called centre of gravity) of a number of material particles situated in a straight line is a point such that were the whole mass placed there, the value of M by L would be the same as before. The distance along the straight line from the origin to the centre of mass may be called the equivalent distance. (Compare Art. 31.) When the particles are situated in one plane, then the centre of mass is a point which satisfies the above condition for two independent axes; and when they are in space, for three independent axes. The vector from the origin to the centre of mass may in a similar manner be called the equivalent-vector. The mass-vector due to the equivalent-vector and the whole mass is the resultant of the several component mass-vectors. ART. 137. When a body of uniform density is symmetrical with respect to a plane, the centre of mass is somewhere in the plane of symmetry; when it is symmetrical with respect to two planes, the centre of mass lies in the axis of symmetry; and when it is symmetrical with respect to three planes, the centre of mass coincides with the centre of symmetry. CENTRE OF MASS. (The body being of uniform density.) Triangle. From a vertex along two thirds of the line to the middle point of the opposite side. Semicircle.-From the vertex along 5756 of the radius. Pyramid or Cone.-From the apex along three fourths of the axis. Hemisphere. From the vertex along five eighths of the radius. EXAMPLES Ex. 1. At the corners of a cube weights are placed of 1, 2, 3, 4, 5, 6, 7, 8 lbs. respectively; determine their centre of mass. Let the side of the cube be L (Fig. 18.) Then for the direction Lalong X, L along Y, 13 L along Z. The length of the vector from the corner 0 to the centre of mass is √92+11+13, L, i.e., 1·07 L; and its direction-cosines are 9 L along XL along vector, etc. 19.3 Ex. 2. A uniform rod, 10 feet long, is bent at right angles at a point 4 feet from its end. Find the perpendicular distances of the centre of mass of the rod from the two straight portions of it. Let the line-density of the rod, which is uniform, be denoted by 1 M per foot; then in the shorter piece there is 4 M, and in the longer piece 6 M. Since either piece is uniform and symmetrical, its centre of mass is at its mid-point. We have now reduced the mass to two masses of 4 M and 6 M situated at the two mid-points. Their centre of mass is in the joining line, and at a distance of 6/10 of the line from the midpoint of the shorter piece. The component of the vector from the corner to the centre of mass along the shorter piece is 4/5 feet, and the component along the longer piece is 9/5 feet. EXERCISE XXV. 1. Find the centre of mass of two spheres of brass, of 1 inch and 2 inches diameter, placed at a distance of 5 inches, the distance being measured from the centres. 2. Where is the centre of mass of a square tin plate? If the plate weighs 5 oz. and a small body weighing 2 oz. is placed at one corner of the plate, where will the centre of mass of the whole be? 3. Find the centre of mass of the figure A when the pieces are of uniform material, and the central piece is half a side piece in length, and is joined at the mid-points of the sides. 4. Find the centre of mass of a T square, the two pieces being of the same material, and equal in length, breadth, and thickness. 5. Find the centre of mass in the case of a wooden F, the principal pieces being of the same length, and the central piece of half that length. Also for an E. 6. Find the centre of mass of the letter Y, the three pieces being uniform, and each one inch in length, and the two upper inclined at an angle of 60°. 7. A wooden vessel, 6 inches square and 6 inches in height, with a neck 2 inches square and 3 inches in height, is full of water. Find the position of the centre of mass of the water. SECTION XXVI.-MOMENTUM. ART. 138.-Unit of Momentum. The idea of momentum is derived from the idea of velocity by introducing the idea of mass. The momentum of a body is proportional to its mass and to its velocity; the general unit is M by (L per T). This unit is equivalent to (M by L) per T, when it is understood that the mass remains constant during change of time. Hence the bracket may be dispensed with, and either of these interpretations put upon M by L per T. Momentum is a directed quantity, its direction being the same as that of the velocity (or mass-vector) on which it depends. Hence it is resolved and compounded after the manner of directed quantities. If the speed only of a body is considered, then we consider only its speed-momentum. There is no special name for the unit of momentum in any of the systems of units. The unit may be denoted at length as lb. by ft. per sec., kilogramme by metre per sec., gm. by cm. per sec. ART. 139.-Impulse. By an impulse is meant the cause which produces a change of momentum in a body, whether the change takes place in the magnitude or in the direction, or in both the magnitude and the direction. The second law of motion states that the impulse is measured by the change of momentum produced. Let I denote the systematic unit of impulse, then ART. 140.-Momentum per Volume and Current. Liquid bodies, such as water and mercury, when at a uniform temperature throughout, are also uniformly dense. Suppose that the density of such a liquid is pM per L3, and that its velocity is v L per T, then by multiplying together these rates we get pr M by L per T = L; that is, pv units of momentum per unit of volume of the liquid. Consider a cross-section perpendicular to the direction of the velocity. The density can be expressed as P M per L2 cross-section per L normal, Р and the velocity is v L normal per T ; hence, by eliminating the common unit, pv M per L2 cross-section per T. This is the idea of current per unit cross-section, and it is ultimately equivalent to the idea of momentum per unit of volume. Let the cross-section be a L2, then apv M per T. This is the idea of current. EXAMPLE. Ex. A shell of 60 lbs. weight is moving before explosion at the rate of 400 feet per second. In consequence of the explosion, a piece weighing 14 lbs. is projected forward in the direction of motion with an additional velocity of 300 feet per second. What will now be the velocity of the remainder ? The mass of the piece is 14 lbs., and its additional velocity is 300 feet forward per second, therefore its additional momentum is per second. 14 × 300 lb. by foot forward X By the third law of motion the additional momentum of the remainder of the shell has the same magnitude but the opposite direction, therefore it is 14 x 300 lb. by foot backward per second; but the mass of the remainder is 60 – 14 lbs.; its additional velocity is Hence the new velocity of the remainder is 400-91.3 feet forward per second, 1. Of two bodies moving with constant velocities, one describes 36 miles in 1 h. 20 m., the other 55 feet in 14 sec.; the former weighs 50 lb., the latter 72 lb. Compare their momenta. 2. How far would a cannon-ball weighing 10 lb. travel in one minute, supposing it to possess the same momentum as a rifle bullet of 2 oz. moving with the velocity of 1,000 ft. per sec. 3. A man whose weight is 12 stones falls freely from a height of 64 feet. Calculate, neglecting the resistance of the air, the velocity and momentum acquired on reaching the ground. 4. A mass of snow, 28 lbs. in weight, falls from the roof of a house to the ground, a distance of 40 feet. Calculate the momentum. 5. A ball weighing 10 lbs. is projected vertically upwards with an initial velocity of 1,660 feet per second. Find its velocity and its momentum after 30 seconds and after 60 seconds. 6. Find the momentum per cubic foot in the case of a stream flowing at 2 miles an hour. Express in terms of the F.P.S. unit. |