Why Do We Use \(\ell + r\theta\) in Polar Coordinates for Analytical Mechanics?

[TheDestroyer]

Thanks for integral, he made the potential.pdf analyse for an analytical mechanic but i still have 2 questions,

1- Why does the length up between the tengency point and top of (h) equals:

[tex]\ell + r\theta[/tex] ?

2- Why the kinetic energy here equals:

[tex]T = \frac{1}{2} m(\ell + r\theta)^2 \dot{\theta}^2[/tex] ?

I mean why we replaced R with [tex]\ell + r\theta[/tex] in the polar coordinates?

Thanks,

TheDestroyer

 

[Integral]

[itex]r \theta [/itex] is the arc length corresponding to the angle [itex]\theta[/itex] Notice that if you let [itex] \theta = 2 \pi [/itex] you get the circumference of the circle.

As you unwind the rope moving the point of tangency through an angle [itex] \theta [/itex] you add the corresponding arc length ( [itex] r \theta[/itex] to the length of rope l which is initially hanging stright down.

I need some time to look at the kinetic engery term. I concentrated on the potential energy term and have not looked into the kinetic energy. I'll get back to you, if no one else contributes.

 

[TheDestroyer]

about the [tex]r\theta[/tex] i know it's the arc length (LOL I'm a second year university physics student), but the question is why does it equal to [tex]\ell + r\theta[/tex] in the tangent,

I didn't understand integral, I'm very sorry, please explain it as a math geometric laws, And try using a simple language (I don't mean you language was complicated),

And thanks very much

 

[TheDestroyer]

Why no body is answeringgggg?

 

[Integral]

You say you are a second year university student. Did you ever learn to ride a bike? Do you remember the the first time you were given a push and told to pedal.

There is nothing left in this problem which you should not be able to figure out on your own. Please study the diagram I drew for you and think about it. YOU CAN figure it out. Get 'er done

 

[TheDestroyer]

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