Guest

a particle starts from the point (0m,8m) and moves with uniform velocity of 3i m/s.after 5 seconds the angular velocity of the particle about the origin will be

a particle starts from the point (0m,8m) and moves with uniform velocity of 3i m/s.after 5 seconds the angular velocity of the particle about the origin will be

Grade:11

3 Answers

Arun
25750 Points
6 years ago
Particle starts from (0m , 8m) from rest with uniform velocity 3 m/s in the x-direction so after 5 sec the particle will traverse a distance along x-axis is 3*5=15 m so new position of the particle is ​(15m , 8m). So the distance of this new position to the origin is 
\sqrt(15- 0)2 + (8 - 0)2
= 17
Angular velocity about origin is given by
W = v/r
= 3/17 rad/sec
Zameer Ansari
19 Points
6 years ago
By conservation of angular momentum: mvr=mVR vr=VRwhere V is final velocity after 5 sec and R is radius at that moment i.e the position vector at that time. 3×8=V×17 V=24/17 ω=V/R ω=24/17*17 ω=24/289
Rishi Sharma
askIITians Faculty 646 Points
3 years ago
Dear Student,
Please find below the solution to your problem.

Particle starts from (0m , 8m) from rest with uniform velocity 3 m/s in the x-direction so after 5 sec the particle will traverse a distance along x-axis is 3*5=15 m so new position of the particle is​(15m,8m). So the distance of this new position to the origin is
√(15- 0)^2+ (8 - 0)^2
= 17
Angular velocity about origin is given by
W = v/r
= 3/17 rad/sec

Thanks and Regards

Think You Can Provide A Better Answer ?

ASK QUESTION

Get your questions answered by the expert for free