Thus n mv2 r mg cos 8 0 ifgbcoso now coso has

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Unformatted text preview: he absence of friction. the conservation of requires KA + VA = KB + VB, or 2:'.': 1 2 2 ;+mgYA = '2mv B + mgYB. Therefore, x = +2g(YB -YA~ (m/k)(gR + 2g(2R = >()rx 2: [5(840 kg}(9.8mjs2)(6.2 m) + ImJjl/2 = 2.87 m. 139 p :j swing to point 2, V2 = mgY2, and K2 = jmv~. Energy conservation implies K2 = VI - V2, or v2 = 2g(YI ~ Y2)= 2g1sin45 =V2g1, where we used the trigonometry apparent in Fig. 8-45 to express the difference in height in terms of the length of the string and angle of swing. (b) The string tension is in the direction of the centripetal acceleration (also shown in Fig. 8-45 at point 2), so the radial component of Newton's second law gives T2 - mg sin 45 = mvVl. Using the result of part (a), we find T2 = (mvVf) + mgsin45 = (m-l2gejf) + (mgj-J2) = 3mgjV2. Problem 70. A particle of mass m slides down a frictionless quarter-circular track of radius R as shown in Fig. 8-46.. If it starts from rest at the top of the track, find (a) its speed and (b) the magnitude of the normal force exerted by the track when the radius vector from the...
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This note was uploaded on 07/27/2009 for the course PHYSICS 101 taught by Professor Wormer during the Spring '08 term at NYU Poly.

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