Answer :
Certainly! Let's derive the equations of motion for the given Lagrangian using the Euler-Lagrange equation. Given the Lagrangian:
[tex]\[ L = \frac{1}{2} m b^2 \dot{\theta}^2 + \frac{1}{2} m b^2 \dot{\phi}^2 \sin^2 \theta + m g b \cos \theta \][/tex]
where [tex]\( m \)[/tex] is the mass, [tex]\( b \)[/tex] is a constant length, [tex]\( g \)[/tex] is the gravitational acceleration, [tex]\( \theta \)[/tex] is the polar angle, and [tex]\( \phi \)[/tex] is the azimuthal angle.
The Euler-Lagrange equation is given by:
[tex]\[ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}_i} \right) - \frac{\partial L}{\partial q_i} = 0 \][/tex]
where [tex]\( q_i \)[/tex] represents the generalized coordinates which are [tex]\( \theta \)[/tex] and [tex]\( \phi \)[/tex] in our case.
### For [tex]\( \theta \)[/tex]:
1. Compute [tex]\(\frac{\partial L}{\partial \dot{\theta}}\)[/tex]:
[tex]\[ \frac{\partial L}{\partial \dot{\theta}} = \frac{\partial}{\partial \dot{\theta}} \left( \frac{1}{2} m b^2 \dot{\theta}^2 \right) = m b^2 \dot{\theta} \][/tex]
2. Compute [tex]\(\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{\theta}} \right)\)[/tex]:
[tex]\[ \frac{d}{dt} \left( m b^2 \dot{\theta} \right) = m b^2 \ddot{\theta} \][/tex]
3. Compute [tex]\(\frac{\partial L}{\partial \theta}\)[/tex]:
[tex]\[ \frac{\partial L}{\partial \theta} = \frac{\partial}{\partial \theta} \left( \frac{1}{2} m b^2 \dot{\phi}^2 \sin^2 \theta + m g b \cos \theta \right) = \frac{1}{2} m b^2 \dot{\phi}^2 \cdot 2 \sin \theta \cos \theta - m g b \sin \theta = m b^2 \dot{\phi}^2 \sin \theta \cos \theta - m g b \sin \theta \][/tex]
Thus, the Euler-Lagrange equation for [tex]\( \theta \)[/tex] is:
[tex]\[ m b^2 \ddot{\theta} - \left( m b^2 \dot{\phi}^2 \sin \theta \cos \theta - m g b \sin \theta \right) = 0 \][/tex]
Simplifying, we get:
[tex]\[ \boxed{\ddot{\theta} - b \dot{\phi}^2 \sin \theta \cos \theta + \frac{g}{b} \sin \theta = 0} \][/tex]
### For [tex]\( \phi \)[/tex]:
1. Compute [tex]\(\frac{\partial L}{\partial \dot{\phi}}\)[/tex]:
[tex]\[ \frac{\partial L}{\partial \dot{\phi}} = \frac{\partial}{\partial \dot{\phi}} \left( \frac{1}{2} m b^2 \dot{\phi}^2 \sin^2 \theta \right) = m b^2 \dot{\phi} \sin^2 \theta \][/tex]
2. Compute [tex]\(\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{\phi}} \right)\)[/tex]:
[tex]\[ \frac{d}{dt} \left( m b^2 \dot{\phi} \sin^2 \theta \right) = m b^2 \left( \ddot{\phi} \sin^2 \theta + 2 \dot{\phi} \sin \theta \cos \theta \dot{\theta} \right) \][/tex]
3. Compute [tex]\(\frac{\partial L}{\partial \phi}\)[/tex]:
[tex]\[ \frac{\partial L}{\partial \phi} = 0 \quad (\text{since } L \text{ does not depend on } \phi) \][/tex]
Thus, the Euler-Lagrange equation for [tex]\( \phi \)[/tex] is:
[tex]\[ m b^2 \left( \ddot{\phi} \sin^2 \theta + 2 \dot{\phi} \sin \theta \cos \theta \dot{\theta} \right) = 0 \][/tex]
Simplifying, we get:
[tex]\[ \boxed{\ddot{\phi} \sin^2 \theta + 2 \dot{\phi} \sin \theta \cos \theta \dot{\theta} = 0} \][/tex]
In conclusion, we have derived the equations of motion for the given Lagrangian:
[tex]\[ \boxed{\ddot{\theta} - b \dot{\phi}^2 \sin \theta \cos \theta + \frac{g}{b} \sin \theta = 0} \][/tex]
and
[tex]\[ \boxed{\ddot{\phi} \sin^2 \theta + 2 \dot{\phi} \sin \theta \cos \theta \dot{\theta} = 0} \][/tex]
[tex]\[ L = \frac{1}{2} m b^2 \dot{\theta}^2 + \frac{1}{2} m b^2 \dot{\phi}^2 \sin^2 \theta + m g b \cos \theta \][/tex]
where [tex]\( m \)[/tex] is the mass, [tex]\( b \)[/tex] is a constant length, [tex]\( g \)[/tex] is the gravitational acceleration, [tex]\( \theta \)[/tex] is the polar angle, and [tex]\( \phi \)[/tex] is the azimuthal angle.
The Euler-Lagrange equation is given by:
[tex]\[ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}_i} \right) - \frac{\partial L}{\partial q_i} = 0 \][/tex]
where [tex]\( q_i \)[/tex] represents the generalized coordinates which are [tex]\( \theta \)[/tex] and [tex]\( \phi \)[/tex] in our case.
### For [tex]\( \theta \)[/tex]:
1. Compute [tex]\(\frac{\partial L}{\partial \dot{\theta}}\)[/tex]:
[tex]\[ \frac{\partial L}{\partial \dot{\theta}} = \frac{\partial}{\partial \dot{\theta}} \left( \frac{1}{2} m b^2 \dot{\theta}^2 \right) = m b^2 \dot{\theta} \][/tex]
2. Compute [tex]\(\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{\theta}} \right)\)[/tex]:
[tex]\[ \frac{d}{dt} \left( m b^2 \dot{\theta} \right) = m b^2 \ddot{\theta} \][/tex]
3. Compute [tex]\(\frac{\partial L}{\partial \theta}\)[/tex]:
[tex]\[ \frac{\partial L}{\partial \theta} = \frac{\partial}{\partial \theta} \left( \frac{1}{2} m b^2 \dot{\phi}^2 \sin^2 \theta + m g b \cos \theta \right) = \frac{1}{2} m b^2 \dot{\phi}^2 \cdot 2 \sin \theta \cos \theta - m g b \sin \theta = m b^2 \dot{\phi}^2 \sin \theta \cos \theta - m g b \sin \theta \][/tex]
Thus, the Euler-Lagrange equation for [tex]\( \theta \)[/tex] is:
[tex]\[ m b^2 \ddot{\theta} - \left( m b^2 \dot{\phi}^2 \sin \theta \cos \theta - m g b \sin \theta \right) = 0 \][/tex]
Simplifying, we get:
[tex]\[ \boxed{\ddot{\theta} - b \dot{\phi}^2 \sin \theta \cos \theta + \frac{g}{b} \sin \theta = 0} \][/tex]
### For [tex]\( \phi \)[/tex]:
1. Compute [tex]\(\frac{\partial L}{\partial \dot{\phi}}\)[/tex]:
[tex]\[ \frac{\partial L}{\partial \dot{\phi}} = \frac{\partial}{\partial \dot{\phi}} \left( \frac{1}{2} m b^2 \dot{\phi}^2 \sin^2 \theta \right) = m b^2 \dot{\phi} \sin^2 \theta \][/tex]
2. Compute [tex]\(\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{\phi}} \right)\)[/tex]:
[tex]\[ \frac{d}{dt} \left( m b^2 \dot{\phi} \sin^2 \theta \right) = m b^2 \left( \ddot{\phi} \sin^2 \theta + 2 \dot{\phi} \sin \theta \cos \theta \dot{\theta} \right) \][/tex]
3. Compute [tex]\(\frac{\partial L}{\partial \phi}\)[/tex]:
[tex]\[ \frac{\partial L}{\partial \phi} = 0 \quad (\text{since } L \text{ does not depend on } \phi) \][/tex]
Thus, the Euler-Lagrange equation for [tex]\( \phi \)[/tex] is:
[tex]\[ m b^2 \left( \ddot{\phi} \sin^2 \theta + 2 \dot{\phi} \sin \theta \cos \theta \dot{\theta} \right) = 0 \][/tex]
Simplifying, we get:
[tex]\[ \boxed{\ddot{\phi} \sin^2 \theta + 2 \dot{\phi} \sin \theta \cos \theta \dot{\theta} = 0} \][/tex]
In conclusion, we have derived the equations of motion for the given Lagrangian:
[tex]\[ \boxed{\ddot{\theta} - b \dot{\phi}^2 \sin \theta \cos \theta + \frac{g}{b} \sin \theta = 0} \][/tex]
and
[tex]\[ \boxed{\ddot{\phi} \sin^2 \theta + 2 \dot{\phi} \sin \theta \cos \theta \dot{\theta} = 0} \][/tex]
