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. (v = 4 m/s, R = 1 m) Find the angular speed of point of intersection P with respect to centre of the circle (in rad/s).
Physics · Mixed Chapter
JEE Mains

Question

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. (v = 4 m/s, R = 1 m) Find the angular speed of point of intersection P with respect to centre of the circle (in rad/s).

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: Let θ be the polar angle of intersection P. The vertical rod has horizontal position x=R cosθ. Step 2: Differentiate: dx/dt=−R sinθ·dθ/dt. Hence |ω|=|dx/dt|/√(R²−x²). Step 3: x=3R/5 and R=1 m give √(R²−x²)=4R/5=0.8 m. With rod speed v=4 m/s, ω=4/0.8=5 rad/s. Answer: 5.