Question

the ring which can slide along the rod are kept at midpoint of a smooth rod of length L.The rod is rotated with constant angular velocity w about vertical axis passing throgh its one end . The ring is released from midpoint . The velocity of the ring,when it just leaves the rod is

Answer 1

The velocity of the ring,when it just leaves the rod is $v^, = \frac{\sqrt{7}}{2}\, \omega L$

Explanation:

Centrifugal force formula,

$m\omega^2x=ma$

$\omega^2 x= \frac{vdv}{dx}$

Take integral both sides,

$\int_{L/2}^{L}\omega^2xdx= \int_{0}^{v}vdv$

$\omega^2(\frac{x^2}{2}) = (\frac{v^2}{2})\omega^2(\frac{L^2}{2}-\frac{L^3}{8}) = \frac{v^2}{2}$

$v=\frac{\sqrt{3}}{2} \, \omega L$

Velocity at time of leaving the rod,

$v^,= \sqrt {(\omega L)^2+ (\frac{\sqrt{3}}{2} \omega L)^2)}$

$v^, = \frac{\sqrt{7}}{2}\, \omega L$

Answer 2

Given that,

Length = L

Angular velocity = ω

We need to calculate the velocity of the ring

Using centrifugal force

$m\omega^2x=ma$

$\omega^2x= \dfrac{vdv}{dx}$

$\omega^2 x dx=vdv$

On integrate both side

$\int_{\frac{L}{2}}^{L}\omega^2 xdx=\int_{0}^{v}{vdv}$

$\omega^{2}(\dfrac{x^2}{2})_{\frac{L}{2}}^{L}=(\dfrac{v^2}{2})_{0}^{v}$

$\omega^2(\dfrac{L^2}{2}-\dfrac{L^2}{8})=\dfrac{v^2}{2}$

$v=\dfrac{\sqrt{3}}{2}\omega L$

We need to the calculate the velocity at time of leaving rod

Using formula for velocity

$v'^2=(\omega L)^2+(\dfrac{\sqrt{3}}{2}\omega L)^2$

$v'=\sqrt{(\omega L)^2+(\dfrac{\sqrt{3}}{2}\omega L)^2}$

$v'=\dfrac{\sqrt{7}}{2}\omega L$

Hence, The velocity at time of leaving rod is $\dfrac{\sqrt{7}}{2}\omega L$