Cambridge Senate-house Problems and Riders for the Year 1860: With SolutionsMacmillan, 1860 - 198 sider |
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Resultat 1-5 av 43
Side 44
... dť dw dt + + n cos + w2 sin dt 。) = 0 . If a be the length of the rod without weight , na that of the rod whose angular velocity is w at the time t , The angular velocity of the particle about the point of junction de = + w . dt The ...
... dť dw dt + + n cos + w2 sin dt 。) = 0 . If a be the length of the rod without weight , na that of the rod whose angular velocity is w at the time t , The angular velocity of the particle about the point of junction de = + w . dt The ...
Side 60
... dT ds = F cos 4 , + F sin & = R ; • . R = F'sin & + √Fcos øds .. P ... ( 1 ) . Letv velocity of the particle freely describing the curve , = then dv2 2F'cos , ds v2 = F sin & ; Ρ Fsin , fFcos pds — c ... 0 = + ( 2 ) . 2 Р Subtracting ...
... dT ds = F cos 4 , + F sin & = R ; • . R = F'sin & + √Fcos øds .. P ... ( 1 ) . Letv velocity of the particle freely describing the curve , = then dv2 2F'cos , ds v2 = F sin & ; Ρ Fsin , fFcos pds — c ... 0 = + ( 2 ) . 2 Р Subtracting ...
Side 61
... dT = -μrdr , Tp = c ; Р 1 P 2 .. ρ 1 2 dr μη dpi .. Toc ∞ √p . Ρ - 11. If any uniform arc of an equiangular spiral attract a particle , placed at the pole with a force varying inversely as the square of the distance , prove that the ...
... dT = -μrdr , Tp = c ; Р 1 P 2 .. ρ 1 2 dr μη dpi .. Toc ∞ √p . Ρ - 11. If any uniform arc of an equiangular spiral attract a particle , placed at the pole with a force varying inversely as the square of the distance , prove that the ...
Side 62
... line at any instant , and & the azimuth of the plane containing r , we have , by taking moments round the axis , 2.2 do dt = C. Also if be the corresponding velocity of the particle , 62 [ Jan. 17 , SENATE - HOUSE PROBLEMS.
... line at any instant , and & the azimuth of the plane containing r , we have , by taking moments round the axis , 2.2 do dt = C. Also if be the corresponding velocity of the particle , 62 [ Jan. 17 , SENATE - HOUSE PROBLEMS.
Side 63
... dt r dt de ar a2 sin e √μ.cos a ( μ . cos a π a2 a2 = sin ede = √μ cos a 0 .. t = Also the normal pressure Αμ cos a = difference between the force resolved along the radius and vel . μ α = sin 0 - арг = 2 0 , whence it follows that ...
... dt r dt de ar a2 sin e √μ.cos a ( μ . cos a π a2 a2 = sin ede = √μ cos a 0 .. t = Also the normal pressure Αμ cos a = difference between the force resolved along the radius and vel . μ α = sin 0 - арг = 2 0 , whence it follows that ...
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Cambridge Senate-house Problems and Riders for the Year 1860: With Solutions Henry William Watson,Edward John Routh Uten tilgangsbegrensning - 1860 |
Cambridge Senate-house Problems and Riders for the Year 1860: With Solutions Henry William Watson,Edward John Routh Uten tilgangsbegrensning - 1860 |
Cambridge Senate-house Problems and Riders for the Year 1860: With Solutions Henry William Watson,Edward John Routh Uten tilgangsbegrensning - 1860 |
Vanlige uttrykk og setninger
acceleration angular velocity Arabic numbers asymptotes attraction axes axis bisects Cambridge centre of force centre of gravity chord circle circular cloth co-ordinates cone conic conic section constant cos² Crown 8vo curvature cylinder described diameter differential equation direction directrix distance drawn dt dt ellipse ellipsoid equal angles equiangular spiral equilibrium extremity Fcap fixed point fluid foci focus given point Hence horizontal hyperbola inclined intersection latus rectum length locus MACMILLAN & CO.'S meridian motion normal parabola parallel particle perpendicular point of contact position pressure prove ratio respectively right angles Roman numbers SENIOR MODERATOR shew sides sin² singular solution smooth sphere spherical straight line string subtend surface tangent tension triangle Trinity College uniform vertex vertical Vis Viva weight аф
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