A uniform disc of mass m and radius `R=(80)/(23pi^2)`m is pivoted smoothly at `P`. If a uniform disc 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)

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 quentities one by one as follows: Finding `I_P:I_P=(I_P)_(disc)+(I_P)_(ri ng)`
Where `(I_P)_(disc)=I_A+m(PA)^2=(mR^2)/(2)+mR^2=(3mR^2)/(2)`
and `(I_P)_(ri ng)=I_b(PB)^2=mR^2+m(3R)=10mR^2`
Then, we have `I_P=(3)/(2)mR^2=10mR^2=(23)/(2)mR^2`
Finding `r`.
`r=vecr_C=(m_1vecr_1+m_2+vecr_(2C))/(m_1+m_2)`
where `m_1=m`,`m_2=m`
`vecr_(1C)=-R` and `r_(2C)=-3R`
This gives `r_C=2R`
Finding M:
`M=(m)_(disc)+(m)_(ri ng)=m+m=2m`
Substituting `I_P=(23)/(2)mR^2`,`M=2m` and `r=2R` in the expression
`T=2pisqrt((I_P)/(Mgr))` we have `T=pisqrt((23R)/(g))`
After subsituting the value we get `t=2s`.