Question

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 1

Here, `omega = 0.1 rad//s`, So `= 3 m, theta = 45^(@)`, vel. Of spot, `v = ?` At any time, let the spot of light `P` be at a distance `x` from `O`, Fig. Let `/_OSP = phi` and` /_OPS = theta`. Clearly,
`v = (dx)/(dt)` and `omega = (d phi)/(dt) = 0.1 rad//s`
In `DeltaOSP, tan phi = (OP)/(OS) = (x)/(3)`
`x = 3 tan phi`
Differentiating, we get,
`(dx)/(dt) = 3 sec^(2) phi = (d phi)/(dt)`
`v = 3 sec^(2) phi (omega)` ..(i)
As `phi = 90^(@) - theta - 45^(@) = 45^(@)`,
`:. sec phi = sec 45^(@) = sqrt(2)`
From (i), `v = 3(sqrt(2))^(2) xx 0.1 = 0.6 m//s`