Radial & Cross-Radial Acceleration: A problem

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You have (or should have) $$\underline{a}=rB^2(\mu^2-1)\hat{\underline{r}}+2B^2\mu r\hat{\underline{n}},$$ where $\hat{\underline{n}}$ is the transverse unit vector. Note the incorrect expression in your answer.

$$|\underline{a}|=rB^2\sqrt{(\mu^2-1)^2+4\mu^2}=rB^2(\mu^2+1),$$ which is proportional to $r$.

Now considering scalar product, $$\underline{a}\cdot\hat{\underline{r}}=rB^2(\mu^2-1)$$

So if $\phi$ is the angle between the acceleration vector and the radial unit vector, $$\cos \phi=\frac{rB^2(\mu^2-1)}{rB^2(\mu^2+1)}=\frac{(\mu^2-1)}{(\mu^2+1)}$$ and this is constant.

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rama_ran
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Updated on August 01, 2022

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  • rama_ran
    rama_ran over 1 year

    A particle moves along $r=Ae^{\mu\theta}$ where $\theta=Bt$, prove that its acceleration is proportional to $r$ and makes a constant angle with the radius vector.

    Approach: $\dot{\theta}=B$ then $\dot{r}=B\mu r$, $\ddot{r}=B^2\mu^2r$, $r{\dot{\theta}}^2=rB^2$

    Radial Acceleration = $A_R=\ddot{r}-r{\dot{\theta}}^2=B^2r(\mu^2-1)$ Also, Cross-Radial Acceleration = $A_C=\frac{1}{r}\frac{d}{dt}{r^2\dot{\theta}}=2B^2\mu r^2$ Resultant acceleration = $\sqrt{A_r^2+A_c^2}=B^2r\sqrt{\mu^2+(\mu^2-1)^2}$

    Therefore, Resultant acceleration is proportional to $r$.

    I am unable to get the 2nd part i.e. "Prove that its acceleration makes a constant angle with the radius vector"