Circular MotionHard
Question
A wheel rotates with an angular acceleration given by α = 4at3 - 3bt2, where t is the time and a and b are constants. If wheel has initial angular speed ω0, then the equation for the angular speed will be :-
Options
A.ω = at4 - bt3 - ω0
B.ω = at3 - bt4 + ω0
C.ω = ω0 + at4 - bt3
D.ω = 12at2 - 6 bt + ω0
Solution
α = 
= 4at3 - 3bt2
⇒
(4at3 - 3bt2)dt
⇒ ω - ω0 = (at4 - bt3)0t
⇒ ω = ω0 + at4 - bt3
= 4at3 - 3bt2⇒
(4at3 - 3bt2)dt⇒ ω - ω0 = (at4 - bt3)0t
⇒ ω = ω0 + at4 - bt3
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