1.

A particle of mass m is moving along a circle of radius R such that its tangential acceleration a_tvaries with distance coveredx 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

Answer»

`beta =(malpha)/3,c=3`
`beta =(malpha)/4 ,c=4`
`beta = (m alpha)/2, c=4`
`beta = (malpha)/2 , c=3`

Solution :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) `
`rArrF*R = mR^2 (ax^2)/R (becauseI = 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`


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