A uniform dics of mass m and radius `R=(80)/(23pi^2)`m is pivoted smoothly at `P`. If a uniform ring of mass `m` and radius `R` is welded at the lowest point of the disc, find the period of `SHM` of the system (disc `+` ring). (in seconds)

Answer

2

Explanation

The time period of a physical pendulum is `T=2pisqrt((I_(p))/(Mgr))`
Here we have three quantities `I_(p)`,m and r.
Let us calculate the quantities one by one as follows:
Finding `I_(p):I_(p)=(I_(p))_(ri ng)`
`where (I_(p))_(ring)=I_(A)+m(PA)_(2)=(mR^(2))/(2)+mR^(2)=(3mR^(2))/(2)` and `(I_(p))_(ring)=I_(B)+m(PB)^(2)=mR^(2)+m(3R^(2))=10mR^(2)` Then, we have `I_(p)=(3)/(2)mR^(2)=10mR^(2)=(23)/(2)mR^(2)`
Finding r, `r=2`