Question

A disc of radius $R = 10\, cm$ oscillates as a physical pendulum about an axis perpendicular to the plane of the disc at a distance $r $ from its centre If $r=\frac{R}{4}$, the approximate period of oscillation is (Take $g=10\, m\, s^{-2})$

• A. $0.84\, s$
• B. $0.94 \,s$
• C. $1.26\, s$
• D. $1.42\, s$

Answer 1

Answer: (B)

Time period of a physical pendulum is
$T=2\pi \sqrt{\frac{I}{mgh}}$

where $I$ is the moment of inertia of the pendulum about an axis through the pivot, $m$ is the mass of the pendulum and $h$ is the distance from the pivot to the centre of mass
In this case, a solid disc of $R$ oscillates as a physical pendulum about an axis perpendicular to the plane of the disc at a distance $r$ from its centre
$\therefore I=\frac{mR^{2}}{2}+mr^{2}=\frac{mR^{2}}{2}+m\left(\frac{R}{4}\right)^{2} =\frac{mR^{2}}{2}+\frac{mR^{2}}{16}$
$=\frac{9mR^{2}}{16} \left(\because r=\frac{R}{4}\right)$
Here, $R=10\,cm=0.1\, m, h=\frac{R}{4}$
$\therefore T=2\pi\sqrt{\frac{\frac{9mR^{2}}{16}}{\frac{mgR}{4}}}=2\pi \sqrt{\frac{9R}{4g}}$
$=2\pi \sqrt{\frac{9\times0.1}{4\times10}}$
$2\pi\times\frac{3}{2}\times\frac{1}{10}$
$=0.94\,s$