How to break up kinetic energy for circular motion?

[Nate Stevens]

Homework Statement

Homework Equations

L = T-V
For constant frequency tangential velocity is (radius)*(w)

The Attempt at a Solution

I need to find r(t) using the Langrangian L = T-V

I just was not sure whether I am on the right track for calculating the total kinetic energy for the above system correctly.

Have I done okay so far?

 

[haruspex]

$\dot r$ should be positive when r is increasing. You seem to be using it the other way.
In calculating "v_tan,+" you lost ω somewhere. As a result your final equation is dimensionally inconsistent.

 

[Nate Stevens]

$\dot r$ should be positive when r is increasing. You seem to be using it the other way.
In calculating "v_tan,+" you lost ω somewhere. As a result your final equation is dimensionally inconsistent.

Didn't I account for r dot's negative sign when adding "vtan, +" and "vtan, -"? (you are right though that initially I didn't attach a negative sign, which is incorrect) And good catch on the ω. I just dropped it when plugging into (vtan, +)-(vtan, -).

 

[haruspex]

Didn't I account for r dot's negative sign when adding "vtan, +" and "vtan, -"? (you are right though that initially I didn't attach a negative sign, which is incorrect) And good catch on the ω. I just dropped it when plugging into (vtan, +)-(vtan, -).

You had not defined that subscript notation. I was reading the first sentence of the second image: "the spring will pull the mass towards ..."
Also, that is not what you mean. It has nothing to do with the spring's pull. It is simply that changing r leads to a component of the particle's velocity in the tangential direction.

What I did not pick up is that you also have a sign problem with the other contributor. The way you have defined θ and ω in the diagram, $\dot\theta=-\omega$.

Bottom line, increasing r contributes a leftward motion and positive (anticlockwise) ω does likewise, so your final equation should have +, not -.

 

[Ray Vickson]

Homework Statement

View attachment 235085
View attachment 235086

Homework Equations

L = T-V
For constant frequency tangential velocity is (radius)*(w)

The Attempt at a Solution

I need to find r(t) using the Langrangian L = T-V

I just was not sure whether I am on the right track for calculating the total kinetic energy for the above system correctly.

View attachment 235091

View attachment 235092
View attachment 235094
View attachment 235095
Have I done okay so far?

I like to do these things in the most straightforward way, avoiding too much "thinking" and fancy geometry. So, I would write the position vector for the center of the long rod as $\vec{x}_c(t) = ( l \cos (\omega t), l \sin(\omega t))$ ---- assuming that the center is at $(l,0)$ when $t = 0.$ The perpendicular direction along the long rod from its center is $\vec{d} = (-\sin(\omega t), \cos(\omega t)),$ so the position vector of the mass at time $t$ is
$$\vec{r}_m (t) = \vec{x}_c(t) + r(t) \vec{d} = (l \cos(\omega t) - r(t) \sin(\omega t), \ \sin(\omega t) + r(t) \cos(\omega t)) \hspace{3ex}(1)$$
You can find the velocity $\vec{V}_m (t)$ of the mass by differentiating wrt $t$ in equation (1). Then you can expand and simplify to obtaing a pretty nice expression for $V_m^2,$. The resulting total kinetic energy is quite simple. I did not bother to try to separate the kinetic energy into radial and tangential components, because I could not see any reason for doing so.