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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: he absence of friction. the conservation of ~3J.energy 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...
View Full Document

Ask a homework question - tutors are online