Поиск по базе сайта:
System of distributed forces of inertia. Total vector and Total moment of the system icon

System of distributed forces of inertia. Total vector and Total moment of the system




Скачати 24.11 Kb.
НазваSystem of distributed forces of inertia. Total vector and Total moment of the system
Дата конвертації18.02.2013
Розмір24.11 Kb.
ТипДокументи

System of distributed forces of inertia.
Total vector and Total moment of the system.




Inertial forces of a moving body are distributed over the body mass.

In Fig. 1 Inertial forces for several cases of rigid body motion are presented. In Fig 1a rod AB translates, in Fig 1b ring rotates about fixed axis perpendicular to the plane of ring with a constant angular velocity, in the Fig 1c the rod rotates about vertical axis with a constant angular velocity.



a) b) c)

Fig. 1

in accordance with the main theorem of statics the system of inertial forces may be substituted for equivalent system, consisting of one force applied at the chosen center of reduction, and one couple with the moment .

The force is total vector of the inertial forces, so it is determined as the geometrical sum of all forces of the system:

, (1)

where is force of inertia acting on element of the rigid body (Fig. 2);



Fig. 2

, и are mass and acceleration of mass center of element of the rigid body correspondingly.

For continuous distribution of body mass the sum (1) becomes integral:



where is position vector of the mass center of element,

is position vector of the body mass center,

и are mass and acceleration of the body mass center correspondingly

.

. (2)

For any reduction center the total vector of inertial forces of a rigid body is equal to the negative product of mass and body’s mass center acceleration.


The couple moment is equal to the total moment of inertial forces about the chosen reduction center, that is geometrical sum of all forces moments about the center O:

, (3)

or

, (4)

where is force of inertia acting on elementary mass of the body.

Magnitude and direction of the total moment are dependent on the reduction center position.


Examine the total moment of inertial forces for the simplest types of rigid body motion.

Translational motion.

consider the body АВD of mass , that is in translation (Fig. 3а). Choosing the center of mass as reduction center (Fig. 3b), we get:

,



where is position vector of the body point with respect to the moving reference with origin in the point С; for translation the accelerations of all points are the same, that is

.



а) б)

Рис.3.


Integral is numerator of the equation for position vector of body mass center with respect to the moving reference:

.

Mass center is origin of the moving reference, so



and

. (5)


For rigid body in translation the system of inertial forces is equivalent to the single force applied at the body mass center equal to the the negative product of mass and body’s mass center acceleration. The force is resultant of inertial forces.


Rigid body rotation about fixed axis

Consider a rigid body that has the plane of symmetry and rotates about the axis perpendicular to this plane (Fig. 4a).




а) b)




с)

Fig.4.

For any point acceleration is

,

that is, force of inertia acting on elementary mass of the body (Fig. 4b) is:



Choosing the center of reduction in the point O where the axis of rotation intersects the plane of body symmetry, we obtain:

(6)

(7)

The total vector of inertial forces is applied at the point O

The total moment of forces of inertia about axis of rotation

,

where is normal force of inertia of elementary mass, that points to the center O (see Fig. 4b) therefore the moment about axis z of the normal force of inertia is zero;

is tangential force of inertia of elementary mass, that is perpendicular to the normal force in direction of angular acceleration (see Fig. 4b);

therefore, we have



. (8)

Seeing the body has plane of symmetry

. (9)

As result we obtain the equivalent system: one force applied at the center O, and one couple with the moment (see Fig. 4b).


Now we consider the simplest equivalent of inertial forces (Fig. 5a), that is resultant of inertial forces. The moment may be substituted for couple of forces , such as , the couple arm is determined as:

,

, (10)

(see Fig. 5b).

The forces of the couple and the total moment of inertial forces rotate the plane in the same direction.



а) б)



с)

Fig.5

Forces form the balanced system, so , the normal force of inertia may be shifted along the line OC to the point В, so the force with the line of action passes through the point B is inertial forces resultant.

If body rotates about axis passes through the mass center than

,

,

.


Problems


1. 1p. Determine the force of gravity of the circular plate of radius R=20 cm rotating about fixed axis in accordance with the law . The axis passes through the disc mass center perpendicular to the plane of disc, the total moment of forces of inertia with respect to the axis of rotation is 4 Nm.

2. 1p. Thin uniform rod of the length l and mass M rotates about axis that passes through the end of the rod and is perpendicular to the rod. The law of rotation is . Determine the magnitudes and directions of the normal and tangential components of the resultant of inertial forces.

3. 1p. The wheel of radius R and mass M rolls without slipping along the straight horizontal rail. Determine the total vector and the total moment of inertial forces about the axis passes through the wheel mass center perpendicular to the plane of wheel. Assume that the wheel is thin uniform disc. The mass center moves in accordance with the law ,  is constant positive value. The axis  is directed along the rail.

4. 1p. In gear mechanism (Fig. 1) the wheel 1 is fixed, the wheel 2 is driven to roll without slipping by rod OC. The rod rotates with a constant angular velocity . The radiuses of the wheels are , the second wheel mass is M. Determine the total vector and the total moment of the second wheel inertial forces about the axis passes through the wheel mass center perpendicular to the plane of wheel.



Fig. 1


5. 3p. Determine the reactions at the supports A and B of crane (Fig.2) if 3000-kg load E is lifted with the acceleration  . The mass of the crane is 2000-kg, its center of mass is situated at the point C. The mass of the vehicle D is 500-kg. The crane and the vehicle are fixed. All dimensions needed are on the Fig. 2.

6. 4p. Determine the reactions at the supports A and B of crane (Fig.2) if 3000-kg load E is lifted with the acceleration  . The mass of the crane is 2000-kg, its center of mass is situated at the point C. The mass of the vehicle D is 500-kg. The crane is fixed and the vehicle moves to the right with the acceleration . All dimensions needed are on the Fig. 2.

Fig. 2

7. 5p. The end A of the thin uniform rod AB of the length  and mass M moves uniformly along the horizontal rail with help of body E. Velocity of the body B is , the rod is all time supported by the edge D. Determine the total vector and the total moment of the rod inertial forces about the axis passes through the rod mass center perpendicular to the plane of motion as function of the angle . Determine the value of dynamical reactions at edge D.



Fig. 3



Схожі:




База даних захищена авторським правом ©lib.exdat.com
При копіюванні матеріалу обов'язкове зазначення активного посилання відкритою для індексації.
звернутися до адміністрації