All Questions
7 questions
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Derivative of the product of a scalar function and a vector valued function
According to Berkeley Physics Course, Volume 1 Mechanics,
The time derivative of a vector valued function can be derived from the formula:
$$
\mathbf{r}(t) = r(t)\mathbf{\hat{r}}(t)
$$
where the ...
- 3,623
1
vote
2
answers
105
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Why must a constraint force be normal?
If we impose that a particle follows a holonomic constraint, so that it always remains on a surface defined by some function $f(x_1,x_2,x_3)=0$ with $f:\mathbb{R^3}\rightarrow\mathbb{R}$, we get a ...
- 213k
2
votes
1
answer
103
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Time derivative of a "general" vector $\vec A$ in an accelerating frame: what about e.g. velocity $\vec v$?
According to Morin "Classical Mechanics" (Section 10.1, page 459), the derivative of a general vector $\vec A$ in an accelerating frame may be given as
$$\frac{d\vec A}{dt}=\frac{\delta \vec ...
- 1,763
2
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5
answers
348
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Why does $\vec{r}\cdot\dot{\vec{r}}=r\dot{r}$?
Why is $$\vec{r}\cdot\dot{\vec{r}}=r\dot {r}$$ true? Before saying anything, I have seen the proofs using spherical coordinates for $$\dot{\vec {r}}= \dot{r}\vec{u_r}+r\dot{\theta}\vec{u_\theta}+r\sin\...
- 12.9k
2
votes
1
answer
384
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Having trouble deriving the exact form of the Kinematic Transport Theorem
The Kinematic transport theorem is a very basic theorem relating time derivatives of vectors between a non rotating frame and another one that's rotating with respect to it with a uniform angular ...
- 3,358
1
vote
1
answer
113
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How to define differentiation of a time-dependent vectors with respect to a specific reference frame in a coordinate-free manner?
It is usual in classical mechanics to introduce the derivative of a time-dependent vector with respect to a reference frame. This is accomplished through the use of a basis that is fixed with respect ...
1
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3
answers
324
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Derivative with respect to vector of a function depending on vectors
I've been trying to understand this concept for hours without any success. I found similar questions on this forum (Derivative with respect to a vector is a gradient?) but I still don't understand.
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- 12.2k