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

Answer

d

Explanation

(d) Given, mass of particle `m =0.01 kg`
Radius of circle along which particle is moving, `r=6.4 cm`.
`therefore` Kinetic energy of particle, `KE=8xx10^(-4)`J
`implies 1/(2)mv^(2)=8xx10^(-4)`J
`impliesv^(2)=(16xx10^(-4))/(0.01)=16xx10^(-2)` .....(i)
As it is given that KE of particle is equal to `8xx10^(-4)`J by the end of second revolution after the beginning of motion of particle. It means, its intial velocity (u) is 0m/s at this moment.
`therefore` By Newton's 3rd equation of motion,
`v^(2)=u^(2)+2a_(t)s`
`implies v^(2)=2a_(t)s or" "v^(2)=2a_(t)(4pir)`
(`therefore` particle covers 2 revolutions)n
`a_(t)=(v^(2))/(8pir)=(16xx10)/(8xx3.14xx6.4xx10^(-4))`
`(therefore` from eq. (i), `V^(2)=16xx10^(-2)`)
`therefore a_(t)=0.1m//s^(2)`