Question 1.16: A particle moves in a plane and polar coordinates are employ......

Q_{1} ≡ Q_{r} =F \cdot \frac{\partial r}{\partial r} = F \cdot \left(\cos \theta \widehat{x} + \sin \theta \widehat{y} \right) = F \cdot \widehat{r} = F_{r} , (1.88)

Q_{2} ≡ Q_{\theta} =F \cdot \frac{\partial r}{\partial \theta} = r F \cdot \left(-\sin \theta \widehat{x} + \cos \theta \widehat{y} \right) = r F \cdot \widehat{\theta} =r F_{\theta} , (1.89)

where \widehat{r} = \cos \theta \widehat{x} + \sin \theta \widehat{y} and \widehat{\theta} = -\sin \theta \widehat{x} + \cos \theta \widehat{y} are the radial and angular unit vectors depicted in Fig. 1.6. From x = r cos θ and y = r sin θ it follows that

\dot{x}= \dot{r} \cos \theta – r \dot{\theta} \sin \theta , \dot{y}= \dot{r} \sin \theta + r \dot{\theta} \cos \theta .(1.90)

Thus, the kinetic energy expressed in terms of the polar coordinates becomes

T= \frac{m}{2} \left(\dot{x}^{2} + \dot{y}^{2}\right)= \frac{m}{2} \left(\dot{r}^{2} + {r}^{2}\dot{\theta}^{2}\right). (1.91)

Therefore, Eqs. (1.83) take the form

\frac{d}{dt}\left(\frac{\partial T}{\partial \dot{q}_{k}}\right)-\frac{\partial T}{\partial q_{k}} = Q_{k}, k = 1, . . . , n , (1.83)

\frac{d}{dt}\left(\frac{\partial T}{\partial \dot{r}}\right)- \frac{\partial T}{\partial r} = Q_{r} ⇒ m\ddot{r}− mr \dot{\theta}^{2} = F_{r} , (1.92)

\frac{d}{dt}\left(\frac{\partial T}{\partial \dot{\theta}}\right)- \frac{\partial T}{\partial \theta} = Q_{\theta} ⇒ \frac{d}{dt} \left(mr^{2} \dot{\theta}\right)=r F_{\theta} , (1.93)

It is readily checked that rF_{\theta} is the torque with respect to the origin and mr^{2} \dot{\theta} is the corresponding angular momentum. Explicitly taking the time derivative, the above equations become

m\ddot{r} − mr \dot{\theta}^{2} = F_{r} ,     mr \ddot{\theta} + 2m\dot{r} \dot{\theta} = F_{\theta} , (1.94)

which are the polar components of Newton’s equations of motion (Symon, 1971).