(85)
Thus is determined the position of the centre of percussion C, corresponding to the axis of rotation O. It is obvious from this equation that, for an axis of rotation parallel to O traversing C, the centre of percussion is at the point where the perpendicular OG meets O.
§ 125.* To find the moment of inertia of a body about an axis through its centre of gravity experimentally.—Suspend the body from any conveniently selected axis O (fig. 48) and hang near it a small plumb bob. Adjust the length of the plumb-line until it and the body oscillate together in unison. The length of the plumb-line, measured from its point of suspension to the centre of the bob, is for all practical purposes equal to the length OC, C being therefore the centre of percussion corresponding to the selected axis O. From equation (85)
R2 = CG × OG = (OC − OG) OG.
The position of G can be found experimentally; hence OG is known, and the quantity R2 can be calculated, from which and the ascertained weight W of the body the moment of inertia about an axis through G, namely, W/g × R2, can be computed.
| Fig. 134. |
§ 126.* To find the force competent to produce the instantaneous acceleration of any link of a mechanism.—In many practical problems it is necessary to know the magnitude and position of the forces acting to produce the accelerations of the several links of a mechanism. For a given link, this force is the resultant of all the accelerating forces distributed through the substance of the material of the link required to produce the requisite acceleration of each particle, and the determination of this force depends upon the principles of the two preceding sections. The investigation of the distribution of the forces through the material and the stress consequently produced belongs to the subject of the [Strength of Materials] (q.v.). Let BK (fig. 134) be any link moving in any manner in a plane, and let G be its centre of gravity. Then its motion may be analysed into (1) a translation of its centre of gravity; and (2) a rotation about an axis through its centre of gravity perpendicular to its plane of motion. Let α be the acceleration of the centre of gravity and let A be the angular acceleration about the axis through the centre of gravity; then the force required to produce the translation of the centre of gravity is F = Wα/g, and the couple required to produce the angular acceleration about the centre of gravity is M = IA/g, W and I being respectively the weight and the moment of inertia of the link about the axis through the centre of gravity. The couple M may be produced by shifting the force F parallel to itself through a distance x. such that Fx = M. When the link forms part of a mechanism the respective accelerations of two points in the link can be determined by means of the velocity and acceleration diagrams described in § 82, it being understood that the motion of one link in the mechanism is prescribed, for instance, in the steam-engine’s mechanism that the crank shall revolve uniformly. Let the acceleration of the two points B and K therefore be supposed known. The problem is now to find the acceleration α and A. Take any pole O (fig. 49), and set out Ob equal to the acceleration of B and Ok equal to the acceleration of K. Join bk and take the point g so that KG: GB = kg : gb. Og is then the acceleration of the centre of gravity and the force F can therefore be immediately calculated. To find the angular acceleration A, draw kt, bt respectively parallel to and at right angles to the link KB. Then tb represents the angular acceleration of the point B relatively to the point K and hence tb/KB is the value of A, the angular acceleration of the link. Its moment of inertia about G can be found experimentally by the method explained in § 125, and then the value of the couple M can be computed. The value of x is found immediately from the quotient M/F. Hence the magnitude F and the position of F relatively to the centre of gravity of the link, necessary to give rise to the couple M, are known, and this force is therefore the resultant force required.
| Fig. 135. |
§ 127.* Alternative construction for finding the position of F relatively to the centre of gravity of the link.—Let B and K be any two points in the link which for greater generality are taken in fig. 135, so that the centre of gravity G is not in the line joining them. First find the value of R experimentally. Then produce the given directions of acceleration of B and K to meet in O; draw a circle through the three points B, K and O; produce the line joining O and G to cut the circle in Y; and take a point Z on the line OY so that YG × GZ = R2. Then Z is a point in the line of action of the force F. This useful theorem is due to G. T. Bennett, of Emmanuel College, Cambridge. A proof of it and three corollaries are given in appendix 4 of the second edition of Dalby’s Balancing of Engines (London, 1906). It is to be noticed that only the directions of the accelerations of two points are required to find the point Z.
For an example of the application of the principles of the two preceding sections to a practical problem see Valve and Valve Gear Mechanisms, by W. E. Dalby (London, 1906), where the inertia stresses brought upon the several links of a Joy valve gear, belonging to an express passenger engine of the Lancashire & Yorkshire railway, are investigated for an engine-speed of 68 m. an hour.