A frame of reference that is accelerated with respect to an inertial frame of reference is called a non-inertial frame of reference. A coordinate system fixed on a circular disc rotating about a fixed axis with a constant angular velocity `omega` is an example of non=inertial frame of reference. The relationship between the force `vecF_(rot)` experienced by a particle of mass m moving on the rotating disc and the force `vecF_(in)` experienced by the particle in an inertial frame of reference is
`vecF_(rot)=vecF_(i n)+2m(vecv_(rot)xxvec omega)+m(vec omegaxx vec r)xxvec omega`.
where `vecv_(rot)` is the velocity of the particle in the rotating frame of reference and `vecr` is the position vector of the particle with respect to the centre of the disc.
Now consider a smooth slot along a diameter fo a disc of radius R rotating counter-clockwise with a constant angular speed `omega` about its vertical axis through its center. We assign a coordinate system with the origin at the center of the disc, the x-axis along the slot, the y-axis perpendicular to the slot and the z-axis along the rotation axis `(vecomega=omegahatk)`. A small block of mass m is gently placed in the slot at `vecr(R//2)hati` at `t=0` and is constrained to move only along the slot.

The distance r of the block at time is

Options

A. (a) `R/4(e^(omegat)+e^(-omegat))`
B. (b) `R/2cos omega t`
C. (c) `R/4(e^(2omegat)+e^(-2omegat))`
D. (d) `R/2 cos 2omegat`

Answer

A

Explanation

Force on the block along slat `=mr omega^2=mv(dv)/(dr)`
`:.` `int_o^vVdv=int_(R//2)^romega^2rdrimpliesV=omegasqrt(r^2-(R^2)/(4))=(dr)/(dt)`
`:.` `int_(R//4)^r(dr)/(sqrt(r^2-(R^2)/(4)))=int_o^t omegadt`
On solving we get
`r+sqrt(r^2-(R^2)/(4))=(R)/(2)e^(wt)`
or `r^2-(R^2)/(4)=(R^2)/(4)e^(2wt)+r^2-2r(R)/(2)e^(wt)`
`:.` `r=(R)/(4)(e^(wt)+e^(-wt))`