Answer to Question #184762 in Classical Mechanics for sinethemba

Question

.A bead slides on a smooth rod which is rotating about one end in a vertical plane with uniform angular velocity Omega .form the equation of motion ?

Answer 1

Let the mass of the bead be m and it is rotating with angular velocity $\omega$ .

Applying the conservation of energy,

$\Rightarrow L= \frac{m v^2}{2}+\frac{mr\omega^2}{2}-mgr\sin\theta$

Now, applying Lagrangian equation,

$\frac{\delta }{\delta t}(\frac{\delta L}{\delta v})-\frac{\delta L}{\delta r}=0$

$\Rightarrow \frac{\delta L}{\delta r}=mr (\dfrac{\delta\omega}{\delta t})^2 -mg\sin\theta$

$\Rightarrow \frac{\delta L}{\delta v}= m\frac{dr}{dt}$

$\Rightarrow \frac{\delta }{\delta t}(\frac{\delta L}{\delta v})=m\frac{d^2r}{dt^2}$

$m\dot{\dot{r}}=mr\omega^2-mg\sin\omega t$