Find time period of uniform disc of mass m and radius r suspended through point `r//2` away from centre, oscillating in a plane parallel to its plane.

Answer

(a) `T = 2pisqrt((7l)/(12g))` , (b) `2pi sqrt((2r)/(g))`
(c) `2pisqrt((sqrt(8)a)/(3g)) , 2pisqrt((3r)/(2g))`

Explanation

Time period of compound pendulum is `T = 2pisqrt((I_(cm) + md^(2))/(mgl))`
For rod of length `l, T = 2pisqrt(((ml^(2))/(12)+m((l)/(4))^(2))/(mgl//4)) = 2pisqrt(((ml^(2)(4+3))/(48))/(mgl//4))`
`T = 2pisqrt((7l)/(12g))` Ans.
(b) `T=2pisqrt((mr^(2)+mr^(2))/(mgr)) : T=2pisqrt((2r)/(g))` Ans.
(c) `T=2pisqrt(((mg)/(6)+m((a)/(sqrt(2)))^(2))/(mgsqrt((a)/sqrt(2)))) : T=2pisqrt((sqrt(8a))/(3g))` Ans.
(d) `T=2pisqrt(((mr^(2))/(2)+m((r)/(2))^(2))/(mg'(r)/(2))) : T = 2pi sqrt((3r)/(2g))`