The angular position of a point on the rim of a rotating wheel is given by θ=4t−3t2+t3, where θ is in radians and t is in seconds. What is the instantaneous angular acceleration at t=2sec?
A
4rad/s2
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B
2rad/s2
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C
6rad/s2
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D
3rad/s2
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Solution
The correct option is C6rad/s2 Given, θ=4t−3t2+t3 ω=dθdt & α=dωdt ⇒ω=dθdt=4−6t+3t2
& α=dωdt=ddt(4−6t+3t2) =−6+6t
At t=2s, α=−6+6×2 =6rad/s2