A disc is free to rotate about an axis passing through its centre and perpendicular to its plane. The moment of inertia of the disc about its rotation axis is I. A light ribbon is tightly wrapped over it in multiple layers. The end of the ribbon is pulled out at a constant speed of u. Let the radius of the ribboned disc be R at any time and thickness of the ribbon be `d (lt lt R)`. Find the force (F) required to pull the ribbon as a function of radius R.

Options

A. `(Idu^(2))/(2piR^(4))`
B. `(Idu^(2))/(3piR^(4))`
C. `(2Idu^(2))/(piR^(4))`
D. `(3Idu^(2))/(piR^(4))`

Answer

A

Explanation

`omegar = u = a` constant
`lnomega + lnR = lnu rArr (1)/(omega)(domega)/(dt) + (1)/(R)(dR)/(dt) = 0`
`:. (domega)/(dt) = (omega)/(R) (dR)/(dt)`
Angular acceleration `alpha = (omega)/(R)(dR)/(dt)".............."`
It we consider an interval `'dt'` in which radius decreases by
`dR`, then `2pi RdR = (udt)d`
`:. (dR)/(dt) = (u.d)/(2piR) = (u^(2)d)/(2piR^(3))`
Put in `(1) alpha = (omega)/(R).(ud)/(2piR) = (u^(2)d)/(2piR^(3))`
`rArr F.R = l. (u^(2)d)/(2piR^(3))rArr F = (lu^(2)d)/(2piR^(4))`