The disk has a weight of `100 N`and rolls without slipping on the horizontal surface as it oscillates about its equilibrium position. If the disk is displaced, by rolling it counterclockwise `0.4rad`, determine the equation which describes its oscillatory motion when it is released.

Answer

A, C, D

Explanation

In the displaced position,
` E = (1)/(2)mv^(2) + (1)/(2) Iomega ^(2) + (1)/(2)k(2x)^(2)`
`I = (1)/(2) mR^(2)` and `omega = (v)/(R)`
`:. E = (3)/(4)mv^(2) + 2kx^(2)`
`E =` constant
`:. (dE)/(dt) = 0`
or `(3)/(2)mv (dv)/(dt) + 4kx (dx)/(dt) = 0`
Substituting, `(dx)/(dt) = v` and `(dv)/(dt) = a`
`a = - (8k)/(3m).x`
Comparing with, `a = - omega^(2)x`
We have
`omega = sqrt ((8k)/(3m)) = sqrt(((8 xx 1000)/(3 xx 100))/(9.8)) = 16.16rad//s`
`:. theta = theta_(0)cos omega t`
or `theta = 0.4cos (16.16t)`.