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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.

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One dot is \dot{q} ($\dot{q}$). Two dots is \ddot{q} ($\ddot{q}$). – Rahul Feb 17 '12 at 20:50
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}}] $ – user89336 Aug 7 '13 at 4:42
Presumably $\bf k$ is a constant. I'm no physicist so I can't tell what the original equation "means" physically, though. – anon Aug 7 '13 at 4:45
Yes, well, that would be the implication to which I am referring–at least with respect to $t$. – user89336 Aug 7 '13 at 4:47
A vector cross product itself is zero. – Shuhao Cao Aug 7 '13 at 5:04
up vote 3 down vote accepted

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 $\mathbf{u}\times\mathbf{u}=0$.

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