Let M denote the mass of the body, and k' its radius of gyration; then will the moment of inertia of the concentrated mass with respect to the axis, be equal to Mk'2; but this must, by definition, be equal to the moment of inertia with respect to the same axis, or Σ(mr2); hence,

That is, the radius of gyration is equal to the square root of the quotient obtained by dividing the moment of inertia with respect to the same axis, by the entire mass.

Since M is constant for the same body, it follows that the radius of gyration will be the least possible when the moment of inertia is the least possible, that is, when the axis passes through the centre of gravity. This minimum radius is called the principal radius of gyration. If we denote the principal radius of gyration by k, we shall have, from the examples of Article (131), the following results:

CHAPTER VI.

CURVILINEAR AND ROTARY MOTION.

Motion of Projectiles.

133. If a body is projected obliquely upwards in vacuum, and then abandoned to the force of gravity, it will be continually deflected from a rectilinear path, and, after describing a curvilinear trajectory, will finally reach the horizontal plane from which it started.

The starting point is called the point of projection; the distance from the point of projection to the point at which the projectile again reaches the horizontal plane, through the point of projection, is called the range, and the time occupied is called the time of flight. The only forces to be considered, are the initial impulse and the force of gravity. Hence, the trajectory will lie in a vertical plane passing through the line of direction of the initial impulse. Let CAB represent this plane, A the point A of projection, AB the range, and AC a vertical line through

K

H

Fig. 115.

G B

A. Take AB and AC as co-ordinate axes; denote the angle of projection DAB, by a, and the velocity due to the initial impulse, by v. Resolve the velocity v into two components, one in the direction AC, and the other in the direction AB. We shall have, for the former, v sina, and, for the latter, v cosa.

The velocities, and, consequently, the spaces described in the direction of the co-ordinate axes, will (Art. 18) be entirely independent of each other. Denote the space

described in the direction AC, in any arbitrary time t, by y. The circumstances of motion in this direction, are those of a body projected vertically upwards with an initial velocity v sina, and then continually acted upon by the force of gravity. Hence, Equation (78) is applicable. Making, in that equation, hy, and v' v sina, we have,

Denote the space described in the direction of the axis AB, in any arbitrary time t, by x. The only force acting in the direction of this axis, is the component of the initial impulse. Hence, the motion in the direction of the axis of a will be uniform, and Equation (55) is applicable. Making sx, and v = v cosa, we have,

If we suppose t to be the same in Equations (113) and (114), they will be simultaneous, and, taken together, will make known the position of the projectile at any instant. From (144), we have,

an equation which is entirely independent of t. It, therefore, expresses the relation between x and y for any value of t whatever, and is, consequently, the equation of the trajectory. Equation (115) is the equation of a parabola whose axis is vertical. Hence, the required trajectory is a parabola.

To find an expression for the range, make y = 0, in (115), and deduce the corresponding value of x. Placing the value of y equal to 0, we have,

The first value of a corresponds to the point of projection, and the second is the value of the range, AB.

From trigonometry, we have,

2sina cosa sin2a.

If we denote the height due to the initial velocity, by h, we shall have,

v2 = 2gh.

Substituting these in the second value of x, and denoting the range by r, we have,

The greatest value of r will correspond to the value a = 45°, in which case, 2a = 90°, and sin 2a = 1. Hence, we have, for the greatest range,

r = 2h.

That is, it is equal to twice the height due to the initial velocity.

If, in (116), we replace a by 90°-a, we shall have,

r2h sin(180° - 2a) 2h sin2a,

=

the same value as before.

Hence, we conclude that there

are two angles of projection, complements of each other,

which give the same range. The trajectories in the two cases are not the same, as may be shown by substituting the values of a, and 90° a, in Equation (115). The greater angle of projection gives a higher elevation, and, consequently, the projectile descends more vertically. It is for this reason that the gunner selects the greater of the two angles of elevation when he desires to crush an object, and the lesser one when he desires to batter, or overturn the object. If a = 90°, the value of r becomes 0. That is, if a body be projected vertically upwards, it will return to the point of projection.

To find the time of flight, make xr, in Equation (114), and deduce the corresponding value of t. This gives,

The range being the same, the time of flight will be greatest when a is greatest. Equation (114) also gives the time required for the body to describe any distance in the direction of the horizontal line AB.

In Equation (117) there are four quantities, t, r, v, and a, and from it, if any three are given, the remaining one may be determined.

As an application of the principles just deduced, let it be required to determine the angle of projection, in order that the projectile may strike a point H, at a horizontal distance AG = ' from the point of projection, and at a height GH y' above it.

=

Since the point I lies on the

trajectory, its co-ordinates must

K

H

A

G B

Fig. 116.

satisfy the equation of the curve, giving

y' = x' tana

gx'2

2v2 cos1a