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.
(i) 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.

Explanation

As a rod AB moves, the point P will always lie on the circle. Hence its velocity will be along the circle as shoen by `V_(P)` is 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)sintheta=VimpliesV_(P)=Vcosectheta`
Here `costheta=(x)/(R)=(1)/(R)`.`(3R)/(5)=(3)/(5)`
`sintheta=(4)/(5)impliescosectheta=(5)/(4)`
`implies=V_(P)=(5)/(4)V`
(b) And angular velocity `omega=(V_(P))/(R)=(5V)/(4R)`