The position vector r of a particle relative to an origin satisfies the vector differential equation r = ω × r , where ω is a constant vector. i) By taking the scalar product of this equation with r, show that the path of the particle lies on a sphere. ii) By taking the scalar product of the

Question

The position vector r of a particle relative to an origin satisfies the vector differential equation [latex]\dot{r} = ω × r [/latex], where ω is a constant vector.
i) By taking the scalar product of this equation with r, show that the path of the particle lies on a sphere.
ii) By taking the scalar product of the equation with the unit vector [latex]\hat{ω}[/latex] show that the path must also lie in a plane.
iii) Hence deduce that the path is a circle and show that the particle moves with constant speed.

Accepted Answer

i) [latex]\dot{r} = ω × r [/latex]

⇒ [latex]r \cdot \dot{r} = r \cdot (ω × r ) = 0[/latex] (two equal vectors in a triple scalar product)

The left-hand side is [latex]\frac{1}{2} \frac{d}{dt} (r \cdot r)[/latex] by the product rule for differentiating dot products.

So [latex]r \cdot r = ext{constant} [/latex] i.e. |r| is constant, so the particle lies on a sphere.

ii) [latex]\hat{ω} \cdot \dot{r} = \hat{ω} \cdot (ω × r) = 0[/latex]

[latex]\hat{ω}[/latex] and ω are parallel, so the triple scalar product is zero. The left-hand side is [latex]\frac{d}{dt} (\hat{ω} \cdot r),[/latex] again by the product rule ([latex]\hat{ω}[/latex] is constant). So [latex]\hat{ω} \cdot r[/latex] = constant.

This is the standard vector equation of a plane which is normal to [latex]\hat{ω}[/latex].

iii) The particle lies on both a sphere and a plane, and must therefore lie on the intersection of these, i.e. a circle.
Differentiating [latex]\dot{r} = ω × r [/latex] gives

[latex]\ddot{r} = ω × \dot{r}[/latex] (by the product rule for vector products: ω is constant)

⇒ [latex]\dot{r} \cdot \ddot{r} = \dot{r} \cdot ( ω × \dot{r}) = 0[/latex].

The left-hand side is [latex]\frac{1}{2} \frac{d}{dt} (\dot{r} \cdot \dot{r})[/latex]. Hence [latex]\dot{r} \cdot \dot{r} = ext{constant} = |\dot{r}|^{2} ext{so the speed } |\dot{r}|[/latex] is constant.

Question 17.10: The position vector r of a particle relative to an origin sa...

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