A wheel rotates around a stationary axis so that the rotation angle θ varies with time as θ=kt2 , where k=1.5 rad s−2 . Find the magnitude of net acceleration of a point P on the rim at the moment t=2/3s , if the linear velocity of the point P at this moment is v=2 ms−1 .
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a
3 ms−2
b
4 ms−2
c
5 ms−2
d
6 ms−2
answer is C.
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Detailed Solution
ω=dθdt=ddt(kt2) =2kt=2(1.5)t=3t Instantaneous angular acceleration, α=dωdt=3 rad s−2 .Note that this is time independent. (ω)t=23s=αt=3(23)=2rad s−1 v=rω ⇒r=vω=22=1m an=v2r=(2)21=4ms−2 And at=rα=(1)(3)=3 ms−2 anet=an2+ar2=42+32=5 ms−2