On establishing equations of motion of mechanical vibration systems placed on moving bases

Abstract

The first author has been teaching the postgraduate course, “The Dynamics of Mechanical Systems” in The ITU Faculty of Mechanical Engineering for nearly 20 years. He has observed that students frequently have problems in obtaining the equations of motion of the vibrating systems which were placed on moving bases. Starting from this observation, he has found that the homework stated below, which was given to the students occasionally, was very helpful in learning the subject. The main idea of the homework is the derivation of the equations of motion, with the help of formulating the Lagrange’s equations with respect to a moving set of axis for a vibration system with two degrees of freedom which consists of a horizontal table rotating with a constant angular velocity around a vertical axis. The students were also asked to solve the same problem with a different method of their choice and to determine the reaction forces as well. We want to share this problem with the reader, which we have assessed as very instructive and appropriate from the viewpoint of applicability of different methods.

Keywords

Lagrange’s equations with respect to moving reference systems Lagrangian form of the D’Alembert’s principle Hamilton’s principle Hamilton’s canonical equations

Introduction

The first author, who has been giving the ‘Dynamics of Mechanical System’ course for years, gave the homework which involved the derivation of the equations of motion of the mechanical system shown in Figure 1.

The mechanical vibration system is placed on a table which rotates with a constant ω angular velocity around a vertical z axis, bedded at point O. There is a channel on the table, as it seen in Figure 1, at distance from the bedding point O and perpendicular to the corresponding radius. Within the channel, there is a moving guide which can move in the channel. This moving guide is connected to the table with two springs k₁. Within the moving guide, there is a moving point mass m₂ which is connected to the guide with two springs k₂.

The students are asked to derive the equations of motion of the two degrees of freedom system subject to small oscillations around the equilibrium position $O'$ on a moving background and then to find the corresponding reaction forces (constraint forces).

The first author, through long years of teaching in this course, has observed that students find it hard to derive the equations of motion of vibrating systems on a moving background. As a first step in this direction, the first author published an article many years ago in this journal.¹ Pursuing the same idea, the students were asked to derive the equations of motion by using two different methods, one of which is “the formulation of the Lagrange’s equations with respect to moving reference systems.” Within the context of this homework, it was also asked that the students find the constraint forces applied by the table to the moving guide and the constraint force applied by the moving guide to the mass m₂.

The homework was given most recently in the 2015/2016 Fall term, and the number of students in the graduate class was 25.

Besides the method which they were asked to obtain the equations of motion, the other methods that the students chose are as follows: Lagrange’s equations (10 students), Lagrangian form of the D’Alembert’s principle (nine students), Hamilton’s principle (four students) and Hamilton’s canonical equations (two students).

It was observed with satisfaction that those students who solved the homework problem largely overcame the problems that were mentioned at the beginning and grasped the subject better.

The $Oxyz$ set of axis is a set of axis which is attached to the table rotating with an angular velocity ω.

Part of the students (those with odd numbers, namely 13) chose the distance x₁ of the center of mass of the moving guide from the $Oy$ axis, and the relative distance of the mass m₂ to this center of mass x₂ as the generalized coordinates. Rest of the students (those with even numbers, namely 12) chose x₁ as above but chose x₂ as the distance of mass m₂ from the $Oy$ axis.

We are going to select the generalized coordinate system given in Figure 2(a) as the basis of the solution given below. Towards the end, we shall obtain the set of equations given by Figure 2(b) by a simple coordinate transformation, where the coordinates are denoted as u₁ and u₂ for the sake of clearness.

Derivation of equations of motion of the mechanical system via formulation of Lagrange’s equations relative to moving reference axes

The differential equations to be used are the Lagrange’s equations

\frac{d}{dt} (\frac{\partial T_{r}}{\partial {\overset{\cdot}{x}}_{1}}) - \frac{\partial T_{r}}{\partial x_{1}} = Q_{1}

(1)

\frac{d}{dt} (\frac{\partial T_{r}}{\partial {\overset{\cdot}{x}}_{2}}) - \frac{\partial T_{r}}{\partial x_{2}} = Q_{2}

(2)

Here, T_r is the relative kinetic energy of the vibrating two degrees of freedom system with respect to the rotating base, i.e. relative to an observer fixed to the rotating table. Q₁ and Q₂ denote the generalized forces corresponding to the generalized coordinates x₁ and x₂.

It is easy to write the relative kinetic energy of the system as

T_{r} = \frac{1}{2} m_{1} {\overset{\cdot}{x}}_{1}^{2} + \frac{1}{2} m_{2} ({\overset{\cdot}{x}}_{1} + {\overset{\cdot}{x}}_{2}) 2

(3)

As is known, when the “Lagrange’s equations of the second kind” (hereafter shortly, Lagrange’s equations) are formulated with respect to an inertial coordinate system, the generalized forces consider only the active forces acting on the system. However, if they are formulated with respect to a noninertial reference system, then the generalized forces have to include also the transport and Coriolis inertia forces.¹

In order to obtain the transport inertia forces, the first one has to derive the transport velocities and then the transport accelerations of the corresponding points.

To this end, let that point of the rotating plate coinciding with the point mass m₁, i.e. the midpoint of the moving guide at time t, be A₁. The absolute velocity, i.e. the velocity relative to the inertial reference system of the point A₁, is known as the transport velocity²

{\vec{v}}_{A_{1}} = (x_{1} \vec{i} + a \vec{j}) \cdot = x_{1} \overset{\cdot}{\to} i + a \overset{\cdot}{\to} j

(4)

where

\vec{i}

and

\vec{j}

represent the unit vectors of the rotating coordinate system fixed to the moving table and a dot denotes the time derivative. Taking into account that the coordinate x₁ should be considered as constant during this derivation process and

\overset{\cdot}{\to} i = \vec{ω} \land \vec{i} = ω \vec{k} \land \vec{i} = ω \vec{j} \overset{\cdot}{\to} j = \vec{ω} \land \vec{j} = ω \vec{k} \land \vec{j} = - ω \vec{i}

(5)

{\vec{v}}_{A_{1}} = x_{1} ω \vec{j} - a ω \vec{i}

(6)

is obtained.³

Here, “∧” denotes vector (or cross) product of the corresponding vectors.

The derivative of ${\vec{v}}_{A_{1}}$ with respect to time t is

{\vec{a}}_{A_{1}} = - ω^{2} (x_{1} \vec{i} + a \vec{j})

(7)

where equation (5) is considered again. This is the transport acceleration of the mass m₁.

The Coriolis acceleration of m₁ is simply

{\vec{a}}_{{Cor}_{1}} = 2 \vec{ω} \land {\vec{v}}_{r_{1}} = 2 (ω \vec{k}) \land {\overset{\cdot}{x}}_{1} \vec{i} = 2 ω {\overset{\cdot}{x}}_{1} \vec{j}

(8)

with

{\vec{v}}_{r_{1}}

being the relative velocity of the point mass m₁ with respect to the rotating frame. Hence, the transport inertia and Coriolis inertia forces of m₁ are

{\vec{T}}_{A_{1}} = - m_{1} {\vec{a}}_{A_{1}} = m_{1} ω^{2} (x_{1} \vec{i} + a \vec{j})

(9)

{\vec{T}}_{{Cor}_{1}} = - m_{1} {\vec{a}}_{{Cor}_{1}} = - 2 m_{1} ω {\overset{\cdot}{x}}_{1} \vec{j}

(10)

In a similar manner, the corresponding vector quantities for the second point mass m₂ can be obtained as

{\vec{v}}_{A_{2}} = [(x_{1} + x_{2}) \vec{i} + a \vec{j}] \cdot = - a ω \vec{i} + ω (x_{1} + x_{2}) \vec{j}

(11)

where A₂ denotes that point of the rotating table, coinciding with m₂ at time t. Hence,

{\vec{v}}_{A_{2}}

is the transport velocity of m₂. The transport acceleration of m₂ is

{\vec{a}}_{A_{2}} = \overset{\cdot}{\to} v_{A_{2}} = - ω^{2} (x_{1} + x_{2}) \vec{i} - a ω^{2} \vec{j}

(12)

where x₁ and x₂ are to be considered to be constant during the differentiation.

The Coriolis acceleration of point mass m₂ is

{\vec{a}}_{{Cor}_{2}} = 2 \vec{ω} \land {\vec{v}}_{r_{2}} = 2 (ω \vec{k}) \land ({\overset{\cdot}{x}}_{1} + {\overset{\cdot}{x}}_{2}) \vec{i} = 2 ω ({\overset{\cdot}{x}}_{1} + {\overset{\cdot}{x}}_{2}) \vec{j}

(13)

Here, ${\vec{v}}_{r_{2}}$ represents the relative velocity of m₂ with respect to the rotating frame. The transport inertia force of m₂ and the Coriolis inertia force are as follow

{\vec{T}}_{A_{2}} = - m_{2} {\vec{a}}_{A_{2}} = m_{2} ω^{2} [(x_{1} + x_{2}) \vec{i} + a \vec{j}]

(14)

{\vec{T}}_{{Cor}_{2}} = - m_{2} {\vec{a}}_{{Cor}_{2}} = - 2 m_{2} ω ({\overset{\cdot}{x}}_{1} + {\overset{\cdot}{x}}_{2}) \vec{j}

(15)

In order to be able to evaluate the Lagrange’s equations in (1) and (2), we need to obtain the generalized forces Q₁ and Q₂ corresponding to the generalized coordinates x₁ and x₂.

Q₁ is composed of three parts

Q_{1} = Q_{{sp}_{1}} + Q_{{tr}_{1}} + Q_{{Cor}_{1}}

(16)

Here, $Q_{{sp}_{1}}$ denotes the generalized force corresponding to the spring force due to k₁ spring, which can be written as

Q_{{sp}_{1}} = - \frac{\partial V}{\partial x_{1}} = - 2 k_{1} x_{1}

(17)

where the potential energy V reads

V = \frac{1}{2} (2 k_{1}) x_{1}^{2} + \frac{1}{2} (2 k_{2}) x_{2}^{2} = k_{1} x_{1}^{2} + k_{2} x_{2}^{2}

(18)

$Q_{{tr}_{1}}$ and $Q_{{Cor}_{1}}$ represent the generalized forces corresponding to the transport and Coriolis inertia forces.

Now, let us apply a virtual displacement $δ {\vec{r}}_{1} = δ x_{1} \vec{i}$ to the point mass m₁ and then calculate the sum of the virtual works of the transport inertia forces of m₁ and m₂

{\vec{T}}_{A_{1}} \cdot δ {\vec{r}}_{1} + {\vec{T}}_{A_{2}} \cdot δ {\vec{r}}_{1} = m_{1} ω^{2} (x_{1} \vec{i} + a \vec{j}) \cdot (δ x_{1} \vec{i}) + m_{2} ω^{2} [(x_{1} + x_{2}) \vec{i} + a \vec{j})] \cdot (δ x_{1} \vec{i}) = [m_{1} ω^{2} x_{1} + m_{2} ω^{2} (x_{1} + x_{2})] δ x_{1} = Q_{{tr}_{1}} δ x_{1}

(19)

This leads to

Q_{{tr}_{1}} = m_{1} ω^{2} x_{1} + m_{2} ω^{2} (x_{1} + x_{2})

(20)

“ċ” denotes the usual scalar (or dot) product of the corresponding vectors.

Due to the same virtual displacement $δ x_{1}$ , the sum of the virtual works of the Coriolis inertia forces gives

{\vec{T}}_{{Cor}_{1}} \cdot δ {\vec{r}}_{1} + {\vec{T}}_{{Cor}_{2}} \cdot δ {\vec{r}}_{1} = (- 2 m_{1} ω {\overset{\cdot}{x}}_{1} \vec{j}) \cdot (δ x_{1} \vec{i}) + [- 2 m_{2} ω ({\overset{\cdot}{x}}_{1} + {\overset{\cdot}{x}}_{2}) \vec{j}] \cdot (δ x_{1} \vec{i}) = 0

(21)

That is

Q_{{Cor}_{1}} = 0

(22)

Hence, from equation (16), one can obtain the first generalized force, via equations (17), (20), and (22) as

Q_{1} = - 2 k_{1} x_{1} + m_{1} ω^{2} x_{1} + m_{2} ω^{2} (x_{1} + x_{2})

(23)

Now, let us imagine that we apply a virtual displacement $δ {\vec{r}}_{2} = δ x_{2} \vec{i}$ to the point mass m₂ and then calculate the total virtual work of the transport and Coriolis inertia forces. The second generalized force Q₂ is, like equation (16), in the form

Q_{2} = Q_{{sp}_{2}} + Q_{{tr}_{2}} + Q_{{Cor}_{2}}

(24)

The partial generalized force $Q_{{sp}_{2}}$ corresponding to spring forces is

Q_{{sp}_{2}} = - \frac{\partial V}{\partial x_{2}} = - 2 k_{2} x_{2}

(25)

The virtual work of the transport inertia forces of both masses is

{\vec{T}}_{A_{2}} \cdot δ {\vec{r}}_{2} = m_{2} ω^{2} [(x_{1} + x_{2}) \vec{i} + a \vec{j})] \cdot (δ x_{2} \vec{i}) = m_{2} ω^{2} (x_{1} + x_{2}) δ x_{2}

(26)

recognizing that the first mass m₁ does not make a virtual displacement. Hence, equating the right side of equation (26) to

Q_{{tr}_{2}} δ x_{2}

, one can obtain

Q_{{tr}_{2}} = m_{2} ω^{2} (x_{1} + x_{2})

(27)

Finally

{\vec{T}}_{{Cor}_{2}} \cdot δ {\vec{r}}_{2} = Q_{{Cor}_{2}} δ x_{2}

(28)

yields via equation (15)

- 2 m_{2} ω ({\overset{\cdot}{x}}_{1} + {\overset{\cdot}{x}}_{2}) \vec{j} \cdot (δ x_{2} \vec{i}) = 0

leading to

Q_{{Cor}_{2}} = 0

(29)

Hence, one can obtain from equation (24) the second generalized force Q₂, via equations (25), (27), and (29) as

Q_{2} = - 2 k_{2} x_{2} + m_{2} ω^{2} (x_{1} + x_{2})

(30)

Now, it is an easy matter to obtain from equation (3) the equations of motion of the mechanical system by carrying out the necessary differentiations in equations (1) and (2) and considering equations (23) and (30)

(m_{1} + m_{2}) {\overset{··}{x}}_{1} + m_{2} {\overset{··}{x}}_{2} + 2 k_{1} x_{1} - m_{1} ω^{2} x_{1} - m_{2} ω^{2} (x_{1} + x_{2}) = 0

(31)

m_{2} ({\overset{··}{x}}_{1} + {\overset{··}{x}}_{2}) + 2 k_{2} x_{2} - m_{2} ω^{2} (x_{1} + x_{2}) = 0

(32)

The equations of motion read in matrix notation

[m_{1} + m_{2} m_{2} m_{2} m_{2}] [{\overset{··}{x}}_{1} {\overset{··}{x}}_{2}] + [2 k_{1} - (m_{1} + m_{2}) ω^{2} - m_{2} ω^{2} - m_{2} ω^{2} 2 k_{2} - m_{2} ω^{2}] [x_{1} x_{2}] = [00]

(33)

Equations of motion via Lagrange’s equations

Referring to Figure 3, it is easy to write the positon vectors of the masses m₁ and m₂ as

{\vec{r}}_{1} = x_{1} \vec{i} + a \vec{j}

(34)

{\vec{r}}_{2} = (x_{1} + x_{2}) \vec{i} + a \vec{j}

(35)

The absolute velocities of m₁ and m₂ can easily be obtained by deriving the above vectors with respect to time and recognizing that the derivatives of the moving unit vectors $\vec{i}$ and $\vec{j}$ are given by equation (5)

{\vec{v}}_{1} = \overset{\cdot}{\to} r_{1} = ({\overset{\cdot}{x}}_{1} - a ω) \vec{i} + x_{1} ω \vec{j}

(36)

{\vec{v}}_{2} = \overset{\cdot}{\to} r_{2} = [({\overset{\cdot}{x}}_{1} + {\overset{\cdot}{x}}_{2}) - a ω] \vec{i} + ω (x_{1} + x_{2}) \vec{j}

(37)

From the above relationships, the (absolute) kinetic energies of the masses m₁ and m₂ can be obtained as

T_{1} = \frac{1}{2} m_{1} v_{1}^{2} = \frac{1}{2} m_{1} [({\overset{\cdot}{x}}_{1} - a ω) 2 + x_{1}^{2} ω^{2}]

(38)

T_{2} = \frac{1}{2} m_{2} v_{2}^{2} = \frac{1}{2} m_{2} {[({\overset{\cdot}{x}}_{1} + {\overset{\cdot}{x}}_{2}) - a ω] 2 + ω^{2} (x_{1} + x_{2}) 2}

(39)

with

v_{1} = | {\vec{v}}_{1} |

and

v_{2} = | {\vec{v}}_{2} |

It is obvious that the elastic potential energies are simply

V_{1} = \frac{1}{2} (2 k_{1}) x_{1}^{2}, V_{2} = \frac{1}{2} (2 k_{2}) x_{2}^{2}

(40)

Hence, the so-called “kinetic potential” (or Lagrange’s function) of the mechanical system is

L = (T_{1} + T_{2}) - (V_{1} + V_{2}) = \frac{1}{2} m_{1} [({\overset{\cdot}{x}}_{1} - a ω) 2 + ω^{2} x_{1}^{2}] + \frac{1}{2} m_{2} {[({\overset{\cdot}{x}}_{1} + {\overset{\cdot}{x}}_{2}) - a ω] 2 + ω^{2} (x_{1} + x_{2}) 2} - k_{1} x_{1}^{2} - k_{2} x_{2}^{2}

(41)

Using the Lagrange’s function above, the Lagrange’s equations

\frac{d}{dt} (\frac{\partial L}{\partial {\overset{\cdot}{x}}_{i}}) - \frac{\partial L}{\partial x_{i}} = 0 (i = 1, 2)

(42)

give the equations of motion of the system as

(m_{1} + m_{2}) {\overset{··}{x}}_{1} + m_{2} {\overset{··}{x}}_{2} - (m_{1} + m_{2}) ω^{2} x_{1} - m_{2} ω^{2} x_{2} + 2 k_{1} x_{1} = 0

(43)

m_{2} {\overset{··}{x}}_{1} + m_{2} {\overset{··}{x}}_{2} - m_{2} ω^{2} x_{1} - m_{2} ω^{2} x_{2} + 2 k_{2} x_{2} = 0

(44)

which are the same equations as given in equations (31) and (32).

Equations of motion via Lagrangian form of D’Alembert’s principle

In order to apply the Lagrangian form of D’Alembert’s principle,⁴ we need first of all the absolute accelerations ${\vec{a}}_{1}$ and ${\vec{a}}_{2}$ of the masses m₁ and m₂, such that the corresponding inertia forces can be calculated.

Each of the absolute accelerations consists of three components, i.e. relative, transport, and Coriolis accelerations

{\vec{a}}_{1} = {\vec{a}}_{r_{1}} + {\vec{a}}_{{tr}_{1}} + {\vec{a}}_{{Cor}_{1}}

(45)

{\vec{a}}_{2} = {\vec{a}}_{r_{2}} + {\vec{a}}_{{tr}_{2}} + {\vec{a}}_{{Cor}_{2}}

(46)

The transport accelerations ${\vec{a}}_{{tr}_{1}}$ and ${\vec{a}}_{{tr}_{2}}$ were given formerly as ${\vec{a}}_{A_{1}}$ and ${\vec{a}}_{A_{2}}$ by equations (7) and (12). Hence

{\vec{a}}_{{tr}_{1}} = - ω^{2} (x_{1} \vec{i} + a \vec{j})

(47)

{\vec{a}}_{{tr}_{2}} = - ω^{2} (x_{1} + x_{2}) \vec{i} - a ω^{2} \vec{j}

(48)

Further, the Coriolis accelerations ${\vec{a}}_{{Cor}_{1}}$ and ${\vec{a}}_{{Cor}_{2}}$ can be taken directly from equations (8) and (13)

{\vec{a}}_{{Cor}_{1}} = 2 ω {\overset{\cdot}{x}}_{1} \vec{j}

(49)

{\vec{a}}_{{Cor}_{2}} = 2 ω ({\overset{\cdot}{x}}_{1} + {\overset{\cdot}{x}}_{2}) \vec{j}

(50)

It is obvious that the relative accelerations, i.e. the accelerations with respect to an observer rotating with the table, are

{\vec{a}}_{r_{1}} = {\overset{··}{x}}_{1} \vec{i}, {\vec{a}}_{r_{2}} = ({\overset{··}{x}}_{1} + {\overset{··}{x}}_{2}) \vec{i}

(51)

Now, inserting equations (47) to (51) into equations (45) and (46) yields

{\vec{a}}_{1} = {\overset{··}{x}}_{1} \vec{i} + [- ω^{2} (x_{1} \vec{i} + a \vec{j})] + 2 ω {\overset{\cdot}{x}}_{1} \vec{j}

(52)

{\vec{a}}_{2} = ({\overset{··}{x}}_{1} + {\overset{··}{x}}_{2}) \vec{i} + [- ω^{2} (x_{1} + x_{2}) \vec{i} - a ω^{2} \vec{j}] + 2 ω ({\overset{\cdot}{x}}_{1} + {\overset{\cdot}{x}}_{2}) \vec{j}

(53)

The corresponding inertia forces are simply

{\vec{T}}_{1} = - m_{1} {\vec{a}}_{1}

(54)

{\vec{T}}_{2} = - m_{2} {\vec{a}}_{2}

(55)

After these preliminaries, we can apply the D’Alembert’s principle in the form of Lagrange to the x direction as follows: if a virtual displacement $δ x_{1}$ is applied to the mass m₁ and a virtual displacement $δ x_{2}$ to the mass m₂, the total virtual work done by all the active forces and inertia forces acting on both masses must be zero, where the only active forces are the spring forces

(- 2 k_{1} x_{1}) δ x_{1} + [- m_{1} {\overset{··}{x}}_{1} + m_{1} ω^{2} x_{1}] δ x_{1} + [- m_{2} ({\overset{··}{x}}_{1} + {\overset{··}{x}}_{2}) + m_{2} ω^{2} (x_{1} + x_{2})] δ x_{1} + (- 2 k_{2} x_{2}) δ x_{2} + [- m_{2} ({\overset{··}{x}}_{1} + {\overset{··}{x}}_{2}) + m_{2} ω^{2} (x_{1} + x_{2})] δ x_{2} = 0

(56)

Equating the coefficients of $δ x_{1}$ and $δ x_{2}$ to zero ( $δ x_{1}$ and $δ x_{2}$ are arbitrary and independent from each other) and rearrangements lead to the following two differential equations which are nothing else but equations (43) and (44)

(m_{1} + m_{2}) {\overset{··}{x}}_{1} + m_{2} {\overset{··}{x}}_{2} - (m_{1} + m_{2}) ω^{2} x_{1} - m_{2} ω^{2} x_{2} + 2 k_{1} x_{1} = 0 m_{2} {\overset{··}{x}}_{2} + m_{2} {\overset{··}{x}}_{1} - m_{2} ω^{2} x_{1} - m_{2} ω^{2} x_{2} + 2 k_{2} x_{2} = 0

Equations of motion via the Hamilton’s principle

As is known, according to the Hamilton’s principle,⁵ a mechanical system moves in a time interval t₁ to t₂ such that

\int_{t_{1}}^{t_{2}} Ldt = extremum

(57)

This in turn means that the variation of the “functional” above should vanish

δ \int_{t_{1}}^{t_{2}} Ldt = 0

(58)

where L above denotes the kinetic potential of the system, previously given by equation (41).

Although we are aware of the fact that starting with equation (58), the equations of motion could be written more directly as the “Euler–Lagrange Equations” of the functional given by equation (57), we want to obtain the equations of motion from equation (58) by applying the rules of the variational calculus.

In the first step, one comes to the following equation by carrying out the necessary variations

\int_{t_{1}}^{t_{2}} {m_{1} [({\overset{\cdot}{x}}_{1} - ω a) δ {\overset{\cdot}{x}}_{1} + ω^{2} x_{1} δ x_{1}] + m_{2} [({\overset{\cdot}{x}}_{1} + {\overset{\cdot}{x}}_{2} - ω a) (δ {\overset{\cdot}{x}}_{1} + δ {\overset{\cdot}{x}}_{2}) + ω^{2} (x_{1} + x_{2}) (δ x_{1} + δ x_{2})] - 2 k_{1} x_{1} δ x_{1} - 2 k_{2} x_{2} δ x_{2}} dt = 0 .

(59)

Partial integration calculations necessary to obtain from the first, third and fourth terms, expressions with $δ x_{1}$ and $δ x_{2}$ as factors are given below separately

\int_{t_{1}}^{t_{2}} m_{1} ({\overset{\cdot}{x}}_{1} - ω a) \underset{\underset{\frac{d}{dt} (δ x_{1})}{\underset{︸}{δ (\frac{d x_{1}}{d t})}}}{\underset{︸}{δ {\overset{\cdot}{x}}_{1}}} d t = \int_{t_{1}}^{t_{2}} \underset{u}{\underset{︸}{m_{1} ({\overset{\cdot}{x}}_{1} - ω a)}} \underset{dv}{\underset{︸}{d (δ x_{1})}} \underset{0}{\underset{︸}{= m_{1} ({\overset{\cdot}{x}}_{1} - ω a) δ x_{1} |_{t_{1}}^{t_{2}}}} - \int_{t_{1}}^{t_{2}} m_{1} {\overset{··}{x}}_{1} δ x_{1} d t = - \int_{t_{1}}^{t_{2}} m_{1} {\overset{··}{x}}_{1} δ x_{1} d t

(60a)

because

δ x_{1} (t_{2}) = δ x_{1} (t_{1}) = 0

and the order of derivation and variation operations are interchangeable.

\int_{t_{1}}^{t_{2}} m_{2} ({\overset{\cdot}{x}}_{1} + {\overset{\cdot}{x}}_{2} - ω a) \underset{\underset{\frac{d}{dt} (δ x_{1})}{\underset{︸}{δ (\frac{{dx}_{1}}{dt})}}}{\underset{︸}{δ {\overset{\cdot}{x}}_{1}}} d t = \underset{0}{\underset{︸}{m_{2} ({\overset{\cdot}{x}}_{1} + {\overset{\cdot}{x}}_{2} - ω a) δ x_{1} |_{t_{1}}^{t_{2}}}} - \int_{t_{1}}^{t_{2}} m_{2} ({\overset{··}{x}}_{1} + {\overset{··}{x}}_{2}) δ x_{1} d t = - \int_{t_{1}}^{t_{2}} m_{2} ({\overset{··}{x}}_{1} + {\overset{··}{x}}_{2}) δ x_{1} d t

(60b)

Similarly

\int_{t_{1}}^{t_{2}} m_{2} ({\overset{\cdot}{x}}_{1} + {\overset{\cdot}{x}}_{2} - ω a) \underset{\underset{\frac{d}{dt} (δ x_{2})}{\underset{︸}{δ (\frac{{dx}_{2}}{dt})}}}{\underset{︸}{δ {\overset{\cdot}{x}}_{2}}} d t = \underset{0}{\underset{︸}{m_{2} ({\overset{\cdot}{x}}_{1} + {\overset{\cdot}{x}}_{2} - ω a) δ x_{2} |_{t_{1}}^{t_{2}}}} - \int_{t_{1}}^{t_{2}} m_{2} ({\overset{··}{x}}_{1} + {\overset{··}{x}}_{2}) δ x_{2} d t = - \int_{t_{1}}^{t_{2}} m_{2} ({\overset{··}{x}}_{1} + {\overset{··}{x}}_{2}) δ x_{2} d t

(60c)

δ x_{2} (t_{2}) = δ x_{2} (t_{1}) = 0

If equations 60(a) to (c) are substituted into equation (59) and the terms are rearranged such that they have the coefficients $δ x_{1}$ or $δ x_{2}$ , then

\int_{t_{1}}^{t_{2}} {[- m_{1} {\overset{··}{x}}_{1} + m_{1} ω^{2} x_{1} - m_{2} ({\overset{··}{x}}_{1} + {\overset{··}{x}}_{2}) + m_{2} ω^{2} (x_{1} + x_{2}) - 2 k_{1} x_{1}] δ x_{1} + [- m_{2} ({\overset{··}{x}}_{1} + {\overset{··}{x}}_{2}) + m_{2} ω^{2} (x_{1} + x_{2}) - 2 k_{2} x_{2}] δ x_{2}} dt = 0 .

(61)

Now, as the variations $δ x_{1}$ and $δ x_{2}$ are arbitrary and independent, according to the well-known “Fundamental Lemma of the Calculus of Variations,”⁶ each of the brackets can be equated to zero, giving

- m_{1} {\overset{··}{x}}_{1} + m_{1} ω^{2} x_{1} - m_{2} ({\overset{··}{x}}_{1} + {\overset{··}{x}}_{2}) + m_{2} ω^{2} (x_{1} + x_{2}) - 2 k_{1} x_{1} = 0

(62)

- m_{2} ({\overset{··}{x}}_{1} + {\overset{··}{x}}_{2}) + m_{2} ω^{2} (x_{1} + x_{2}) - 2 k_{2} x_{2} = 0

(63)

A little bit rearrangement shows that these are nothing else but the equations of motion given previously by equations (43) and (44).

Equations of motion via the Hamilton’s canonical equations

As it plays a central role in obtaining the Hamilton’s canonical equations,⁵ we repeat here the expression of the Lagrange function L, previously given by equation (41)

L = (T_{1} + T_{2}) - (V_{1} + V_{2}) = \frac{1}{2} m_{1} [({\overset{\cdot}{x}}_{1} - a ω) 2 + ω^{2} x_{1}^{2}] + \frac{1}{2} m_{2} {[({\overset{\cdot}{x}}_{1} + {\overset{\cdot}{x}}_{2}) - a ω] 2 + ω^{2} (x_{1} + x_{2}) 2} - k_{1} x_{1}^{2} - k_{2} x_{2}^{2}

Hence, the so-called “generalized momenta” are

p_{1} = \frac{\partial L}{\partial {\overset{\cdot}{x}}_{1}} = (m_{1} + m_{2}) {\overset{\cdot}{x}}_{1} + m_{2} {\overset{\cdot}{x}}_{2} - (m_{1} + m_{2}) a ω

(64)

p_{2} = \frac{\partial L}{\partial {\overset{\cdot}{x}}_{2}} = m_{2} [({\overset{\cdot}{x}}_{1} + {\overset{\cdot}{x}}_{2}) - a ω]

(65)

The Hamiltonian (or, Hamilton’s function) is defined as

H = \sum_{i = 1}^{2} p_{i} {\overset{\cdot}{x}}_{i} - L

(66)

Let us rearrange equations (64) and (65) in the form

(m_{1} + m_{2}) {\overset{\cdot}{x}}_{1} + m_{2} {\overset{\cdot}{x}}_{2} = p_{1} + (m_{1} + m_{2}) a ω

(67)

m_{2} {\overset{\cdot}{x}}_{1} + m_{2} {\overset{\cdot}{x}}_{2} = p_{2} + m_{2} a ω

(68)

If these set of linear equations are solved for the generalized velocities ${\overset{\cdot}{x}}_{1}$ and ${\overset{\cdot}{x}}_{2}$ in terms of p₁ and p₂

{\overset{\cdot}{x}}_{1} = \frac{p_{1}}{m_{1}} - \frac{p_{2}}{m_{1}} + a ω

(69)

{\overset{\cdot}{x}}_{2} = - \frac{p_{1}}{m_{1}} + \frac{p_{2}}{m_{2}} + \frac{p_{2}}{m_{1}}

(70)

are obtained.

Now, if these last two expressions are substituted into equation (66), considering equation (41), after some calculations and rearrangements, the Hamiltonian is obtained as

H = \frac{p_{1}^{2}}{2 m_{1}} + \frac{p_{2}^{2}}{2 m_{1}} + \frac{p_{2}^{2}}{m_{2}} - \frac{p_{1} p_{2}}{m_{1}} - \frac{p_{2}^{2}}{2 m_{2}} + a ω p_{1} - \frac{m_{1}}{2} ω^{2} x_{1}^{2} - \frac{m_{2}}{2} ω^{2} (x_{1} + x_{2}) 2 + k_{1} x_{1}^{2} + k_{2} x_{2}^{2}

(71)

Let us calculate further the following partial derivatives

\frac{\partial H}{\partial x_{1}} = - m_{1} ω^{2} x_{1} - m_{2} ω^{2} (x_{1} + x_{2}) + 2 k_{1} x_{1}

(72)

\frac{\partial H}{\partial x_{2}} = - m_{2} ω^{2} (x_{1} + x_{2}) + 2 k_{2} x_{2}

(73)

\frac{\partial H}{\partial p_{1}} = \frac{p_{1}}{m_{1}} - \frac{p_{2}}{m_{1}} + a ω

(74)

\frac{\partial H}{\partial p_{2}} = \frac{p_{2}}{m_{1}} + \frac{2 p_{2}}{m_{2}} - \frac{p_{1}}{m_{1}} - \frac{p_{2}}{m_{2}} = - \frac{p_{1}}{m_{1}} + \frac{p_{2}}{m_{2}} + \frac{p_{2}}{m_{1}}

(75)

We remember that the Hamilton’s canonical equations are as follow⁵

{\overset{\cdot}{p}}_{k} = - \frac{\partial H}{\partial x_{k}}; (k = 1, 2)

(76)

{\overset{\cdot}{x}}_{k} = \frac{\partial H}{\partial p_{k}}; (k = 1, 2)

(77)

Hence, inserting equation (72) to equation (75) into the equations above yields

{\overset{\cdot}{p}}_{1} = m_{1} ω^{2} x_{1} + m_{2} ω^{2} (x_{1} + x_{2}) - 2 k_{1} x_{1}

(78)

{\overset{\cdot}{p}}_{2} = m_{2} ω^{2} (x_{1} + x_{2}) - 2 k_{2} x_{2}

(79)

{\overset{\cdot}{x}}_{1} = \frac{p_{1}}{m_{1}} - \frac{p_{2}}{m_{1}} + a ω

(80)

{\overset{\cdot}{x}}_{2} = - \frac{p_{1}}{m_{1}} + \frac{p_{2}}{m_{2}} + \frac{p_{2}}{m_{1}}

(81)

which represent the Hamilton’s canonical equations of the mechanical vibration system in Figure 1.

Transition from Hamilton’s canonical equations to the classical equations of motion

It is not difficult, starting from the equations above, to obtain the classical equations of motion of the system.

Actually, differentiation of both sides of equations (80) and (81) with respect to time and then summing up side by side leads to

{\overset{\cdot}{p}}_{2} = m_{2} ({\overset{··}{x}}_{1} + {\overset{··}{x}}_{2})

(82)

which when equated to equation (80) yields

m_{2} ({\overset{··}{x}}_{1} + {\overset{··}{x}}_{2}) - m_{2} ω^{2} x_{1} - m_{2} ω^{2} x_{2} + 2 k_{2} x_{2} = 0

Further, if both sides of equation (81) are differentiated with respect to time t and equation (82) is considered

{\overset{\cdot}{p}}_{1} = m_{1} {\overset{··}{x}}_{1} + m_{2} ({\overset{··}{x}}_{1} + {\overset{··}{x}}_{2})

(83)

is found.

At the last step, if the equation above is equated to the right side of equation (78)

(m_{1} + m_{2}) {\overset{··}{x}}_{1} + m_{2} {\overset{··}{x}}_{2} - (m_{1} + m_{2}) ω^{2} x_{1} - m_{2} ω^{2} x_{2} + 2 k_{1} x_{1} = 0

is obtained.

Hence, as claimed at the beginning, starting with the Hamilton’s canonical equations, we succeeded to obtain again the classical equations of motion of the mechanical vibration system.

Having obtained the equations of motion of the mechanical system in Figure 1 in terms of the generalized coordinates shown in Figure 2(a), using different methods, it is quite in order to write them in matrix-vector notation as

[m_{1} + m_{2} m_{2} m_{2} m_{2}] [{\overset{··}{x}}_{1} {\overset{··}{x}}_{2}] + [2 k_{1} - (m_{1} + m_{2}) ω^{2} - m_{2} ω^{2} - m_{2} ω^{2} 2 k_{2} - m_{2} ω^{2}] [x_{1} x_{2}] = [00]

(84)

which was already given in equation (33).

Transformation of the coordinates

A comparison of Figure 2(a) and (b) reveals that the transformation between both pairs of generalized coordinates is of the form

[x_{1} x_{1}] = [10 - 1 0] [u_{1} u_{2}]

(85)

If this is substituted into equation (84) and the resulting equation is multiplied from the left by the transpose of the transformation matrix in equation (85), for having symmetrical mass and stiffness matrices

[m_{1} 00 m_{2}] [{\overset{··}{u}}_{1} {\overset{··}{u}}_{1}] + [2 k_{1} + 2 k_{2} - m_{1} ω^{2} - 2 k_{2} - 2 k_{2} 2 k_{2} - m_{2} ω^{2}] [u_{1} u_{2}] = [00]

(86)

is obtained.

This is just the equation of motion which students with “even” numbers had to establish. It is seen clearly that for this set of generalized coordinates, the equations of motion are “inertially uncoupled,” whereas the previous equation (84) was “inertially coupled”, where both were “elastically coupled.”⁷

Obtaining the constraint forces

This section is devoted to obtaining the constraint forces (also called reaction forces) exerted from the rotating table to the moving guide and from the moving guide to the point mass m₂. The forces shown in Figure 4 are as follow:

${\vec{N}}_{1}$ : constraint force exerted from the rotating table to the moving guide on its midpoint;

${\vec{N}}_{2}$ : constraint force exerted from the moving guide to the point mass m₂ (both forces are perpendicular to the x- axis because the whole system is assumed to be frictionless).

${\vec{T}}_{1}$ : inertia force of the moving guide, given by equations (52) and (54) as

{\vec{T}}_{1} = - m_{1} {{\overset{··}{x}}_{1} \vec{i} + [- ω^{2} (x_{1} \vec{i} + a \vec{j})] + 2 ω {\overset{\cdot}{x}}_{1} \vec{j}}

(87)

${\vec{T}}_{2}$ : inertia force of the point mass m₂, given by equations (53) and (55) as

{\vec{T}}_{2} = - m_{2} {({\overset{··}{x}}_{1} + {\overset{··}{x}}_{2}) \vec{i} + [- ω^{2} (x_{1} + x_{2}) \vec{i} - a ω^{2} \vec{j}] + 2 ω ({\overset{\cdot}{x}}_{1} + {\overset{\cdot}{x}}_{2}) \vec{j}}

(88)

Finally, ${\vec{F}}_{{sp}_{1}}$ and ${\vec{F}}_{{sp}_{2}}$ denote the spring forces exerted to the moving guide and point mass m₂.

According to the well-known D’Alembert’s principle,⁵ the mechanical vibration system under investigation will be in “dynamical equilibrium,” if the inertia forces are taken into account.

The sum of the moments of the forces with respect to the point $O ″$ gives

(x_{2} \vec{i}) \land (- N_{2} \vec{j}) + (x_{2} \vec{i}) \land {\vec{T}}_{2} = \vec{0}

(89)

If this equation is evaluated, considering equation (88)

{- x_{2} N_{2} + x_{2} m_{2} [a ω^{2} - 2 ω ({\overset{\cdot}{x}}_{1} + {\overset{\cdot}{x}}_{2})]} \vec{k} = \vec{0}

(90)

is obtained, yielding

N_{2} = m_{2} [a ω^{2} - 2 ω ({\overset{\cdot}{x}}_{1} + {\overset{\cdot}{x}}_{2})]

(91)

In order to calculate the reaction force ${\vec{N}}_{1}$ , let us take the sum of the moments of the forces with respect to the point mass m₂ (i.e. point $O ″'$ ) and equate it to zero

(- x_{2} \vec{i}) \land (- N_{1} \vec{j}) + (- x_{2} \vec{i}) \land {\vec{T}}_{1} = \vec{0}

(92)

Substitution of equation (87) here and carrying out vector cross products results in

[x_{2} N_{1} + m_{1} x_{2} (- a ω^{2} + 2 ω {\overset{\cdot}{x}}_{1})] \vec{k} = \vec{0}

(93)

Hence

N_{1} = m_{1} (a ω^{2} - 2 ω {\overset{\cdot}{x}}_{1})

(94)

is obtained.

At this point, it is quite in order to make a control whether the vectorial equation of the dynamical equilibrium in y-direction is satisfied

{\vec{N}}_{1} + {\vec{N}}_{2} + {\vec{T}}_{1_{Y}} + {\vec{T}}_{2_{Y}} = ? \vec{0}

(95)

Considering that ${\vec{N}}_{1}$ and ${\vec{N}}_{2}$ are assumed to be in the $- y$ direction, one can write

- m_{1} (a ω^{2} - 2 ω {\overset{\cdot}{x}}_{1}) - m_{2} [a ω^{2} - 2 ω ({\overset{\cdot}{x}}_{1} + {\overset{\cdot}{x}}_{2})] - m_{1} (- a ω^{2} + 2 ω {\overset{\cdot}{x}}_{1}) + m_{2} [a ω^{2} - 2 ω ({\overset{\cdot}{x}}_{1} + {\overset{\cdot}{x}}_{2})] = 0

(96)

It is clearly seen that this control equation is actually satisfied.

Conclusions

The first author has observed that the students had difficulty in obtaining the equations of motion of mechanical systems on moving grounds, in the graduate course “Dynamics of Mechanical Systems” that he has given for many years. The author has also experienced that a homework he gave, was very helpful in overcoming these difficulties. The homework, in essence, involves the derivation of the equations of motion, with the aid of formulating the Lagrange’s equations with respect to a moving set of axis for a vibrating system with two degrees of freedom, which is placed on a horizontal table rotating with a constant angular velocity around a vertical axis.

Besides this method which was compulsory for all students, the students were asked to obtain the same equations by a second method of their choice.

The feedback from the students is that this problem was very helpful especially in understanding the formulation of Lagrange’s equations with respect to a moving set of axis, and further, the mechanical system constitutes a good example for the application of the Lagrange’s equations, D’Alembert principle in the style of Lagrange, Hamilton’s principle, and Hamilton’s canonical equations.

Footnotes

Declaration of Conflicting Interests

The author(s) declared no potential conflicts of interest with respect to the research, authorship, and/or publication of this article.

Funding

The author(s) received no financial support for the research, authorship, and/or publication of this article.

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