Solutions of the Cambridge Problems, from 1800 to 1820 Volume 2

Solutions of the Cambridge Problems, from 1800 to 1820 Volume 2

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This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1825 edition. Excerpt: ...of the contrary. The magnet consisting of two equal arms, the action of the ship's, motion upon the one is counteracted by its opposite action upon the other. 615. Generally, w being' the height of the cylinder, and p the power or weight which puts it in motion, f the radius of the base, and r the distance, then tie accelerating force on p is F=--1H pr +--t-: -.-. v = gYt = X t Now by the question 616. Generally, required the length of a pendulum that would oscillate seconds at the distance of n radii of the Earth from its centre. If F be the force which accelerates the pendulum, whose length is L, then the time of an oscillation is got from (see Bridge, vol. II.) Hence, if I be the length of a second's pendulum at the surface of the earth, where g is the accelerating force, we hare r: t-: A Jt But by the question T = 1," and we have: ' F: ?: i-:: i: -'.. (hr) R1 9 9 orL = Li-If n = 2. Then L = JL = S.9 $ inches = 9.8inches. 4 4 Within the surface F = jr. JL. whence p is easily found. XV 617. Let be the distance of the particle fM from the axis of suspension, a the length of the rod, D its density at the lower extremity, L the distance between the point of suspension and centre of oscillation; then But the density at the distance x being = D.-. dM = iDx dx a 618. Generally let AC (Fig. 102, ) 6e any curve whatever revolving uniformly round the vertical axis AB, and suppose the body P descending along the curve by the force of gravity; required the velocity of the body at any given epoch t. Let the PM = y, AM = x, BC = 0, AB =, and V the known velocity of the point C; then resolving the centrifugal force RP into the tangential and normal ones PQ, and RQ, we have PQ = PR X cos. RPQ = PR x ds and RQ, = PR x sin. RPQ = PR x--ds which...show more

Product details

  • Paperback | 94 pages
  • 189 x 246 x 5mm | 181g
  • Rarebooksclub.com
  • Miami Fl, United States
  • English
  • Illustrations, black and white
  • 123661562X
  • 9781236615626