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A particle of unit mass with position vector $\vec{r}(t)$ at time $t$ is moving in space under the actions of certain forces, where $\vec{v}$ and $\vec{a}$ are the velocity and acceleration vector.

(a) Prove that $~\vec{r} \times \vec{a} = \vec{0}~$ implies $~\vec{r} \times \vec{v} = \vec{c}$, where $\vec{c}$ is a constant vector.

Differentiating $~\vec{r} \times \vec{v} = \vec{c}~$ both sides with respect to time implies $~\vec{r} \times \vec{a} = \vec{0}~$, meaning, if we integrate $~\vec{r} \times \vec{a} = \vec{0}~$, we get $~\vec{r} \times \vec{v} = \vec{c}$.

(b) If $~\vec{r} \times \vec{v} = \vec{c}$, prove that the motion takes place in a plane. Consider both $\vec{c} = 0$ and $\vec{c} \neq 0$.

I don't know how to answer this, can anyone give me a hint?

(c) If the net force acting on the particle is always directed toward the origin, prove that the particle moves in a plane.

This follows from part (a), since $~\vec{r} \times \vec{a} = \vec{0}~$, $~\vec{a}~$ is parallel to $~\vec{r}~$ and it is directed to the origin. Also from part (b), the particle moves in a plane.

(d) Is $~\vec{r} \times \vec{v}$ necessarily constant if a particle moves in a plane?

I also don't know how to answer this.

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2 Answers 2

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Answering (b)

$$ \vec r \times \dot{\vec r} = c \Rightarrow < \vec r,\vec r \times \dot{ \vec r} > = < \vec r, \vec c > = 0 $$

so the movement pertains to the plan $< \vec r, \vec c > = 0$ or

$$ c_1 x(t)+ c_2 y(t)+ c_3 z(t) = 0 $$

Here $< \cdot,\cdot >$ represents the scalar product of two vectors.

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Answering (d)

No, because $\vec{r} \times \vec{v} = \vec{c}$ implies $\vec{r} \cdot \vec{c}=0$, so the particle is confined to a plane passing through the origin. If the particle moves in a plane not passing through the origin then $\vec{r} \times \vec{v} \neq \vec{c}$.

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