A thin uniform 15 kg post, 1.75 m long is held vertically using a cable and is a

Question

A thin uniform 15 kg post, 1.75 m long is held vertically using a cable and is attached to a 5kg mass and a pivot at its bottom end. The string attached to the 5kg mass passes over a massless, frictionless pulley and pulls perpendicular to the post. suddenly cable breaks. a) Find the angular acceleration of the post about the pivot just after the cable breaks. b) will the angular acceleration in part a remains same as the post falls.(before it hits the pulley)? why? c). what is the acceleration of the 5kg mass the instant after the cable breaks? please show steps

Explanation / Answer

(a) The moment of inertia of the rod about one of its end is ML^2/3 you have to take in to consideration the moment of inertia of the 5 kg mass as well. hence the moment of inertia of the post and the 5 kg mass system will be (ML^2)/3 + 5(1.75-0.5)^2 = 15*1.75*1.75/3 + 5*1.25*1.25 = 23.125 kgm^2 a = T/I = 61.25/23.125=2.65 b) No c)The moment cable breaks, the angular acclns of the post and that of the pully will be same. Now linear accln a' = ra ( r= distance of 5 kg mass from the pivot) =1.25*2.65 = 3.31 m/s/s

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