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A light rod carries three equal masses A...

A light rod carries three equal masses A, B and C as shown in figure. Find the velocity of B in vertical position of rod, if it is released from horizontal position. As shown in figure.

A

`sqrt(2gl)`

B

`sqrt((18gl)/7)`

C

`sqrt((4gl)/(3))`

D

`sqrt((8gl)/7)`

Text Solution

Verified by Experts

d) Loss in PE = Gain in angular kinetic energy
or `mgl/3 + mg((2l)/3) + mgl = 1/2 lomega^(2)`………………(i)
Here, l= Moment of inertia about fixed point
= `m(l/3)^(2) + m((2l)/3)^(2) + ml^(2) = 14/9 ml^(2)`
Form Eq (i),
`2mgl = 1/2(14/9ml^(2))omega^(2)` `therefore` `omega = sqrt((36g)/(14l))`
`therefore` `v_(B) = rw = (2l)/(3) sqrt((36g)/(14l))` = `sqrt(8gl)/7)`
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