Answer

Verified

395.7k+ views

**Hint:**When an object is rotating in a circle its direction of velocity keeps on changing. When velocity changes there will be an acceleration and this contributes for the force. Now the force which is responsible for the change in direction in case of circular motion is called a centripetal force. We will solve the question by using this.

**Formula used:**

${a_c} = \dfrac{{{v^2}}}{R} = R{\omega ^2}$

**Complete answer:**

Centripetal force is given by the formula ${F_C} = \dfrac{{m{v^2}}}{R}$

Where ‘v’ is the magnitude of velocity of rotation and ‘R’ is the radius of the circle of the circular motion and ‘m’ is the mass of the object which is rotating in a circle.

Force will be the product of mass and acceleration. So the centripetal acceleration will be

${a_c} = \dfrac{{{v^2}}}{R}$

We know that $v = R\omega $

So the centripetal acceleration will be ${a_c} = \dfrac{{{v^2}}}{R} = R{\omega ^2}$

If we refer the diagram below we can find that

It is given that the bead is at rest. So forces will be balanced in both radial and tangential directions.

In the above diagram $mg\cos \theta $ component will be balanced by the normal reaction and component of centripetal force.

Along the tangential direction $mg\sin \theta $ component will be balanced only by $mR{\omega ^2}\cos \theta $ component.

where $R = \dfrac{r}{2}$

So by equating them we will get

$\eqalign{

& mR{\omega ^2}\cos \theta = mg\sin \theta \cr

& \Rightarrow \sin \theta = \dfrac{{\dfrac{r}{2}}}{r} = \dfrac{1}{2} \cr

& \Rightarrow \theta = {30^0} \cr

& \Rightarrow \tan \theta = \tan {30^0} = \dfrac{1}{{\sqrt 3 }} \cr

& \Rightarrow mR{\omega ^2}\cos \theta = mg\sin \theta \cr

& \Rightarrow {\omega ^2} = \dfrac{{g\tan \theta }}{R}{\text{ and R = }}\dfrac{r}{2} \cr

& \Rightarrow {\omega ^2} = \dfrac{{2g\tan {{30}^0}}}{r} \cr

& \therefore {\omega ^2} = \dfrac{{2g}}{{\sqrt 3 r}} \cr} $

**Hence option D is the correct answer.**

**Note:**

Here they had mentioned that the bead is at rest with respect to the rotating wire. This means that if we observe from a wire frame of reference the bead is stationary. The motion of wire will be the same as motion of bead i.e. circular. From the ground frame of reference the bead is rotating in a circle but from the wire frame of reference the bead is in equilibrium so we equated the forces in tangential direction.

Recently Updated Pages

Cryolite and fluorspar are mixed with Al2O3 during class 11 chemistry CBSE

Select the smallest atom A F B Cl C Br D I class 11 chemistry CBSE

The best reagent to convert pent 3 en 2 ol and pent class 11 chemistry CBSE

Reverse process of sublimation is aFusion bCondensation class 11 chemistry CBSE

The best and latest technique for isolation purification class 11 chemistry CBSE

Hydrochloric acid is a Strong acid b Weak acid c Strong class 11 chemistry CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

Define limiting molar conductivity Why does the conductivity class 12 chemistry CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Difference Between Plant Cell and Animal Cell

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Name 10 Living and Non living things class 9 biology CBSE

The Buddhist universities of Nalanda and Vikramshila class 7 social science CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE