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 net reaction of the disc on the block is

Options

A. `m omega^(2) R sin omega t hatj - m ghatk`
B. `1/2 m omega^(2) R (e^(omega t) -e^(-omega t))hatj + mg hatk`
C. `1/2m omega^(2)R(e^(2omegat) - e^(-2omegat))hatj - mg hatk`
D. `-momega^(2) R cos omega hatj - m g hatk`

Answer

B

Explanation

From previous question, term `2m(vecV_(rxt) xx vecomega)`
gives normal reaction from edge `(N_(e))`
`V_(rxt) = (dr)/(dt) = R/4 [e^(omegat) - e^(-omegat)] =(Romega)/4 [e^(omega t) - e^(-omegat)]`
So, `N_(e) = 2m.(Romega)/4 e^(omegat) - e^(-omegat).omega , therefore vecN_(e) =(MRomega^(2))/2(e^(omegat) -e^(-omegat))hatj`
Normal reaction from bottom of slot `=mghatk , therefore` Net reaction from slot.
`vecR = (MRomega^(2))/2(e^(omegat) -e^(-omegat))hatj + mghatk` Hence (B)