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Problem 5: (Bead on a rotating hoop) A bead of mass $m$ is constrained to move on a frictionless, massless hoop of radius $a$. The hoop lies in a vertical plane and is forced to rotate about the vertical with constant angular speed $\omega$, as shown in the figure below. Write down the Lagrangian and find the equations of motion, assuming the only external force is that of gravity. Show that if $\omega$ is greater than a critical value $\omega_0$, there can be a solution in which the particle remains stationary on the hoop at a point other than at the bottom, but that if $\omega < \omega_0$, the only stationary point of the particle is at the bottom of the hoop. What is the value of $\omega_0$?

          Problem 5: (Bead on a rotating hoop) A bead of mass $m$ is constrained to move on a frictionless, massless hoop of radius $a$. The hoop lies in a vertical plane and is forced to rotate about the vertical with constant angular speed $\omega$, as shown in the figure below. Write down the Lagrangian and find the equations of motion, assuming the only external force is that of gravity. Show that if $\omega$ is greater than a critical value $\omega_0$, there can be a solution in which the particle remains stationary on the hoop at a point other than at the bottom, but that if $\omega < \omega_0$, the only stationary point of the particle is at the bottom of the hoop. What is the value of $\omega_0$?

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Problem 5: (Bead on a rotating hoop) A bead of mass m is constrained to move on a frictionless, massless hoop of radius a. The hoop lies in a vertical plane and is forced to rotate about the vertical with constant angular speed ω, as shown in the figure below. Write down the Lagrangian and find the equations of motion, assuming the only external force is that of gravity. Show that if ω is greater than a critical value ω0, there can be a solution in which the particle remains stationary on the hoop at a point other than at the bottom, but that if ω < ω0, the only stationary point of the particle is at the bottom of the hoop. What is the value of ω0?

Added by Eric A.

University Physics with Modern Physics

Hugh D. Young 14th Edition

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Solved by Expert Adi S

09/17/2023

Step 1

The Lagrangian for the system can be written as: L = T - V where T is the kinetic energy and V is the potential energy. The kinetic energy of the bead is given by: T = (1/2) m (a^2 \omega^2) where a is the radius of the hoop and \omega is the angular speed. The Show more…

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Problem 5: (Bead on a rotating hoop) A bead of mass m is constrained to move on a frictionless, massless hoop of radius a. The hoop lies in a vertical plane and is forced to rotate about the vertical with constant angular speed omega , as shown in the figure below. Write down the Lagrangian and find the equations of motion, assuming the only external force is that of gravity. Show that if omega is greater than a critical value omega _(0), there can be a solution in which the particle remains stationary on the hoop at a point other than at the bottom, but that if omega <omega _(0), the only stationary point of the particle is at the bottom of the hoop. What is the value of omega _(0) ? Problem 5: (Bead on a rotating hoop) A bead of mass m is constrained to move on a frictionless. massless hoop of radius a. The hoop lies in a vertical plane and is forced to rotate about the vertical with constant angular speed w, as shown in the figure below. Write down the Lagrangian and find the equations of motion, assuming the only external force is that of gravity. Show that if w is greater than a critical value wo, there can be a solution in which the particle remains stationary on the hoop at a point other than at the bottom, but that if w < wo, the only stationary point of the particle is at the bottom of the hoop. What is the value of wo? Z m

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Transcript

00:01 According to the question, this is the bead given.

00:04 This bead is threaded on a frictionless circular wire hoop.

00:10 Now this is the circular wire hoop on which the bead is threaded.

00:15 This is the normal line across the center of the circle to which the bead is threaded.

00:23 The angle it makes is theta while this is the line which connects.

00:29 This value is rho while the radius is given by r.

00:34 It is said that the rotation, angular rotation is omega.

00:40 The vertical angular velocity.

00:43 Angular velocity is represented by omega.

00:49 Similarly we have theta as the angle from vertical.

00:57 Angle from vertical.

01:00 Now according to this, we need to find the lagrangian derivative for the given system.

01:13 So according to this, we know that lagrangian system.

01:17 What is this? this system basically is the system which gives us difference between the kinetic energy and the potential energy of the bead.

01:36 Kinetic energy is the energy due to motion.

01:40 Energy due to motion.

01:43 While potential energy is the energy due to position.

01:50 Now basically lagrangian system gives us the difference between kinetic energy and potential energy.

01:57 So to find it, we know that kinetic energy of the bead must be t is equal to half mv square.

02:06 So in this case it would be half m then r square omega square cos square theta to find v square.

02:18 Right? this would be the value of t.

02:20 That is the kinetic energy...

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