Question

A uniform plate of mass M stays horizontally and symmetrically on two wheels rotating in opposite directions (figure 12-E16). The separation between the wheels is L. The friction coefficient between each wheel and the plate is $\mu .$ Find the time period of oscillation of the plate if it is slightly displaced along its length and released.

Answer 1

Let 'x' be the displacement of the plank towards left. Now the centre of gravity is also displaced through 'x'.

$R_1+R_2=mg$

Taking moment about g, we get,

$R_1(\frac{1}{2}-x)=R_2(\frac{1}{2}-x)$

$=(mg-R_1)(\frac{1}{2}+x)...\left(1\right)$

So, $R_1(\frac{1}{2}-x)=(mg-R_1)(\frac{1}{2}+x)$

$\Rightarrow R_1\frac{1}{2}-R_{1}x=mg\frac{1}{2}-R_{1}x+mgx-R_1\frac{1}{2}$

$\Rightarrow R_1\frac{1}{2}+R_1\frac{1}{2}=mg(x+\frac{1}{2})$

$\Rightarrow R_1(\frac{1}{2}+\frac{1}{2})=mg\left(\frac{2x+1}{2}\right)$

$\Rightarrow R_{1}l=mg(2x+1)$

$\Rightarrow R_1=mg\frac{(1+2x)}{2l}...\left(i\right)$

Now, $F_1=\mu R_1=\mu mg\frac{(1+2x)}{2l}$

Similarly, $F_2=\mu R_2=\mu mg\frac{(1+2x)}{2l}$

Since, $F_1>F_2$

$\Rightarrow F_1-F_2=ma=\left(2\mu \frac{mg}{l}\right)x$

$\frac{a}{x}=2\mu \frac{mg}{l}={\omega }^2$

$\Rightarrow =\omega \mu \sqrt{\frac{g}{l}}$

$\therefore$ Time period $=2\pi \sqrt{\frac{l}{2\mu g}}$