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Q.

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
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