Tangential acceleration at=rα= constant =K
α=rK
At the end of second revoluation angular velocity is ω then
ω2−ω02=2αθ
ω2=2(rK)(4π)
ω2=r8πK
K.E. of the particle is =K.E.=21mv2
K.E.=21mr2ω2
K.E.=21m(r2)(r8πK)=21mr(8πK)
8×10−4=21×10×10−3×6.4×10−2×8×3.14×K
K=6.4×3.142=0.1ms−2