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Figure 7.16 is a bird's-eye view of a smooth horizontal
Chapter 7, Problem 7.35(choose chapter or problem)
Figure 7.16 is a bird's-eye view of a smooth horizontal wire hoop that is forced to rotate at a fixed angular velocity co about a vertical axis through the point A. A bead of mass m is threaded on the hoop and is free to move around it, with its position specified by the angle \(\phi\) that it makes at the center with the diameter AB. Find the Lagrangian for this system using \(\phi\) as your generalized coordinate. (Read the hint in Problem 7.29.) Use the Lagrange equation of motion to show that the bead oscillates about the point B exactly like a simple pendulum. What is the frequency of these oscillations if their amplitude is small?
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QUESTION:
Figure 7.16 is a bird's-eye view of a smooth horizontal wire hoop that is forced to rotate at a fixed angular velocity co about a vertical axis through the point A. A bead of mass m is threaded on the hoop and is free to move around it, with its position specified by the angle \(\phi\) that it makes at the center with the diameter AB. Find the Lagrangian for this system using \(\phi\) as your generalized coordinate. (Read the hint in Problem 7.29.) Use the Lagrange equation of motion to show that the bead oscillates about the point B exactly like a simple pendulum. What is the frequency of these oscillations if their amplitude is small?
Step 1 of 4
The angular displacement of the loop is equal to the product of angular velocity \(\omega\) and time \(t\).
\(\theta = \omega t\)
Let us consider the point A as the origin. After time \(t\), the bead's angular displacement will be \(\omega t + \phi\) and the loop's \(\omega t\).
The position of the bead will be,
\(r = \left( {R\cos \left( {\omega t + \phi } \right),R\sin \left( {\omega t + \phi } \right)} \right) + \left( {R\cos \left( {\omega t} \right),R\sin \left( {\omega t} \right)} \right)\)
\( = R\left( {\cos \left( {\omega t + \phi } \right) + \cos \left( {\omega t} \right),\sin \left( {\omega t + \phi } \right) + \sin \left( {\omega t} \right)} \right)\)
Differentiate the above equation to get the velocity of the loop.
The velocity of the bead will be,
\(v = \mathop r\limits^ \cdot \)
\( = R\left[ { - \sin \left( {\omega t + \phi } \right)\left( {\omega + \mathop \phi \limits^ \cdot } \right) - \left( {\sin \omega t} \right)\omega ,\cos \left( {\omega t + \phi } \right)\left( {\omega + \mathop \phi \limits^ \cdot } \right) + \left( {\cos \omega t} \right)\omega } \right]\;\;\;\;\;...\;\left( 1 \right) \)