A rod of length / is held vertically stationary with its lower end located at a point P, on the horizontal plane. When the rod is released to topple about P, the velocity of the upper end of the rod with which it hits the ground is

Options

A. `sqrt((g)/(l))`
B. `sqrt(gl)`
C. `3sqrt((g)/(l))`
D. `sqrt((3g)/(l))`

Answer

(b)

Explanation

From the law of conservation of energy, the potential energy of the metre stick is converted into rotational kinetic energy.
P.E of metre stick= `(mgl)/(2)`
Rotational K.E of a body = `(1)/(2)Iomega^(2)`
`Rightarrow mg(1)/(2)=(1)/(2)Iomega^(2)`
  Where, I=Moment of inertia of metre stick= `(ml)^(2)/(3)`
` Rightarrow mg((l)/(2))=(1)/(2) (ml)^(2)/(3)((v)/(l))^(2)  (because omega=(v)/(r)=(v)/(l))`
`3 gl=v^(2) Rightarrow v=sqrt3gl`