A rod of length `l` is travelling with velocity `v` and rotating with angular velocity `omega` such that `(omegal)/2=v`. The distance travelled by end `B` of the rod when rod rotates by an angle `(pi)/2` is

Options

A. `sqrt2(2)l`
B. `5/2 l`
C. `3l`
D. `4l`

Answer

A

Explanation

Here the point `B` if following a curvilinear path whose velocity at any instant is given by vector sum of `vec(V)` and `(vec(omega)xxvec(L))/2`. When rod rotates by an angle `theta`, the situation is as shown in the figure. The speed of end point at this instant is

`v_(1)=sqrt(v^(2)+((omegaL)/2)^(2)+2v+(omegaL)/2xxcos(pi-theta))`
`v_(1)=sqrt(2v)(sqrt([1+cos theta]))`
`=sqrt(2v)xxsqrt(2)"cos" theta/2=2v "cos" theta/2`
`v_(1)=(omegaL)/2xx2"cos" theta/2,(ds)/(dt)=(d theta)/(dt)xxL"cos" (theta)/2`
`=int_(0)^(s)ds=Lint_(0)^(pi//2)"cos"(theta)/2 d theta=(2L)/(sqrt(2))=sqrt(2)L`