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Rigid body kinetic energy extra term

0

$begingroup$

I am looking at the derivation of the kinetic energy of a rigid body, with an initial velocity:

$v = v_com + w times r$

I start with:
$T = int frac12 v cdot v dm=int frac12(v_com + w times r) cdot (v_com + w times r) dm$

and after expanding all terms and substituting known variables I end up with

$T = frac12M(v_com cdot v_com)+frac12 (w cdot Iw)+Mv_com cdot w times r $

I was wondering why this last term ends up disappearing since the final solution listed in my textbook is $T = frac12M(v_com cdot v_com)+frac12 (w cdot Iw) $

dynamical-systems cross-product

share|cite|improve this question

asked Mar 24 at 6:13

sladerslader

1

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  isn't $v_com$ perpendicular to $wtimes r$? so dot product is zero.
  $endgroup$
  – Nasser
  Mar 24 at 6:58
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Nasser - no, the direction in which an object is headed has no a priori relationship to how the object is spinning.
  $endgroup$
  – Paul Sinclair
  Mar 24 at 15:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @slader - how can you have $r$ in your final expression for $T$? $r$ is the location of the differential element $dm$ being integrated. It is only defined insdie the integration. Once the integration is finished, there is no longer such a thing as $r$.
  $endgroup$
  – Paul Sinclair
  Mar 24 at 15:20
  
  

  
  
  
  

add a comment | 

0

$begingroup$

I am looking at the derivation of the kinetic energy of a rigid body, with an initial velocity:

$v = v_com + w times r$

I start with:
$T = int frac12 v cdot v dm=int frac12(v_com + w times r) cdot (v_com + w times r) dm$

and after expanding all terms and substituting known variables I end up with

$T = frac12M(v_com cdot v_com)+frac12 (w cdot Iw)+Mv_com cdot w times r $

I was wondering why this last term ends up disappearing since the final solution listed in my textbook is $T = frac12M(v_com cdot v_com)+frac12 (w cdot Iw) $

dynamical-systems cross-product

share|cite|improve this question

asked Mar 24 at 6:13

sladerslader

1

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  isn't $v_com$ perpendicular to $wtimes r$? so dot product is zero.
  $endgroup$
  – Nasser
  Mar 24 at 6:58
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Nasser - no, the direction in which an object is headed has no a priori relationship to how the object is spinning.
  $endgroup$
  – Paul Sinclair
  Mar 24 at 15:18
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @slader - how can you have $r$ in your final expression for $T$? $r$ is the location of the differential element $dm$ being integrated. It is only defined insdie the integration. Once the integration is finished, there is no longer such a thing as $r$.
  $endgroup$
  – Paul Sinclair
  Mar 24 at 15:20
  
  

  
  
  
  

add a comment | 

$begingroup$

I am looking at the derivation of the kinetic energy of a rigid body, with an initial velocity:

$v = v_com + w times r$

I start with:
$T = int frac12 v cdot v dm=int frac12(v_com + w times r) cdot (v_com + w times r) dm$

and after expanding all terms and substituting known variables I end up with

$T = frac12M(v_com cdot v_com)+frac12 (w cdot Iw)+Mv_com cdot w times r $

I was wondering why this last term ends up disappearing since the final solution listed in my textbook is $T = frac12M(v_com cdot v_com)+frac12 (w cdot Iw) $

dynamical-systems cross-product

share|cite|improve this question

asked Mar 24 at 6:13

sladerslader

1

$endgroup$

share|cite|improve this question

asked Mar 24 at 6:13

sladerslader

1

share|cite|improve this question

asked Mar 24 at 6:13

sladerslader

1

asked Mar 24 at 6:13

sladerslader

1

asked Mar 24 at 6:13

sladerslader

1

1 Answer
1

active

oldest

votes

0

$begingroup$

When you expand $T = int frac12(v_com + omega times r) cdot (v_com + omega times r) dm$, you get

$$T = frac 12 int v_com^2 ,dm + frac 12 int |omega times r|^2, dm + int v_comomega times r,dm$$

As you've noted, the first two terms simplify to $frac 12 Mv_com^2 + frac 12(omega cdot Iomega)$. But in the last term, $r$ is not constant, so it does not simplify to $Mv_comomega times r$. And as I said in the comment, $r$ is not even defined outside the integral.

Instead, since $v_com$ and $omega$ are constant, we get
$$int v_comomega times r,dm = v_comomega times int r,dm$$

So the question remains, what is $int r,dm$? This can be a little tricky to recognize. $r$ is the location of the differential element $dm$, relative to the center of mass. If we fix an arbitrary coordinate system and let $R$ be the position vector, then $r = R - R_com$, where $R_com$ is the center of mass. So
$$int r,dm = int (R - R_com),dm = int R,dm - MR_com$$

But the center of mass is defined by $$R_com := fracint R,dmM$$

So you end up with $$int r,dm = 0$$

share|cite|improve this answer

answered Mar 24 at 16:26

Paul SinclairPaul Sinclair

20.8k21543

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  perfect, thank you for the explanation
  $endgroup$
  – slader
  Mar 24 at 17:20
  

  
  
  
  

add a comment | 

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0

$begingroup$

When you expand $T = int frac12(v_com + omega times r) cdot (v_com + omega times r) dm$, you get

$$T = frac 12 int v_com^2 ,dm + frac 12 int |omega times r|^2, dm + int v_comomega times r,dm$$

As you've noted, the first two terms simplify to $frac 12 Mv_com^2 + frac 12(omega cdot Iomega)$. But in the last term, $r$ is not constant, so it does not simplify to $Mv_comomega times r$. And as I said in the comment, $r$ is not even defined outside the integral.

Instead, since $v_com$ and $omega$ are constant, we get
$$int v_comomega times r,dm = v_comomega times int r,dm$$

So the question remains, what is $int r,dm$? This can be a little tricky to recognize. $r$ is the location of the differential element $dm$, relative to the center of mass. If we fix an arbitrary coordinate system and let $R$ be the position vector, then $r = R - R_com$, where $R_com$ is the center of mass. So
$$int r,dm = int (R - R_com),dm = int R,dm - MR_com$$

But the center of mass is defined by $$R_com := fracint R,dmM$$

So you end up with $$int r,dm = 0$$

share|cite|improve this answer

answered Mar 24 at 16:26

Paul SinclairPaul Sinclair

20.8k21543

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  perfect, thank you for the explanation
  $endgroup$
  – slader
  Mar 24 at 17:20
  

  
  
  
  

add a comment | 

0

$begingroup$

When you expand $T = int frac12(v_com + omega times r) cdot (v_com + omega times r) dm$, you get

$$T = frac 12 int v_com^2 ,dm + frac 12 int |omega times r|^2, dm + int v_comomega times r,dm$$

As you've noted, the first two terms simplify to $frac 12 Mv_com^2 + frac 12(omega cdot Iomega)$. But in the last term, $r$ is not constant, so it does not simplify to $Mv_comomega times r$. And as I said in the comment, $r$ is not even defined outside the integral.

Instead, since $v_com$ and $omega$ are constant, we get
$$int v_comomega times r,dm = v_comomega times int r,dm$$

So the question remains, what is $int r,dm$? This can be a little tricky to recognize. $r$ is the location of the differential element $dm$, relative to the center of mass. If we fix an arbitrary coordinate system and let $R$ be the position vector, then $r = R - R_com$, where $R_com$ is the center of mass. So
$$int r,dm = int (R - R_com),dm = int R,dm - MR_com$$

But the center of mass is defined by $$R_com := fracint R,dmM$$

So you end up with $$int r,dm = 0$$

share|cite|improve this answer

answered Mar 24 at 16:26

Paul SinclairPaul Sinclair

20.8k21543

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  perfect, thank you for the explanation
  $endgroup$
  – slader
  Mar 24 at 17:20
  

  
  
  
  

add a comment | 

$begingroup$

When you expand $T = int frac12(v_com + omega times r) cdot (v_com + omega times r) dm$, you get

$$T = frac 12 int v_com^2 ,dm + frac 12 int |omega times r|^2, dm + int v_comomega times r,dm$$

As you've noted, the first two terms simplify to $frac 12 Mv_com^2 + frac 12(omega cdot Iomega)$. But in the last term, $r$ is not constant, so it does not simplify to $Mv_comomega times r$. And as I said in the comment, $r$ is not even defined outside the integral.

Instead, since $v_com$ and $omega$ are constant, we get
$$int v_comomega times r,dm = v_comomega times int r,dm$$

So the question remains, what is $int r,dm$? This can be a little tricky to recognize. $r$ is the location of the differential element $dm$, relative to the center of mass. If we fix an arbitrary coordinate system and let $R$ be the position vector, then $r = R - R_com$, where $R_com$ is the center of mass. So
$$int r,dm = int (R - R_com),dm = int R,dm - MR_com$$

But the center of mass is defined by $$R_com := fracint R,dmM$$

So you end up with $$int r,dm = 0$$

share|cite|improve this answer

answered Mar 24 at 16:26

Paul SinclairPaul Sinclair

20.8k21543

$endgroup$

share|cite|improve this answer

answered Mar 24 at 16:26

Paul SinclairPaul Sinclair

20.8k21543

answered Mar 24 at 16:26

Paul SinclairPaul Sinclair

20.8k21543

answered Mar 24 at 16:26

Paul SinclairPaul Sinclair

20.8k21543