Question

A uniform plate of mass M stays horizontally and symmetrically on two wheels rotating in opposite direction (figure 12−E16). 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.
Figure

Answer 1

Answer:

Time period of oscillation of plate is given as

$T = 2\pi\sqrt{\frac{L}{\mu g}}$

Explanation:

If the plate is slightly displaced by distance "x"

Now we will have

$N_1 + N_2 = Mg$

also by torque balance about COM of plate

$N_1(\frac{L}{2} - x) = N_2(\frac{L}{2} + x)$

now we have

$N_1 = N_2\frac{(\frac{L}{2} + x)}{(\frac{L}{2} - x)}$

$N_2(\frac{(\frac{L}{2} + x)}{(\frac{L}{2} - x)} + 1) = Mg$

$N_2 = Mg(\frac{\frac{L}{2} - x}{L})$

$N_1 = Mg(\frac{\frac{L}{2} + x}{L})$

now we have

$F = \mu (N_1 - N_2)$

$F = \mu(\frac{2Mgx}{L})$

now we have

$a = - \frac{2\mu g}{L} x$

so we have

$T = 2\pi\sqrt{\frac{L}{2\mu g}}$

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Topic : SHM

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