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Question
A uniform plate of mass M stays horizontally and symmetrically on two wheels rotating in opposite direction in Figure . The separation between the wheels is L. The friction coefficient between each wheel and the plate is μ. Find the time period of oscillation of the plate if it is slightly displaced along its length and released.
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Solution
Let x be the displacement of the uniform plate towards left.
Therefore, the centre of gravity will also be displaced through displacement x.
At the displaced position,
R1 + R2 = mg
Taking moment about g, we get:
\[R_1 \left( \frac{L}{2} - x \right) = R_2 \left( \frac{L}{2} + x \right) = \left( Mg - R_1 \right) \left( \frac{L}{2} + x \right) . . . . \left( 1 \right)\]
\[ \therefore R_1 \left( \frac{L}{2} - x \right) = \left( Mg - R_1 \right) \left( \frac{L}{2} + x \right)\]
\[ \Rightarrow R_1 \frac{L}{2} - R_1 x = Mg\frac{L}{2} - R_1 x + Mgx - R_1 \frac{L}{2}\]
\[ \Rightarrow R_1 \frac{L}{2} + R_1 \frac{L}{2} = Mg\left( x + \frac{L}{2} \right)\]
\[ \Rightarrow R_1 \left( \frac{L}{2} + \frac{L}{2} \right) = Mg\left( \frac{2x + L}{2} \right)\]
\[ \Rightarrow R_1 L = Mg\left( 2x + L \right)\]
\[ \Rightarrow R_1 = Mg\frac{\left( L + 2x \right)}{2L}\]
\[\text { Now }, \]
\[ F_1 = \mu R_1 = \mu Mg\frac{\left( L + 2x \right)}{2L}\]
\[\text { Similarly }, \]
\[ F_2 = \mu R_2 = \mu Mg\frac{\left( L - 2x \right)}{2L}\]
\[\text { As } F_1 > F_2 , \text { we can write: }\]
\[ F_1 - F_2 = Ma = \left( 2\mu\frac{Mg}{L} \right)x\]
\[ \frac{a}{x} = 2\mu\frac{g}{L} = \omega^2 \]
\[ \Rightarrow \omega = \sqrt{\frac{2\mu g}{L}}\]
\[\text { Time period }\left( T \right)\text{ is given by }, \]
\[T = 2\pi\sqrt{\frac{L}{2\mu g}}\]
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