Home
Class 12
PHYSICS
A thin circular wire of radius R rotatit...

A thin circular wire of radius `R` rotatites about its vertical diameter with an angular frequency `omega` . Show that a small bead on the wire remain at its lowermost point for `omegalesqrt(g//R)` . What is angle made by the radius vector joining the centre to the bead with the vertical downward direction for `omega=sqrt(2g//R)` ? Neglect friction.

Text Solution

AI Generated Solution

To solve the problem, we will analyze the forces acting on the bead on the rotating wire and derive the conditions for it to remain at the lowermost point. ### Step-by-step Solution: 1. **Understanding the Setup**: - We have a circular wire of radius \( R \) rotating about its vertical diameter with an angular frequency \( \omega \). - A small bead is placed on the wire, and we need to analyze its position. ...
Promotional Banner

Topper's Solved these Questions

  • DYNAMICS OF A PARTICLE

    VMC MODULES ENGLISH|Exercise LEVEL 1|75 Videos

  • DYNAMICS OF A PARTICLE

    VMC MODULES ENGLISH|Exercise LEVEL 2|49 Videos

  • DYNAMICS OF A PARTICLE

    VMC MODULES ENGLISH|Exercise LEVEL 0 (SHORT ANSWER TYPE - ii)|11 Videos

  • DC CIRCUIT

    VMC MODULES ENGLISH|Exercise JEE ADVANCED ARCHIVE|68 Videos

  • ELECTROMAGNETIC INDUCTION & ALTERNATING CURRENT

    VMC MODULES ENGLISH|Exercise IN-CHAPTER EXERCISE-G|10 Videos

Similar Questions

Explore conceptually related problems

A thin circular wire of radius R rotates about its vertical diameter with an angular frequency omega . Show that a small bead on the wire remain at its lowermost point for omegalesqrt(g//R) . What is angle made by the radius vector joining the centre to the bead with the vertical downward direction for omega=sqrt(2g//R) ? Neglect friction.

A thin circular wire of radius R rotates about its vertical diameter with an angular frequency omega . Show that a small bead on the wire remain at its lowermost point for omegalesqrt(g//R) . What is angle made by the radius vector joining the center to the bead with the vertical downward direction for omega=sqrt(2g//R) ? Neglect friction.

Moment of inertia of a circular wire of mass M and radius R about its diameter is

A spherical bowl of radius R rotates about the verical diameter with angular velocity omega .The bowl contains a small object inside and in absence of friction, this object takes up a position inside the bowl such that its radius vector makes an angle theta with the vertical (see figure). Then

A ring of radius r is rotating about a vertical axis along its diameter with constant angular velocity omega.A read of mass m remains at rest w.r.t. ring at the position shown in figure. Then w^(2) is:

A ring of radius r is rotating about a vertical axis along its diameter with constant angular velocity omega.A read of mass m remains at rest w.r.t. ring at the position shown in figure. Then w^(2) is:

A thin circular ring of mass M and radius r is rotating about its axis with an angular speed omega . Two particles having mass m each are now attached at diametrically opposite points. The angular speed of the ring will become

A thin circular ring of mass M and radius r is rotating about its axis with an angular speed omega . Two particles having mass m each are now attached at diametrically opposite points. The angular speed of the ring will become

A thin circular ring of mass M and radius R is rotating about its axis with constant angular velocity omega . The objects each of mass m are attached gently to the ring. The wheel now rotates with an angular velocity.

A solid sphere of radius r and mass m rotates about an axis passing through its centre with angular velocity omega . Its K.E . Is