JEE MAIN - Physics (2019 - 12th April Evening Slot - No. 12)
A smooth wire of length 2$$\pi $$r is bent into a circle and kept in a vertical plane. A bead can slide smoothly on
the wire. When the circle is rotating with angular speed $$\omega $$ about the vertical diameter AB, as shown in figure,
the bead is at rest with respect to the circular ring at position P as shown. Then the value of $$\omega $$2
is equal to -
_12th_April_Evening_Slot_en_12_1.png)
$${{\sqrt 3 g} \over {2r}}$$
$${{2g} \over {\left( {r\sqrt 3 } \right)}}$$
$${{\left( {g\sqrt 3 } \right)} \over r}$$
$${{2g} \over r}$$
Explanation
_12th_April_Evening_Slot_en_12_2.png)
In $$\Delta $$ORP,
sin $$\theta $$ = $${{r/2} \over r}$$ = $${1 \over 2}$$ = sin 30o
$$ \Rightarrow $$ $$\theta $$ = 30o
As the bead is at rest with respect to the circular ring at position P, The net force on bead is zero.
tan $$\theta $$ = $${{m{\omega ^2}r/2} \over {mg}}$$
$$ \Rightarrow $$ tan 30o = $${{{\omega ^2}r} \over {2g}}$$
$$ \Rightarrow $$ $${1 \over {\sqrt 3 }}$$ = $${{{\omega ^2}r} \over {2g}}$$
$$ \Rightarrow $$ $${\omega ^2} = {{2g} \over {r\sqrt 3 }}$$
Comments (0)
