A particle is initially at rest, moves along in a circle of radius R=2m with an angular acceleration α=π8rad/sec2. The magnitude of average velocity of the particle over the time it moves by half of the circle is
A
4m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
3m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
2m/s
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
1m/s
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct option is D1m/s Given, radius of circle R=2m, angular acceleration α=π8rad/s2
Since, particle was initially at rest, ω0=0
Using the formula, θ=ω0t+12αt2 where, θ is the angular displacement ⇒θ=12αt2=πt216...(i)
Also, we have for half circle, θ=π...(ii)
Using (i)&(ii) we have π=πt216⇒t2=16⇒t=4sec
Now, average velocity is given by vavg=total displacementtotal time taken ⇒vavg=2R4=44=1m/s