A uniform disc of radius `R=13cm` can rotate about a horizontal axis perpendicular to its plane and passing through the edge of the disc. Find the period of small oscillations of that disc if the logarithmic damping decrement is eqal to `lambda=1.00`.

Explanation

The resotring coouple is
`T=-mg R sin varphi ~= - mg R varphi`
The moment of inertia is
`I=( 3mR^(2))/(2)`
Thus for undamped oscillations
`(3 m R^(2))/(2) ddot(varphi)+mg R varphi=0`
so, `omega_(0)^(2)=(2g)/(3R)`
Also `lambda=(2pi beta)/(omega)=(2pi beta)/ (sqrt(omega_(0)^(2)-beta^(2))`
Hence `(beta)/(sqrt(omega_(0)^(2)-beta^(2)))=(lambda)/( 2pi)` or `(omega_(0))/( sqrt( omega_(0)-beta^(2)))=sqrt(1+((lambda)/( 2pi))^(2))`
Hence finally the period . `T` of small oscillation comes to
`T=(2pi)/(omega)=(2pi)/(omega_(0))xx(omega_(0))/(sqrt(omega_(0)^(2)-beta^(2)))2pisqrt((2R)/(2g)(1+((lambda)/(2pi))^(2)))`
`=sqrt((3R)/(2g)(sqrt(4pi^(2))+ lambda^(2)))=0.90 sec`.