Question

2. A particle starts on the x axis at (b. 0) at time 0, and moves at constant speed voin a straight line in the direction. Derive explicit expressions (in terms of b. 10, and t) for the polar coordinates r and ф as functions of time, and for r. r, and ф . Finally, calculate the radial and azimuthal acceleration components at and at, as functions of time.

Answer 1

The particle is initially at (b,0) at time t=0, and moves at constant speed $\dpi{100} v_0$ in a straight line in the +y direction.

Therefore the equation of motion for the particle is given by,

$\dpi{100} \frac{dx}{dt}=0\\ \frac{dy}{dt}=v_0$

Integrating both the equation we have,

$\dpi{100} \int_{b}^{x}dx=0 \Rightarrow x=b\\ \int_{0}^{y}dy = v_0 \int_{0}^{t}dt \Rightarrow y = v_0t$

Now in polar coordinates,

$\dpi{100} r=\sqrt{x^2+y^2}=\sqrt{b^2+(v_0t)^2}\\ \phi=tan^{-1}(\frac{y}{x}) = tan^{-1}(\frac{v_0t}{b})$

Thereofore,

$\dpi{100} \dot r = \frac{d}{dt} (\sqrt{b^2+(v_0t)^2}) = \frac{v_0^2 t}{\sqrt{b^2+(v_0t)^2}}\\ \ddot r = \frac{d \dot r}{dt}= \frac{v_0^2}{\sqrt{b^2+(v_0t)^2}} - \frac{v_0^4t^2}{(b^2+v_0^2t^2)^{\frac{3}{2}}}=\frac{b^2v_0^2}{(b^2+v_0^2t^2)^{\frac{3}{2}}}\\ \dot \phi = \frac{d\phi}{dt} = \frac{1}{1+(\frac{v_0t}{b})^2}= \frac{b^2}{b^2+v_0^2t^2}\\ \ddot \phi = \frac{d\dot \phi}{dt}=-\frac{2b^2v_0^2t}{(b^2+v_0^2t^2)^2}$

Now radial acceleration in polar coordinate is given by,

$\dpi{100} a_r = \ddot r - r \dot \phi ^2 = \frac{b^2(v_0^2-1)}{(b^2+v_0^2t^2)^{\frac{3}{2}}}$

and azimuthal acceleration is given by,

$\dpi{100} a_{\phi} = r\ddot \phi +2 \dot r \dot \phi =0$