A light rod is pivoted at one end so that it can swing freely as a pendulum. Two masses $$2m$$ and m are attached to it at distances b and $$3b$$ respectively from the pivot. The rod is held horizontal and then released the angular acceleration at the instant it is released is

Question

Infinity Learn · Accepted Answer

mass $$2m$$ at a distance b and mass m at a distance $$3b$$ form the support. When they are released from the horizontal position the torque acting due to gravitational force  about the pivot is $$2mgb+$$ mg $$3b$$ $$=$$ $$5mgbthen$$ $$5mgb=$$ Iα  here  I $$=2mb2+m3b2$$ $$=$$ $$11mb211mb2$$ α $$=5mgb$$ α $$=5g11b$$

A light rod is pivoted at one end so that it can swing freely as a pendulum. Two masses 2m and m are attached to it at distances b and 3b respectively from the pivot. The rod is held horizontal and then released the angular acceleration at the instant it is released is

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