Find the time period of small oscillations of the following systems.
(a) A uniform rod of mass m and length L is suspended through a pin hole at distance `L//4` from top as shown.
(b) A ring of mass m and radius r suspended through a point on its periphery.
(c) A uniform disc of mass m and radius r suspended through a point `r//2` away from centre.
(d) A uniform square plate of edge a suspended through a corner.

Explanation

(a)

Restoring torque about O
`tau=-mgL/4sintheta`
`tau=-mgL/4theta(sintheta=theta)`
`Ialpha=-mgL/4theta`
`alpha=-(mgL)/(I)theta=-omega^2theta` (case of SHM)
`T=2pisqrt((I)/(mgL//4))`
I: M.I. about OA (OA is `_|_^(are)` to plane of motion)
`I=(mL^2)/(12)+m(L/4)^2=(mL^2)/(12)+(mL^2)/(16)=(mL^2)/(48)`
`T=2pisqrt((7mL^2//48)/(mgL//4))=2pisqrt((7L)/(12g))`
(b)
`tau=-mgrsintheta`
`I=mr^2+mr^2=2mr^2`
`sintheta=theta`
`tau=-mgrtheta`
`Ialpha=-mgrtheta`
`alpha=-(mgr)/(I)theta=-omega^2theta`
`T=2pisqrt((I)/(mgr))=2pisqrt((2mr^2)/(mgr))=2pisqrt((2r)/(g))`
(c)
`tau=-mgr/2sintheta`, `I=1/2mr^2+m(r//2)^2=3/4mr^2`
`tau=-(mgr)/(2)theta` (`:' sin theta=theta`)
`Ialpha=-(mgr)/(2)thetaimpliesalpha=-(mgr)/(2I)`, `theta=-omega^2theta`
`T=2pisqrt((2I)/(mgr))=2pisqrt((2xx3mr^2//4)/(mgr))`
`=2pisqrt((3r)/(2g))`
(d)
`tau=-mga/sqrt2sintheta`
`I=(ma^2)/(6)+m(a/sqrt2)^2=2/3ma^2`
`tau=-(mga)/(sqrt2)theta` (`:' sin theta=theta`)
`Ialpha=-(mga)/(sqrt2)theta`
`alpha=-(mga)/(sqrt2I)theta=-omega^2theta`
`T=2pisqrt((sqrt2I)/(mga))=2pisqrt((sqrt2*2/3ma^2)/(mga))`
`=2pisqrt((2sqrt2a)/(3g))`