Deriving Cross Vectors: Understanding d/dt[a * (v x r)] = a (v x r)

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In summary, the person is trying to prove that d/dt[a * (v x r)] = a (v x r) if r,v and a denote the position, velocity and the acceleration of a particle, but they are not sure how to start. They found that da/dt * (v x r) + a (dv/dt x r + v x dr/dt) = da/dt * (v x r) + a (d/dt[(v x r)]) which means a (dv/dt x r + v x dr/dt) should be = 0, but it is not. They are wondering if they are using '*' incorrectly for the scalar
  • #1
happyparticle
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Homework Statement
derivate a cross vectors product
Relevant Equations
d/dt[a * (v x r)] = a (v x r)
Hi,
I need to prove that d/dt[a * (v x r)] = a (v x r) if r,v and a denote the position, velocity and the acceleration of a particle.

I see someone else posted the same question, but I didn't understand the answer.

Actually, I don't know how to derivate a cross vectors product. I'm not even sure where to begin.

This is what I did.
d/dt[a * (v x r)] = da/dt * (v x r) + a (d/dt[(v x r)])

I'm not sure at all about what I did, but anyway I'm stuck here. I'm wondering if you guys can tell me if I'm wrong and some tips about d/dt[(v x r)].
 
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  • #2
$$\frac{d}{dt}(\mathbf u\times\mathbf v)=\frac{d\mathbf u}{dt}\times\mathbf v+\mathbf u\times\frac{d\mathbf v}{dt}$$
 
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  • #3
d/dt[a * (v x r)] = da/dt * (v x r) + a (d/dt[(v x r)])
= da/dt * (v x r) + a (dv/dt x r + v x dr/dt)

If the final answer is da/dt * (v x r) then a (dv/dt x r + v x dr/dt) should be = 0, right?
 
  • #4
EpselonZero said:
d/dt[a * (v x r)] = da/dt * (v x r) + a (d/dt[(v x r)])
= da/dt * (v x r) + a (dv/dt x r + v x dr/dt)

If the final answer is da/dt * (v x r) then a (dv/dt x r + v x dr/dt) should be = 0, right?
Are you using ##*## for the scalar product? If so, it shouldn't be disappearing from your expressions.
 
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  • #5
Along with what PeroK has said, bear in mind that ##\mathbf u\times\mathbf u=\mathbf 0##, and that ##\frac{d\mathbf r}{dt}=\mathbf v##.
 
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  • #6
EpselonZero said:
I need to prove that d/dt[a * (v x r)] = a (v x r) if r,v and a denote the position, velocity and the acceleration of a particle.

I second PeroK's question about whether '##*##' denotes the scalar product, however I thought I would also point out that in any case, that result is dimensionally inconsistent, so it can't be right.

Are you sure the result isn't ##\dot{\vec{a}} \cdot (\vec{v} \times \vec{r})##, or something?
 
  • #7
Yes, the result should be a ⋅ (v x r), sorry.

I keep working on it. Thanks for you help.
 
  • #8
EpselonZero said:
Yes, the result should be a ⋅ (v x r), sorry.

Surely the question cannot be to prove that ##\frac{d}{dt} \left( \vec{a} \cdot (\vec{v} \times \vec{r}) \right) = \vec{a} \cdot (\vec{v} \times \vec{r})##... because that is not true! Does the ##\vec{a}## have a dot on it, ##\dot{\vec{a}}##?
 
  • #9
I made few mistakes by typing my question. a (point above) like da/dx ⋅ ( a x r).
Is that make more sense?
 
  • #10
EpselonZero said:
I made few mistakes by typing my question. a (point above) like da/dx ⋅ ( a x r).
Is that make more sense?

Hang on, what's ##x## got to do with it now? And why has the ##\vec{v}## in the cross product changed to ##\vec{r}##? I'm inclined to think the question actually asks you to show$$\frac{d}{dt} (\vec{a} \cdot (\vec{v} \times \vec{r})) = \frac{d\vec{a}}{dt} \cdot (\vec{v} \times \vec{r})$$Is that what's written?
 
  • #11
With all the examples and try and error I switched the vectors, but yes it is what you are written. Sorry for the mistakes.
 
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  • #12
Alright, I did few more steps.
da/dt ⋅ (v x r) + a(dv/dt x r + v x dr/dt) = da/dt ⋅ (v x r) + a(dv/dt x r + v x v) = da/dt ⋅ (v x r) + a(dv/dt x r) = da/dt ⋅ (v x r) + a(a x r).Am l on the right way?
 
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  • #13
Edit: Posted in response to #11; response to #12 is below.

OK, good. It's always a good idea to figure out what we're meant to prove, before we start to prove it :wink:. You want to evaluate $$u = \frac{d}{dt} (\vec{a} \cdot (\vec{v} \times \vec{r}))$$Firstly, notice that the derivative of a dot product is$$\frac{d}{dt} (\vec{\alpha} \cdot \vec{\beta}) = \frac{d\vec{\alpha}}{dt}\cdot \vec{\beta} + \vec{\alpha} \cdot \frac{d\vec{\beta}}{dt}$$Can you do this step for the expression in the question, e.g. using that ##\vec{\alpha} = \vec{a}## and ##\vec{\beta} = \vec{v} \times \vec{r}##?

Once you've done this, the second term will be a derivative of a cross product. You can evaluate this using the formula @archaic wrote in #2. Finally, you will need to use the fact that the vector triple product is unchanged under cyclic shifts of the arguments, and also that ##\vec{\gamma} \times \vec{\gamma} = \vec{0}## for any vector ##\vec{\gamma}##. But we can deal with that when we get there!
 
  • #14
EpselonZero said:
Alright, I did few more steps.
da/dt ⋅ (v x r) + a(dv/dt x r + v x dr/dt) = da/dt ⋅ (v x r) + a(dv/dt x r + v x v) = da/dt ⋅ (v x r) + a(dv/dt x r) = da/dt ⋅ (v x r) + a(a x r).Am l on the right way?

Yes, that looks right, except you're forgetting to write in the scalar product between ##\vec{a}## and ##(\vec{a} \times \vec{r})## in the second term. Insert the dot, and it's all good.

So now consider the term on the right. The triple product is unchanged under cyclic shifts of the operators or the arguments, so ##\vec{a} \cdot (\vec{a} \times \vec{r}) = (\vec{a} \times \vec{a}) \cdot \vec{r}##...
 
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  • #15
Alright, so basically, the last steps are da/dt ⋅ (v x r) + a ⋅ ( a x r) = da/dt ⋅ (v x r) + (a x a) ⋅ r = da/dt ⋅ (v x r) + 0 ⋅ r = da/dt ⋅ (v x r).
 
  • #16
Yeah, that looks fine :smile:. You can also notice that ##\vec{a} \times \vec{r}## is a vector perpendicular to ##\vec{a}##, so when we take the dot product with ##\vec{a}## we'll get zero.

N.B. It might be a good idea in the future to learn to typeset in latex, since that will help avoid some of the confusion that occurred earlier on!
 
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  • #17
Alright, thanks alot. Yeah, I'll learn it right now.
 

1. What does the equation "D/dt[a * (v x r)] = a (v x r)" represent?

The equation represents the derivative of the angular momentum of a rotating object with respect to time.

2. What do the variables "a", "v", and "r" stand for in the equation?

The variable "a" represents the angular acceleration of the rotating object, "v" represents the velocity of the object, and "r" represents the distance from the axis of rotation to the point of interest.

3. How is the equation derived?

The equation is derived from the general equation for angular momentum, L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity. By taking the derivative of both sides with respect to time, and using the product rule, the equation D/dt[a * (v x r)] = a (v x r) can be derived.

4. What is the significance of this equation in physics?

The equation is significant in physics because it relates the change in angular momentum to the angular acceleration, velocity, and distance of a rotating object. It is used to study the dynamics of rotating systems and is a fundamental equation in rotational motion.

5. Can this equation be applied to non-rotating systems?

No, this equation is specific to rotating systems and cannot be applied to non-rotating systems. However, a similar equation can be derived for linear motion, D/dt[m * (v x r)] = m (a x r), where m is the mass, v is the velocity, and a is the acceleration of the object.

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