NTA Abhyas JEE Main2020PhysicsLaws of MotionPractice
A particle moves in a circle with a uniform speed. When it goes from a point B to a diametrically opposite point A , the momentum of the particle changes by P → A - P → B = 2 j ^ kg m s -1 and the centripetal force acting on it changes by F → A - F → B = 8 i ^ N . Where i ^ and j ^ are unit vectors along x and y axes respectively. The angular velocity of the particle is
Options
- A8 rad s - 1
- B4 rad s - 1
- C2 r a d   s - 1
- D16   rad s - 1
Correct answer
B. 4 rad s - 1
Step-by-step solution
P → A − P → B = m ( V → A − V → B ) = m v ( j ^ − ( − j ^ ) ) 2 m v j ^ = 2 kg m s − 1 j ^ ..... ( i ) F → A − F → B = m v 2 R ( − i ^ ) − m v 2 R ( + i ^ ) = 2 m v 2 R - i ^ ...(ii) Equating magnitudes (Also using the force relation given in question) 2 m v 2 R = 8 N From the above two equations, we get v R = ω = 4 rad s - 1