Question

A coordinate \(q_i\) is called cyclic (or ignorable) if:

• A. \(\dfrac{\partial L}{\partial q_i} = 0\)
• B. \(\dfrac{\partial L}{\partial \dot{q}_i} = 0\)
• C. \(\dfrac{d}{dt}\!\left(\dfrac{\partial L}{\partial q_i}\right) = 0\)
• D. \(\dfrac{d}{dt}\!\left(\dfrac{\partial L}{\partial \dot{q}_i}\right) = 0\)

Hint:

Cyclic means the coordinate itself does not appear in \(L\); its velocity still can. That is \(\partial L/\partial q_i = 0\).

Answer 1

The Correct Option is A

Step 1: By definition, a generalized coordinate \(q_i\) is cyclic (ignorable) when the Lagrangian does not contain that coordinate explicitly, i.e.
\[\frac{\partial L}{\partial q_i} = 0\]
Step 2: Look at the Euler-Lagrange equation for that coordinate:
\[\frac{d}{dt}\!\left(\frac{\partial L}{\partial \dot{q}_i}\right) - \frac{\partial L}{\partial q_i} = 0\]
Step 3: If \(\partial L/\partial q_i = 0\), then
\[\frac{d}{dt}\!\left(\frac{\partial L}{\partial \dot{q}_i}\right) = 0 \quad\Rightarrow\quad p_i = \frac{\partial L}{\partial \dot{q}_i} = \text{constant}\]
so the conjugate momentum is conserved. The defining condition is therefore option (A).
\[\boxed{\dfrac{\partial L}{\partial q_i} = 0}\]