Answer to Question #163234 in Optics for ajay

Question

A final linear velocity of Com for a sphere is found to be x wr, when it is mode to fall on a surface (coefficient of friction r). Initially the sphere was rotating with an angular velocity of w about its own horizontal axis The instance it is mode to fall on the surface, it begins to stand first and then starts rotating without standing. (a) Find the value of x? (b) Distance covered before reacting their final linear velocity is y = r²omega²/meuo x g  Find the value of y 

Answer 1

Given that,

Final velocity of the center of mass of the object $= x\omega \mu$

Coefficient of friction $=\mu$

$y = r^2\frac{\omega^2}{\mu x g}$

$\Rightarrow \frac{dy}{dt}=\frac{r^2}{\mu g}(\frac{d}{dt}(\frac{\omega^2}{x}))$

$\Rightarrow \frac{dy}{dt}=\frac{r^2}{\mu g }(\frac{2\omega x\frac{d\omega}{dt}-\omega^2\frac{dx}{dt}}{x^2})$

$d= \sqrt{x^2+y^2}=\sqrt{(x\omega \mu)^2+(r^2\frac{\omega^2}{\mu x g})^2}$