Question

A table with smooth horizontal surface is fixed in a cabin that rotates with a uniform angular velocity ω in a circular path of radius R. A smooth groove AB of length L(< Figure

Answer 1

Explanation:

Centripetal acceleration acting on the Table = ω²r

Its Component on AB = ω²r Cosθ

Given in figure

A table with smooth level surface is fixed in a lodge that pivots with a uniform precise speed ω in a roundabout way of span R. A smooth section AB of length L(<<R) is made the outside of the table. The furrow makes an edge θ with the sweep OA of the hover in which the lodge pivots. A little molecule is kept at the point A ready and is discharged to move at the point A ready and is discharged to move along AB

Now, Using second equation of the Motion,  

$\begin{equation}s=u t+\frac{1}{2} a t^{2}$

Initial velocity = u = 0

a= ω²r Cosθ

S = L

$\begin{equation}\begin{aligned}&L=0+\frac{1}{2} \omega^{2} r \cos \theta t^{2}\\&L=\frac{1}{2} \omega^{2} r \cos \theta t^{2}\\&2 L=\omega^{2} r \cos \theta t^{2}\\&\frac{2 L}{\omega^{2} r}=\cos \theta t^{2}\\&\frac{2 L}{\omega^{2} r \cos \theta}=t^{2}\end{aligned}$

$\begin{equation}t=\sqrt{\frac{2 L}{\omega^{2} r \cos \theta}}$

Answer 2

${\bold{\huge{\red{\underline{\green{ANSWER}}}}}}$

Please refer the above attachment