A ring is rotating with an angular acceleration given by α=4t3−32t2, where t is time. Find the equation of angular speed ω if the ring starts from rest.
A
t4−t32
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
t3−t
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
t42−t3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
12t2−3t
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is At4−t32 We know that angular acceleration (α) is the time rate of change of angular velocity (ω)
It is given the angular acceleration is, α=4t3−32t2
and hence for finding angular velocity we integrate (α)
α=dωdt⇒ω=∫αdt ω=∫(4t3−32t2)dt =4t44−32t33+c
As it starts from rest t=0, ω=0⇒c=0
So, ω=t4−t32