wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A particle is moving in a circular path with an angular velocity of 2 rad/s and angular acceleration of α=52ω, where ω is the angular velocity at any particular instant. Find the angular velocity after the particle has moved through an angle of 3π.

A
20.64 rad/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
15.52 rad/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
25.56 rad/s
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
28.25 rad/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 25.56 rad/s
Given angular acceleration is
α=52ω
α=dωdt=52ω

Using chain rule, we get,

dωdθdθdt=52ω

ωdωdθ=52ω

2dω=5 dθ

Integrating with limits on both sides, we get,

2ω2dω=53π0dθ

2(ω2)=5(3π0)

ω=15π+4225.56 rad/s





flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Relative
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon