Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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
show 1 more comment

1 Answer 1

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.