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This historic book may have numerous typos and missing text
Purchasers can usually download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1875 edition. Excerpt: ...be this angular velocity and a the radius of the tube, the pressure of the particle on the tube is constant and equal to Mmau2 M m'' The impressed forces on the tube are its own weight and the normal pressure of the particle. These have no moment round the centre, and therefore the angular velocity of the tube remains constant. Let co be the angular velocity of the particle round the centre of (not relatively to) the tube. Let u, i) be the accelerations of the centre in and at right angles to the direction of the particle at any moment. Also let g, fig be the component accelerations in these directions of the force of gravity on a unit mass. Then the accelerations of the particle are u--aco2 and v aco. Let R be the normal pressure, inwards on the particle, outwards on the tube. Then the tube would be in equilibrium under Mg, R and Mil reversed, and Mfig, and Mis The particle would be in equilibrium under mg,--R and m(u--aco2) reversed, and nifig, and m (v ad) Resolving fig--v = 0, fig--v--aw = 0; whence co = 0, or the particle moves round the tube with constant velocity. Resolving again, Mkg R-If w = 0, and mXg--R--m(u--a&2) = 0. (a) Supposing O the angular velocity of the tube, what will be the time in which the particle will come round to the same point of the tube again? (/3) Shew from first principles that gravity has no effect in altering the relative motion of the tube and particle, or the mutual pressure. (7) In what path does the common centre of inertia move? (8) About what point is the whole angular momentum constant? (e) Prove by means of this principle that the angular velocity of the particle is constant, assuming that the angular velocity of the tube is constant. 5. Two uniform rods AB, BC, of masses m, m'', and...
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