A paritcal of mass `10 g` moves along a circle of radius `6.4 cm` with a constant tangennitial acceleration. What is the magnitude of this acceleration . What is the magnitude of this acceleration if the kinetic energy of the partical becomes equal to `8 xx 10^(-4) J` by the end of the second revolution after the beginning of the motion?

Options

A. `0.2 m//s^(2)`
B. `0.1 m//s^(2)`
C. `0.15 m//s^(2)`
D. `0.18 m//s^(2)`

Explanation

`(1)/(2) mv^(2) = E implies (1)/(2) ((10)/(1000)) v^(2) = 8 xx 10^(-4) m//s`
` implies v^(2) = 16 xx 10^(-2) implies v = 4 xx 10^(-1) = 0.4 m//s`
Now `v^(2) = u^(2) + 2a_(r)s (s = 4 piR)`
`= (16)/(100) = 0^(2) + 2a_(t) (4 xx (22)/(7) xx (6.4)/(100))`
` implies a_(t) = (16)/(100) xx (7 xx 100)/(8 xx 22 xx 6.4) = 0.1 m//s^(2)`