Question

A light rod carries three equal masses A,B,C as shown in the figure.What will be the velocity of B in the vertical position of the rod,if it is released from horizontal position as shown in the figure​

Answer 1

at first , we should find out moment of inertia about fixed point.

so, moment of inertia of system about fixed point , I = m(l/3)² + m(2l/3)² + m(l)²

= ml²/9[ 1 + 4 + 9]

= 14ml²/9

from energy conservation theorem,

change in potential energy = change in kinetic energy

{mgl - mg(2l/3)} + {mgl - mg(l/3)} + {mgl - 0} = 1/2 Iω²

or, {mgl/3} + {2mgl/3} + mgl = 1/2 × 14ml²/9 × ω²

or, 2mgl = 7ml²/9 × ω²

or, ω² = 18gl/7

or, ω = √{18g/7l}

for velocity of B, r = 2l/3

then, v = ωr = √{18g/7l} × 2l/3

= √{(18g × 4l²)/(7l × 9)}

= √{8gl/7}

hence, velocity of B in the vertical position of the rod is √{8gl/7}

Answer 2

plz refer to the attachemnt!!

hope it is helpful!

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cheers!!