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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


3 years ago

Arun
25768 Points
							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= 17Angular velocity about origin is given byW = v/r= 3/17 rad/sec

3 years ago
Zameer Ansari
19 Points
							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

2 years ago
Rishi Sharma
623 Points
							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= 17Angular velocity about origin is given byW = v/r= 3/17 rad/secThanks and Regards

5 months ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions