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Question
Find the time period of small oscillations of the following systems. (a) A metre stick suspended through the 20 cm mark. (b) A ring of mass m and radius r suspended through a point on its periphery. (c) A uniform square plate of edge a suspended through a corner. (d) A uniform disc of mass m and radius r suspended through a point r/2 away from the centre.
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Solution
(a) Moment Of inertia \[\left( I \right)\] about the point X is given by ,
I = IC.G + mh2
\[= \frac{m l^2}{12} + m h^2 \]
\[ = \frac{m l^2}{12} + m \left( 0 . 3 \right)^2 \]
\[ = m\left( \frac{1}{12} + 0 . 09 \right)\]
\[ = m\left( \frac{1 + 1 . 08}{12} \right)\]
\[ = m\left( \frac{2 . 08}{12} \right)\]
The time Period \[\left( T \right)\] is given by,
\[T = 2\pi\sqrt{\frac{I}{mgl}}\] \[\text { where } I = \text{ the moment of inertia, and } \] \[ l= \text{ distance between the centre of gravity and the point of suspension . }\] \[\text {On substituting the respective values in the above formula, we get: }\] \[T = 2\pi\sqrt{\frac{2 . 08 m}{m \times 12 \times 9 . 8 \times 0 . 3}}\]
\[ = 1 . 52 s\]
(b) Moment Of inertia \[\left( I \right)\] about A is given as,
I = IC.G. + mr2 = mr2 + mr2 = 2mr2
The time period (T) will be,
\[T = 2\pi\sqrt{\frac{I}{mgl}}\]
\[\text { On substituting the respective values in the above equation, we have: }\]
\[T = 2\pi\sqrt{\frac{2m r^2}{mgr}}\]
\[ = 2\pi\sqrt{\frac{2r}{g}}\]
(c) Let I be the moment of inertia of a uniform square plate suspended through a corner.
\[I = m\left( \frac{a^2 + a^2}{3} \right) = \frac{2m}{3} a^2\]
In the
\[\bigtriangleup\] ABC , l2 + l2 = a2
\[\therefore l = \frac{a}{\sqrt{2}}\]
\[ \Rightarrow T = 2\pi\sqrt{\frac{I}{mgl}}\]
\[ = 2\pi\sqrt{\frac{2m a^2}{3mgl}}\]
\[ = 2\pi\sqrt{\frac{2 a^2}{3ga\sqrt{2}}}\]
\[ = 2\pi\sqrt{\frac{\sqrt{8}a}{3g}}\]
(d)
\[\text { We know }\]
\[ h = \frac{r}{2}\]
\[\text { Distance between the C . G . and suspension point }, l = \frac{r}{2}\]
Moment of inertia about A will be:
l = IC.G. + mh2
\[= \frac{m r^2}{2} + m \left( \frac{r}{2} \right)^2 \]
\[ = m r^2 \left( \frac{1}{2} + \frac{1}{4} \right) = \frac{3}{4}m r^2\]
Time period (T) will be,
\[T = 2\pi\sqrt{\frac{I}{mgl}}\]
\[ = 2\pi\sqrt{\frac{3m r^2}{4mgl}} = 2\pi\sqrt{\frac{3r}{2g}}\]
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