2
$\begingroup$

Hey I'm doing a course in mechanics and these keep cropping up!

So for this question I'm working in 3d, and so far have

$$m \mathbf{k} \cdot (\mathbf{q} \times \ddot{\mathbf{q}} )=0$$ so I need to integrate this with respect to $t$ to get: $$ m \mathbf{k} \cdot (\mathbf{q} \times \dot{\mathbf{q}} ) =\text{a constant}$$

I know why this is constant but have no idea how you integrate what's in the brackets.

  • 1
    One dot is `\dot{q}` ($\dot{q}$). Two dots is `\ddot{q}` ($\ddot{q}$).2012-02-17
  • 0
    The answer above implies that $\frac{d}{dt} \mathbf{k}= \mathbf{0}$. Is this true? Hey! It's a reasonable question! What is $\mathbf{k}$? Because mathematically, if it isn't constant then $\frac{d}{dt}[m\mathbf{k}\cdot(\mathbf{q}\times\mathbf{\dot{q}})] = m\mathbf{\dot{k}} \cdot (\mathbf{q}\times\mathbf{\dot{q}}) + m\mathbf{k} \cdot [(\mathbf{\dot{q}} \times \mathbf{\dot{q}}) + (\mathbf{q} \times \mathbf{\ddot{q}})] = m\mathbf{\dot{k}} \cdot (\mathbf{q}\times\mathbf{\dot{q}}) + m\mathbf{k} \cdot [\mathbf{q} \times \mathbf{\ddot{q}}] $2013-08-07
  • 0
    Presumably $\bf k$ is a constant. I'm no physicist so I can't tell what the original equation "means" physically, though.2013-08-07
  • 0
    Yes, well, that would be the implication to which I am referring–at least with respect to $t$.2013-08-07
  • 0
    A vector cross product itself is zero.2013-08-07
  • 0
    I know. It is zero after the second equal sign.2013-08-07

1 Answers 1

3

just notice that $$\frac{d}{dt}(\mathbf{q}\times\frac{d}{dt}\mathbf{q})=(\frac{d}{dt}\mathbf{q})\times(\frac{d}{dt}\mathbf{q})+ \mathbf{q}\times\frac{d^2}{dt^2}\mathbf{q}$$ and that the first term is zero, as $\mathbf{u}\times\mathbf{u}=0$.