A rod `AB` is moving on a fixed circle of radius `R` with constant velocity `'v'` as shown in figure. `P` is the point of intersection of the rod and the circle. At an instant the rod is at a distance `x=(3R)/(5)` from centre of the circle. The velocity of the rod is perpendicular to the rod and the rod is always parallel to the diameter `CD`.

`(a)` Find the speed of point of intersection `P`
`(b)` Find the angular speed of point of intersection
`P` with respect to centre of the circle.

Answer

`(a)_(p)=(5)/(4)V (v)V co sectheta`

Explanation

As a rod `AB` moves, the point `'P'` will always lies on the circle.
`:. `Its velcoity will be along the circle as shown by `'V_(P)'` in the figure. If the point `P` has to lie on the rod `'AB'` also then it should have component in `'x'` direction as `'V'`
`:. V_(P) sin theta =V`
`rArr V_(p)=V co se ctheta`
here` cos theta=(x)/(R)=(1)/(R). (3R)/(5)=(3)/(5)`
`....Ans. `
`omega =(V_(P))/(R)=(5V)/(4R)`
ALTERNATE SOLUTION `:`
Sol. `(a)` Let `'P'` have coordinate `(x,y)`
`x=R cos theta, y=R sin theta `
`V_(x)=(dx)/(dt)=-R sin theta (d theta )/(dt)=VrArr (d theta)/(dt)=(-V)/(R sin theta )`
`V_(y)=R cos theta (d theta )/(dt)=R cos theta (-(V)/(R sin theta ))`
`=-V cot theta`
`:. V_(P)=sqrt(V_(x)^(2)+V_(y)^(2))=sqrt(V^(2)+V^(2)cot ^(2) theta )`
`=V c o s ec theta ....Ans. `
`Sol. (b)omega =(V_(P))/(R)=(5V)/(4R)`