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 = $\frac { \mathrm { R } } { 4 }$ , 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)

0.94 s

Answer 2

Time period of a physical pendulum is

$\mathrm { T } = 2 \pi \sqrt { \frac { 1 } { \mathrm { mgh } } }$

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

$\therefore \mathrm { I } = \frac { \mathrm { mR } ^ { 2 } } { 2 } + \mathrm { mr } ^ { 2 } = \frac { \mathrm { mR } ^ { 2 } } { 2 } + \mathrm { m } \left( \frac { \mathrm { R } } { 4 } \right) ^ { 2 } = \frac { \mathrm { mR } ^ { 2 } } { 2 } + \frac { \mathrm { mR } ^ { 2 } } { 16 }$ $\left( \because r = \frac { R } { 4 } \right)$

Here, R = 10 cm $= 0.1 \mathrm {~m} , \mathrm {~h} = \frac { \mathrm { R } } { 4 }$

$= 2 \pi \sqrt { \frac { 9 \times 0.1 } { 4 \times 10 } } = 2 \pi \times \frac { 1 } { 10 } = 0.94 \mathrm {~s}$