Question

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

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

(q) 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) $2\pi \sqrt{\frac{3r}{2g}}$ (iv)$T=2\pi \sqrt{\frac{2R}{g}}$

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

Answer 1

The correct option is A p - (i) ; q - (iii)

(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)

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

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

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

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

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

(q)

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

For small $\theta$

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

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

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

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

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

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

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