Question

LUULE LUNI RW 1 (10/1/ h 11:59 PM) Suspended rod - Timer A rod of length 51.0 cm and mass 1.10 kg is suspended by two strings which are 38.0 cm long, one at each end of the rod. Notes Elte The string on side B is cut. Find the magnitude of the initial acceleration of end B. Submit Answer Tries 0/12 The string on side B is retied and now has only half the length of the string on side A. d/2 Find the magnitude of the initial acceleration of the end B when the string is cut. Sen Submit Answer Tries 0/12

Answer 1

Given, man of rood - sang, length of sod sie Now, when firing B is cut rood will notate about A due to torque of its own weight mxgxd TAX 1943 mexa Here my man of rodo lo length of rod. In moment of Inertia of rod about point AB= I x x as a = 8x) = $x 3 g - 39. & fere rol since rod is rotating Taking g lomis? about end na ao Is m/s2 Now, Here, Tusque will be -- Mrg Case I - TAX. mag coro e mos x x = x= 32 ab= l x x = x3 g Care 9. = 39 Coso = 38 x 47.32 = 13.9 17 m/s² I g=10m/s² 51cm 2019 Cm ng mgco Caso = 9. 49.3 5 51