Answer to Question #164493 in Electricity and Magnetism for ajay

Question

A final linear velocity of COM for a sphere is found to be x Ωr, when it is made to fallon a surface (coefficient of friction μ). Initially the sphere was rotating with an angular

velocity of  about its own horizontal axis The instance it is made to fall on the

surface, it begins to skid first and then starts rotating without skidding.

(a) Find the value of x?

(b) Distance covered before reacting their final linear velocity is y =( r²ω²) / μg

Find the value of y?

Answer 1

Given that,

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

Coefficient of friction $=\mu$

At the instant, when it falls on the surface of earth, it was not moving so

$\mu mg =m.\frac{d v}{dt}$

$\Rightarrow \omega r \frac{dx}{dt}=\mu g$

$\Rightarrow dx=\frac{\mu g}{\omega r}dt$

taking integration of both side

$x=\frac{\mu g}{\omega r}t$

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

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

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

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

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