A particle of mass m is moving along a circle of radius R such that its tangential acceleration `a_t ` varies with distance covered x as `a_t =ax^2` where `alpha` is a constant.
the kinetic energy , K of the particle varies with the distance as `K= beta x^c` , where `beta` and c are constants. The values of `beta` and c are

Options

A. `beta =(malpha)/3,c=3`
B. `beta =(malpha)/4 ,c=4`
C. `beta = (m alpha)/2, c=4`
D. `beta = (malpha)/2 , c=3`

Answer

A

Explanation

Given mass of a particle = m
tangential acceleration of particle , `a_t =ax^2 ` and kinetic energy , `k= beta x^c`
Here as we know that torque , `tau = F*R = I alpha`
`rArr F*R = I (a_t)/R (because alpha =a/R) `
`rArr F*R = mR^2 (ax^2)/R (because I = MR^2)`
` rArr F =max^2 ` .........(i)
Hence, the work done in displacement of dx,
`W = int f. dx =int max^2 dx `
`W = 1/3 max^3 ` ...... (iii)
From work -energy theorem ,
`W = Delta KE`
here ,w = work done and
`Delta KE ` = change in kinetic Energy ,
`(max^3)/3 = beta x^c ( because K= beta x ^c)`
Now , from the above equation we get,
`beta = (ma)/3 and c =3`