Question

Find the time period of small oscillations of the following systems.

(p) A metre stick suspended through the 20 cm mark.

(q) A ring of mass m and radius r suspended through a point on its periphery.

(r) A uniform square plate of edge 'a' suspended through a corner.

(s) A uniform disc of mass m and radius r suspended through a point $\frac{r}{2}$ away from the centre.

(i)$2\pi \sqrt{\frac{2\sqrt{2}a}{3g}}$ (ii) 1.51 sec (iii) $2\pi \sqrt{\frac{3r}{3g}}$ (iv)$T=2\pi \sqrt{\frac{2R}{g}}$

• A. p - (i) ; q - (iii) ; r - (ii) ; s-(iv)
• B. p - (ii) ; q - (iv) ; r - (i) ; s-(iii)
• C. p - (ii) ; q - (i) ; r - (iii) ; s-(iv)
• D. p - (iv) ; q - (iii) ; r - (ii) ; s-(i)

Answer 1

The correct option is B

p - (ii) ; q - (iv) ; r - (i) ; s-(iii)

The ruler is suspended at 20 cm mark.C.O.M is at 50cm so the C.O.M is going in a circle of 30 cm radius.

Circle of 30 cm radius so if ruler is displaced by an angle$\theta$ and left,then torque of mg will try to restore it.

Torque of mg=0.3 mg sin $\theta$ = -l $\alpha$ . . . (1)

$I_{ruler}=\frac{m{L}^2}{12}+m{d}^2$ (L=1m;d=30 cm)
$I_{ruler}=m\left[\begin{array}{c}\frac{1}{12}+(\frac{30}{100}{)}^2\end{array}\right]$
= m (0.173)
$e{q}^n$ (1) becomes 0.3 mg $\theta =-I_r\alpha$
$(\because \theta issmall,sin\theta \simeq \theta )$
$\alpha =-\frac{0.3mg\theta }{I_r}(\alpha =-{\omega }^2\theta )$
$T=\frac{2\pi }{\omega }$
$=2\pi \sqrt{\frac{I_r}{0.3mg}}$
= 1.51 sec

(q)

If displaced by angle $\theta$ (small) restoring torque will be - mgR sin $\theta$ = $l\alpha$
$\alpha =\frac{mgR\theta }{I}(\alpha =-{\omega }^2\theta )$
${\omega }^2=\frac{mgR}{I}$
$T=2\pi \sqrt{\frac{2R}{g}}$

(r)

Distance of Suspension = $\frac{a}{\sqrt{2}}$
$-mg\frac{a}{\sqrt{2}}sin\theta =I\alpha$
$-mg\frac{a}{\sqrt{2}}sin\theta =(Iz+m(\frac{a}{\sqrt{2}}{)}^2)\alpha$ (Parallel axis theorem)

$-mg\frac{a}{\sqrt{2}}sin\theta =(Ix+Iy+m{\left(\frac{a}{\sqrt{2}}\right)}^2)\alpha$ (Perpendicular axis theorem)

$-mg\frac{a}{\sqrt{2}}sin\theta =(\frac{m{a}^2}{12}+\frac{m{a}^2}{12}+m{\left(\frac{a}{\sqrt{2}}\right)}^2)\propto$

$-mg\frac{a}{\sqrt{2}}sin\theta =\frac{2m{a}^2}{3}\propto$

$\propto =-\frac{3g}{a2\sqrt{2}}\theta$

${\omega }^2=\frac{3g}{a2\sqrt{2}}$

$T=2\pi \sqrt{\frac{2\sqrt{2}a}{3g}}$

(s)

$-\frac{mgr}{2}sin\theta =I\propto$

For small $\theta$

$-\frac{mgr}{2}\theta =I\propto$

$-\frac{mgr}{2}\theta =(Ic+m{\left(\frac{r}{2}\right)}^2)\propto$

$-\frac{mgr}{2}\theta =(\frac{m{r}^2}{2}+m{\left(\frac{r}{2}\right)}^2)\propto$

$-\frac{mgr}{2}\theta =\left(\frac{3m{r}^2}{4}\right)\propto$

$\propto =-\frac{2g}{3r}\theta$

${\omega }^2=\frac{2g}{3r}$

$T=2\pi \sqrt{\frac{3r}{2g}}$