A particle of mass 10g moves along a circle of radius 6.4cm with a...

Tangential acceleration ${a}_{t} = r\alpha =$ constant $= K$

$\alpha = \dfrac{K}{r}$

At the end of second revoluation angular velocity is $\omega$ then

${\omega}^{2} - {\omega}_{0}^{2} = 2\alpha\theta$

${\omega}^{2} = 2( \dfrac{K}{r} )( 4\pi )$

${\omega}^{2} = \dfrac{8\piK}{r}$

K.E. of the particle is $= K . E . = \dfrac{1}{2}m{v}^{2}$

$K . E . = \dfrac{1}{2}m{r}^{2}{\omega}^{2}$

$\text{K} . \text{E} . = \dfrac{1}{2}\text{m}( {r}^{2} )( \dfrac{8\text{\pi K}}{r} ) = \dfrac{1}{2}mr( 8\piK )$

$8 \times {10}^{- 4} = \dfrac{1}{2} \times 10 \times {10}^{- 3} \times 6.4 \times {10}^{- 2} \times 8 \times 3.14 \times \text{K}$

$K = \dfrac{2}{6.4 \times 3.14} = 0.1m{s}^{- 2}$