An expression about kinetic energy

[mech-eng]

Hi, all. Here is an expression about kinetic energy from analytical dynamics of Haim Baruh which confuses me.

"Consider a particle and the case which the kinetic energy is only quadratic in terms of velocity of the particle. We take the differential form of work-energy principle and remove from it all velocity and time-dependent terms(including the kinetic energy as well as all time-dependent parts of nonconservative force)"

Can you expand above explanation please. Here I don't understand why the writer says "case which the kinetic energy is only quadratic in terms of velocity ..." because kinetic energy is always in terms of square of namely quadratic of the linear or angular velocities or sum of both and what are the time dependent terms in the work-energy principle.

 

[dauto]

I'll answer the first question. Have you ever seen the relativistic formula for kinetic energy? Is it quadratic in terms of velocity?

 

[mech-eng]

I'll answer the first question. Have you ever seen the relativistic formula for kinetic energy? Is it quadratic in terms of velocity?

Is it about Einstein's general or special relativity theorem? If it is I know only E=m*C^2 here it is still
quadratic in terms terms of (light) velocity. But I don't know E is kinetic energy or something different.

 

[dauto]

That's the formula for rest energy. The kinetic energy formula is K=mc^2\left(\frac{1}{\sqrt{1-{v^2}/{c^2}}}-1\right)

 

[JonnyMaddox]

Hi, the kinetic energy T could be explicit time dependent:

When you have r_{i}=r_{i}(q_{1},q_{2},...,q_{n},t)(old coordinates in terms of new generalized coordinates), important is the explicit time dependence, which can occur for example in a system where a bead is sliding on a rotating wire.Take \dot{r_{i}}=v_{i}= \frac{dr_{i}}{dt} = \frac{\partial r_{i}}{\partial q_{j}}\dot{q_{j}}+ \frac{\partial r_{i}}{\partial t}

T= \frac{1}{2} m_{i} v^{2}_{i}= \frac{1}{2}m_{i}(\frac{\partial r_{i}}{\partial q_{j}}\dot{q_{j}}+ \frac{\partial r_{i} } {\partial t})^{2}

When expandet it is:

Eq.: 1
T=M_{0}+ M_{j}\dot{q_{j}}+\frac{1}{2}M_{jk}\dot{q_{j}}\dot{q_{k}}

Where

M_{0} = \frac{1}{2} m_{i} (\frac{\partial r_{i} }{\partial t})^{2}
M_{j} = m_{i} \frac{\partial r_{i} }{\partial t} \frac{\partial r_{i} }{\partial q_{j} }
M_{jk} = m_{i} \frac{\partial r_{i} }{\partial q_{j} } \frac{\partial r_{i} }{\partial q_{k} }

Now, when you have no explicit time dependence in the coordinates then only the last term in Eq. 1 survives and you have the usual kinetic energy which is quadratic in the generalized velocities :-)

Hope that helps !
(I dropped the summation signs out of layzieness)
Greets

 

[mech-eng]

Hi, JonnyMaddox. Your post is too advanced to understand for me. From what books did you take those examples
and what is the meaning of explicit time dependence please.

 

[mech-eng]

That's the formula for rest energy. The kinetic energy formula is K=mc^2\left(\frac{1}{\sqrt{1-{v^2}/{c^2}}}-1\right)

But here kinetic energy is still quadratic in terms of velocities as it is always form of that. I ask that why does the writer says when it is only quadratic in terms of velocities

 

[UltrafastPED]

Please provide more context ... like what is the section heading, what does he do next?

He may be leading up to the virial theorem ... just scan in the entire page and attach the image.

 

[mech-eng]

This it it.

 

[UltrafastPED]

The conditions for static equilibrium are being sought - and the process is to find the critical points of the work-energy expression by setting the derivative to zero. Since the static condition is assumed, all of the time dependent terms should vanish.

Kinetic energy is very often quadratic inthe velocity .. but not always. To obtain the simplest form the text limits you to the quadratic case.