A spot light `S` rotates in a horizontal plane with a constant angular velocity of `0.1 rad//s`. The spot of light `P` move along the wall at a disatnce `3 m`. What is the velocity of the spot `P` when `theta = 45^(@)` ?

Answer

`(u^(2)) sin 2 theta)/(g cos theta)`

Explanation

`(u^(2) sin 2alpha)/(g cos theta)`
Let u is the velocity of the particle with respect to the box. Considering the motion parallel and perpendicular to inclined plane. `u_x` is the relative velocity of particle with respect to the box is x –direction.
`u_x` is the relative velocity of particle with respect to the box is y –direction.
`u_(y)=+u sin alpha, u_(x)=+u cos alpha`
`overset(-)a PG=-g sin theta hati-g cos theta hatj, overset(-)a_(BG)=-g sin theta hati`
`overset(-)a_(PB)=overset(-)a_(PG)-overset(-)a_(BG)=-g cos theta hatj, a_(y)=-g cos theta, a_(x)=0`
`S_(y)=0` (motion is taken till the time the particle comes back to the box)
`s_(y)=u_(y)t+1/2 a_(y)t^(2) rArr =(u sin alpha )t-1/2 g cos theta xx t^(2) rArr t=0 or t=(2u sin alpha)/(g cos theta)`
`R=u cos alpha t=u cos alpha =(2u sin alpha)/(g cos theta)=(u^(2) sin 2 alpha)/(g cos theta)`