The angular velocity of a point on the rim of a rotating wheel is given by ω=(4−6t+3t2)rad/s. What is the average angular acceleration for the time interval t=2s to t=4s ?
A
12rad/s2
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B
6rad/s2
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C
24rad/s2
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D
3rad/s2
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Solution
The correct option is A12rad/s2 As we know, Average angular acceleration aavg=ΔωΔt=ωf−ωitf−ti =ω(t=4s)−ω(t=2s)4−2 From ω=(4−6t+3t2)rad/s ω(t=2s)=(4−6×2+3×22) =4rad/s ω(t=4s)=(4−6×4+3×42) =28rad/s