Rotational MotionHard
Question
The instantaneous anglular position of a point on a rotating wheel is given by the equation θ(t) = 2t3 - 6t2. The torque on the wheel becomes zero at
Options
A.t = 1 s
B.t = 0.5 s
C.t = 0.25 s
D.t = 2 s
Solution
Given θ(t) = 2t3 - 6t2
∴
= 6t2 - 12t
= 12t - 12
Angular acceleration, α = = 12t - 12
When angular acceleration (α) is zero, than the torque on the wheelbecomes zero (∵
= Iα)
⇒ 12t - 12 = 0 or, t = l s
∴
= 6t2 - 12t = 12t - 12Angular acceleration, α =
= 12t - 12When angular acceleration (α) is zero, than the torque on the wheelbecomes zero (∵
= Iα)⇒ 12t - 12 = 0 or, t = l s
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