- #1
Bling Fizikst
- 69
- 7
- Homework Statement
- refer to image
- Relevant Equations
- refer to image
Source : JEE Advanced , Physics Sir JEE YT
I tried to attempt it using Lagrangian , so according to the coordinate axes given in the diagram , the position of the particle is let's say ##(0,d,-z)##
Let ##r## be the distance between the particle and the axis of rotation such that it subtends an angle of ##\theta## from the y axis .
So , ##-z=d\tan\theta\implies -\dot{z}=d\sec^2\theta \dot{\theta}##
$$\mathcal{L}=\frac{1}{2}m\dot{z}^2=\frac{1}{2}md^2\sec^4\theta \dot{\theta}^2$$
Now , writing the euler-lagrange equation and simplifying gives : $$\ddot{\theta}=-2\tan\theta \dot{\theta}^2$$
I am not sure how to deal with this .
I tried to attempt it using Lagrangian , so according to the coordinate axes given in the diagram , the position of the particle is let's say ##(0,d,-z)##
Let ##r## be the distance between the particle and the axis of rotation such that it subtends an angle of ##\theta## from the y axis .
So , ##-z=d\tan\theta\implies -\dot{z}=d\sec^2\theta \dot{\theta}##
$$\mathcal{L}=\frac{1}{2}m\dot{z}^2=\frac{1}{2}md^2\sec^4\theta \dot{\theta}^2$$
Now , writing the euler-lagrange equation and simplifying gives : $$\ddot{\theta}=-2\tan\theta \dot{\theta}^2$$
I am not sure how to deal with this .
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