L 2/3.
Then its period of oscillation will be equal to:
T = 2 * pi * √ (2 * L / (3 * g)).
Oberbek Pendulum
The figure below shows this kind of pendulum.
It can be seen from the figure that if a load is suspended from a thread, then 4 rods with loads begin to rotate with some angular acceleration.
Oberbek’s pendulum is used to carry out laboratory work in physics in order to verify the equation of rotational motion.
Determination of the moment of inertia of the Oberbek pendulum
To solve this problem, it is necessary to make an important approximation: the weight of the rods and disks to which it is suspended on the strands of overloads is negligible in comparison with the weight of one load m. Given that the size of the cargo is much less than their distance to the axis of rotation, you can use the formula for the moment of inertia of the material point. Since there are 4 cargoes and all of them have the same mass, but are located at different distances from the axis, we obtain the following formula for the moment of inertia of the Oberbek pendulum:
I = I 1 + I 2 + I 3 + I 4 = m * (R 1 2 + R 2 2 + R 3 2 + R 4 2 ).
Since this pendulum allows you to adjust the position of each load on the rod, its moment of inertia can vary.