Does kinetic energy depend on position as well as velocity?

[kq6up]

Just read in Taylor that in general the kinetic energy can depend on position as well as velocity. I believe him, but how is that true? I thought it only depended on velocity.

Chris

 

[berkeman]

Just read in Taylor that in general the kinetic energy can depend on position as well as velocity. I believe him, but how is that true? I thought it only depended on velocity.

Chris

Can you type in the full quote?

 

[kq6up]

I got it now. I see it with the general bead on the wire w/ the Lagrange mechanics.

 

[vanhees71]

Sure, if you have a system described by generalized coordinates,
\vec{x}_j=\vec{x}_j(q_1,\ldots,q_f),
the kinetic energy is
T=\sum_j \frac{m_j}{2} \dot{\vec{x}_j}^2=\sum_j \frac{m_j}{2} \left (\sum_k \frac{\partial \vec{x}_j}{\partial q_k} \dot{q}_k \right)^2,
which shows that the Lagrangian still is a quadratic form in \dot{q}_k but with coefficients that in general depend on the q_k,
i.e.,
T=\frac{1}{2} \sum_{ij} g_{ij}(k) \dot{q}_i \dot{q}_j
with
g_{ij}(q)=\sum_k \frac{m_k}{2} \frac{\partial \vec{x}_k}{\partial q_i} \cdot \frac{\partial \vec{x}_k}{\partial q_j}.

 

[AlephZero]

Or to give a concrete example, for uniform circular motion of a point mass, KE = $\frac 1 2 mr^2\omega^2$.

 

[RazorSphinx]

This is confusing. Are we saying that for the same velocity, just varying position can result in different KE?

I can see that KE can depend on position - but only because velocity depends on position - see AlephZero's example in post #5

I suppose it boils down to what we mean by velocity - if you think of it as time derivative of your coordinates, sure KE can depend on coordinates in addition to the rate of change of coordinate. But if velocity is is the rate of change of position (distance and direction from some origin), I don't see how KE can depend on position and its rate of change.

I second berkeman's request for a full quote (sorry I don't have a copy of the book).

 

[agenthulk]

I like to look at it in basic form first V=d/t (velocity) and KE=1/2mv squared ( kinetic energy) now look at the meaning of both velocity is speed and direction and KE is energy an object has by virtue of motion

The amount of KE in an object has depends upon the objects mass as well as it's (speed) v=d/t

Equations of speed and velocity may look the same but they are not the d in speed is distance from the objects starting position to its position after time has elapsed

The d in velocity is displacement defined as the straight-line distance from the objects starting point to its final position

Now look at mass and the meaning not weight which gravity is the factor.

 

[kq6up]

Edit\

The original quote from page 524:

"Recall that, in general, the kinetic energy can depend on $q$ as well as $\dot{q}$, whereas, for conservative systems, the potential energy depends only on $q$. For example, for the simple pendulum (mass m and length L)""

\Edit

Quoting Taylor page 438 (the part that fixed it for me):

"A bead of mass m is threaded on a frictionless wire that lies in the xy plane (y vertically up), bent in the shape y = f (x) with a minimum at the origin, as shown in Figure 11.12. [just a general function in the XY that has some dip at the origin where the bead can oscillate back and forth] Write down the potential and kinetic energies and their simplified forms appropriate for small oscillations about 0.

This system has just one degree of freedom, and the natural choice of generalized coordinate is just x. With this choice, the potential energy is simply $U=mgy=mgf(x)$. When we confine ourselves to small oscillations, we can Taylor expand $f(x)$. Since $f(0)=f^{ \prime }(0)=0$, this gives $U(x)=mgf(x)\approx \frac { 1 }{ 2 } mg{ f(x) }^{ \prime \prime }x^{ 2 }$

The kinetic energy is $T=\frac { 1 }{ 2 } m(\dot { x } ^{ 2 }+\dot { y } ^{ 2 })$, where, by the chain rule, $\dot { y } =f^{ \prime }(x)\dot { x }$. Therefore, $T=\frac { 1 }{ 2 } [1+{ f }^{ \prime }(x)^{ 2 }]\dot { x } ^{ 2 }$. Notice that the exact expression for T depends on $\dot{x}$ as well as $x$. However, since T already contains the factor $\dot{x}^2$, when we make the small-oscillation approximation we can simply replace the term $f^{\prime}(x)$ by its value at x=0 (namely zero) and we get $T=\frac { 1 }{ 2 } m(\dot { x } ^{ 2 }+\dot { y } ^{ 2 })\approx\frac { 1 }{ 2 } m\dot { x } ^{ 2 }$ As expected, the small-oscillation approximation has reduced both U and T to homogeneous quadratic functions of $x$ (for U) or $\dot{x}$ (for T)."

We can see that the approximation of T for small oscillations does not depend on x, but the exact solution does.

QED.

Chris

 

[RazorSphinx]

Chris,

Thanks for the quotes.

Makes sense, since they are talking about coordinates and their rate of change. In that case, KE can certainly depend on ${q}$ and $\dot{q}$. However, the way I see it, this only happens because the velocity itself depends on $\dot{q}$ and ${q}$, either because of the choice of coordinate system or through some other constraint such as in the bead on the wire case.

 

[USeptim]

Quoting Taylor page 438 (the part that fixed it for me):

"A bead of mass m is threaded on a frictionless wire that lies in the xy plane (y vertically up), bent in the shape y = f (x) with a minimum at the origin, as shown in Figure 11.12. [just a general function in the XY that has some dip at the origin where the bead can oscillate back and forth] Write down the potential and kinetic energies and their simplified forms appropriate for small oscillations about 0.

This system has just one degree of freedom, and the natural choice of generalized coordinate is just x. With this choice, the potential energy is simply $U=mgy=mgf(x)$. When we confine ourselves to small oscillations, we can Taylor expand $f(x)$. Since $f(0)=f^{ \prime }(0)=0$, this gives $U(x)=mgf(x)\approx \frac { 1 }{ 2 } mg{ f(x) }^{ \prime \prime }x^{ 2 }$

The kinetic energy is $T=\frac { 1 }{ 2 } m(\dot { x } ^{ 2 }+\dot { y } ^{ 2 })$, where, by the chain rule, $\dot { y } =f^{ \prime }(x)\dot { x }$. Therefore, $T=\frac { 1 }{ 2 } [1+{ f }^{ \prime }(x)^{ 2 }]\dot { x } ^{ 2 }$. Notice that the exact expression for T depends on $\dot{x}$ as well as $x$. However, since T already contains the factor $\dot{x}^2$, when we make the small-oscillation approximation we can simply replace the term $f^{\prime}(x)$ by its value at x=0 (namely zero) and we get $T=\frac { 1 }{ 2 } m(\dot { x } ^{ 2 }+\dot { y } ^{ 2 })\approx\frac { 1 }{ 2 } m\dot { x } ^{ 2 }$ As expected, the small-oscillation approximation has reduced both U and T to homogeneous quadratic functions of $x$ (for U) or $\dot{x}$ (for T)."

This example shows that the kinetic energy may depend on the position indirectly. Since a coordinate of velocity depends on position, the kinetic energy is expressed as a function of position, but essentially, kinetic energy for a punctual particle depends only on the inertial mass and its speed.Sergio