SECT. 8. Centrifugal Force.

(1) Two unequal weights are connected by a string of given length, which passes through a small ring: to find how many times in a second the lighter must revolve, in order that the heavier may be at rest at a given distance from the ring.

Let P, P', fig. (120), be the positions of the two weights, O the ring, being the inclination of OP' to the vertical.

Let OP'c', and let m, m', denote the masses of P, P', respectively.

The forces acting on P' are the tension of the string, which is equal to mg, the weight m'g, and the centrifugal force m'w'c' sin 0, acting horizontally.

For the equilibrium of P' we have, resolving horizontally, m'w'c' sin 0mg sin 0,

Hence, n denoting the number of revolutions performed by P' in a second,

(2) A string will just bear a weight of 16 lbs. without breaking: if a weight of lb. be attached to it, and whirled round in a horizontal circle, the radius of which is two feet, find the number of revolutions the weight must make in a second, in order to break the string.

2

The required number of revolutions is equal to√g.

CHAPTER IV.

IMPACT.

(1) A PARTICLE is projected, with a given velocity, from one extremity of a diameter of a horizontal circle, and, after reflection at the curve, passes through the other extremity: to find the elasticity, in order that the time of motion may be n times that of describing the diameter with the velocity of projection.

Let AB, fig. (121), be the diameter, APB the path of the particle, C the centre of the circle.

=

Join CP: let r the radius, < PAC-a, V= the velocity of projection.

Since the component of the velocity, at right angles to CP, is the same before and after impact, and since A, B, are equidistant from the diameter PCP', it follows that the time of movement from A to P is equal to that from P to B: let t denote each of these times.

Since the line AP is described, in the time t, with the velocity V,

Considering the component of the motion from P to B, parallel to PC, we have, since eV cos a is the normal component of the velocity after impact on the curve at P,

or

r (1 − cos 2a) = eV cos a. t,

2r sin2 a = eV cos a .t .......

(2).

Again, the time of motion being equal to that of describing AB with the velocity V, we have

From (4) and (5), by the elimination of a, we find that

(3).

(4),

(5).

From the equation (4),

we see that n must be less than 2, and, from (6), that 4-n must be less than n2, or n greater than √2. Thus, that the problem may be possible, it is necessary that n be less than 2 and greater than 2.

(2) An imperfectly elastic ball is projected in a given direction within a fixed horizontal hoop, so as to go on rebounding from the surface of the hoop: to find the limit to which the velocity of the ball will approach; and to shew that it will attain this limit at the end of a finite time.

Let V, V, V, V,... be the velocities of the successive impacts; a, a, a, a, ... the acute angles between the successive directions of impact and the tangents to the circle at the respective points of impact.

Then, the tangential components of the velocities of impact and rebound being equal at each point of impact, we have

Again, resolving normally the velocities of impact and rebound,

Multiplying these last n equations together and rejecting factors common to both sides, we have

tan an= en tan a.

When n∞, we see from this equation that tan a„ = 0, and therefore, by (1), that the terminal velocity of the ball is V cos a.

Again, r denoting the radius of the hoop, the time through the chord between the nth and (n+1)th impact, is equal to

hence, the whole time between the first and (n + 1)th impact is less than

and therefore the ball arrives at its terminal velocity in a time less than

(3) A particle, of given elasticity e, is projected along a horizontal plane, from the middle point of one of the sides of an isosceles right-angled triangle, so as, after reflection at the hypotenuse and remaining side, to return to the same point; to prove that the cotangents of the angles of reflection are e+1 and e+2 respectively.

Let ABC, fig. (122), be the triangle, A being the right angle: let PQRP be the course of the particle: let

Again, by the geometry,

PQ sin (0 +0') = PR sin (4+ 4′),

(2).

whence, AP being equal to BP,

cot '+2 cot = cot (1+cot 0) + 1,

(4) A locomotive is travelling at the rate of 60 miles an hour: shew that, if it were suddenly stopped, the violence of the concussion would be nearly as great as if it fell from a height of 40 yards.

(5) A ball, of given elasticity, impinges against a circular arc: determine the point of impact, in order that it may rebound at right angles to its direction of incidence.

Let A be the point of the arc, the radius of which is parallel to the direction of incidence, and let P be the point of impact: then, r denoting the radius and e the elasticity,

AP=r tan1 (e3).