[Math] Radial and transverse components of velocity and acceleration.

vector analysis

I need help with this problem:

Find (i) the velocity $\dot r(t)$, (ii) the speed $\Vert \dot r(t)\Vert$, (iii) the acceleration $\ddot r(t)$, (iv) the radial and
transverse componentes of velocity and acceleration at $t$, of a
particle whose motion in $\mathbb{R}^2$ is described by:

  1. $r(t)=(a\cos (t^2), a\sin(t^2), t\in\mathbb{R}$
  2. $r(t)=(a\cos\omega t, b\sin\omega t), t\in\mathbb{R}$, $\omega$ constant.
  3. $r(t)=(a(1+\cos t)\cos t, a(1+\cos t)\sin t), t\in\mathbb{R}.$

This is what I did:

  • $r(t)=(a\cos (t^2), a\sin(t^2), t\in\mathbb{R}$

(i) $\dot r(t)=(-a\sin(t^2)2t, a\cos(t^2)2t)$

(ii) $\Vert \dot r(t)\Vert=\sqrt{(-a\sin(t^2)2t)^2 + (a\cos(t^2)2t)^2}=\sqrt{4a^2t^2}=2at$

(iii) $\ddot r(t)=(-2a(t\cos(t^2)2t+\sin (t^2)), 2a(-t\sin(t^2)2t+\cos(t^2))=(-4at^2\cos(t^2)-2a\sin(t^2), -4at^2\sin(t^2)+2a\cos(t^2))$

  • $r(t)=(a\cos\omega t, b\sin\omega t), t\in\mathbb{R}$, $\omega$ constant.

(i) $\dot r(t)=(-a\omega \sin(\omega t), b\omega\cos(\omega t))$

(ii) $\Vert \dot r(t)\Vert=\sqrt{(-a\omega \sin(\omega t))^2 + (b\omega\cos(\omega t))^2}=\omega\sqrt{a^2\sin^2(\omega t)+b^2\cos^2(\omega t)}$

(iii) $\ddot r(t)=(-a\omega^2\cos(\omega t), -b\omega^2\sin(\omega t))$

  • $r(t)=(a(1+\cos t)\cos t, a(1+\cos t)\sin t), t\in\mathbb{R}.$

(i) $\dot r(t)=(-a\sin t-2a\sin t\cos t, a\cos t+ a\cos^2 t-a\sin^2 t)$

(ii) $\Vert \dot r(t)\Vert=\sqrt{a^2+2a^2\sin^2\cos t+2a^2\sin^2 t+\cos^2 t+a^2\cos^4 t+ a^2\sin^4 t+2a^2\cos^3 t}$

(iii) $\ddot r(t)=(-a\cos t+2a\sin^2 t-2a\cos^2 t, -a\sin t-4a\cos t\sin t)$

Am I correct? How do I calculate the radial and transverse components of the velocity and the acceleration? Please help me.

Best Answer

I did not check the math for the last case, but the first two are correct. In order to find the radial and transverse components, you must use the scalar product. Define $$\hat r(t)=\frac {r(t)}{|r(t)|}$$ Then the radial component of a vector $v$ is $$v_r=(v\cdot\hat r(t))\hat r(t)$$ If you care only about the magnitude $$|v_r|=v\cdot\hat r(t)$$ For the transverse component, we use the fact that $$v=v_r+v_t$$ Therefore $$v_t=v-(v\cdot\hat r(t))\hat r(t)$$ So take the case of velocity in the first part: $$\dot r(t)=(-2at\sin t^2,2at\cos t^2)$$ You have $$\hat r(t)=(\cos t^2,\sin t^2)$$ Then $$|\dot r_r(t)|=-2at\sin t^2\cos t^2+2at\cos t^2\sin t^2=0$$ It means that the speed is all transverse, with no radial component. This is not surprising, since the first case is movement along a circle.

Related Question