Question 11.8: Two particle s of equal mass m are attached to the ends of a...

Following the impulse, the center of mass of the system is at (x *, y* ) in  \psi  and the rod makes an angle  \theta  with the y-axis. The three  generalized coordinates, therefore , are  \left(q_1, q_2, q_3\right)=\left(x^*, y^*, \theta\right).  The total kinetic energy of the system in its subsequent general motion is given by (8.53); we find

K(\beta, t)=K^*(\beta, t)  +  K_{r C}(\beta, t)                  (8.53)

T=\frac{1}{2}(2 m)\left(\dot{x}^{* 2}  +  \dot{y}^{* 2}\right)  +  \frac{1}{2}\left(\frac{m \ell^2}{2}\right) \dot{\theta}^2 .                        (11.59a)

All of the generalized variables are ignorable, and hence   \partial T / \partial q_r \equiv 0.  Since the system is at rest initially, the instantaneous change s in the generalized momenta (11.56) are given by

\left.\frac{\partial T}{\partial \dot{q}_k}\right|_{t_0} ^t=\Delta p_k                  (11.56)

\Delta p_1^*=2 m \dot{x}^*, \quad \Delta p_2^*=2 m \dot{y}^*, \quad \Delta p_3^*=\frac{m \ell^2}{2} \dot{\theta}                     (11.59b)

Notice that  \Delta p_1^*  and   \Delta p_2^*  are instantaneous change s in linear momenta due to an impulsive force, and   \Delta p_3^*  is the instantaneous change in the moment of momentum due to a torque-impulse.

Let  \left(\mathscr{Q}_1^*, \mathscr{Q}_2^*, \mathscr{Q}_3^*\right)  denote the corresponding generalized impulsive forces. The position vector  x_p  of the point of application of the impulsive force P = Pi is given by   \mathbf{x}_P=\left(x^*  +  \frac{1}{2} \ell \sin \theta\right)  \mathbf{i}  +  \left(y^*-\frac{1}{2} \ell \cos \theta\right) \mathbf{j}  so its virtual displacement is   \delta \mathbf{x}_P=\left(\delta x^*  +  \frac{1}{2} \ell \delta \theta \cos \theta\right) \mathbf{i}  +  \left(\delta y^*  +  \frac{1}{2} \ell \delta \theta \sin \theta\right) \mathbf{j}.  Hence, at the impulsive instant at which  \theta=0,  the  virtual work of the applied forces and the generalized impulsive  forces is given by

\delta \mathscr{W}^*=\mathscr{Q}_1^* \delta x^*  +  \mathscr{Q}_2^* \delta y^*  +  \mathscr{Q}_3^* \delta \theta=P\left(\delta x^*  +  \frac{1}{2} \ell \delta \theta\right),                  (11.59c)

for all arbitrary virtual displacements. Therefore,

\mathscr{Q}_1^*=P, \quad \mathscr{Q}_2^*=0, \quad \mathscr{Q}_3^*=\frac{P \ell}{2} .                       (11.59d)

We thus see that   \mathscr{Q}_1^*  is the instantaneous impulsive force while  \mathscr{Q}_3^*  is its instantaneous moment about the center of mass. Use of (11. 59b) and (11.59d) in (11.58) yields the instantaneous values of the generalized velocities  \left(\dot{x}_i^*, \dot{y}_i^*, \dot{\theta}_i\right):

\mathscr{D}_k^*=\Delta p_k^* ;                (11.58)

\dot{x}_i^*=\frac{P}{2 m}, \quad \dot{y}_i^*=0, \quad \dot{\theta}_i=\frac{P}{m \ell} .                          (11.59e)

The instantaneous increase in the total energy due to the impulse on the system , initially at rest, follows by substitution of (11.59e) into (11.59a):  T_i=P^2 / 2 m.

The values (11.59e) of the instantaneous translational velocity    \dot{\mathbf{x}}^*=\dot{x}_i^* \mathbf{1}  of the center of mass and the instantaneous angular velocity  \boldsymbol{\omega}=\dot{\theta}_i \mathbf{k}  of the system about the center of mass are the initial conditions for the subsequent motion of the system under no forces . It follows that the motion of the center of mass is uniform with velocity   \mathbf{v}^*=P / 2 m \mathbf{i},  its initial value. Moreover, there are no applied torques, so the moment of momentum for the system about the center of mass is constant:   h_C=m \ell^2 \dot{\theta} / 2 \mathbf{k},  and hence the angular velocity is constant,  \omega=P / m \ell \mathbf{k}.  Alternatively, this being a  conservative system with total kinetic energy (11.59a), and all of whose generalized coordinates are ignorable and whose generalized  forces are zero, (11.47) yields the principles of conservation of generalized momenta:

p_e\left(\dot{q}_r, q_r, t\right)=\frac{\partial L\left(\dot{q}_r, q_r, t\right)}{\partial \dot{q}_e}=\gamma_e, \text { a constant. }                        (11.47)

p_1=2 m \dot{x}^*=\gamma_1, \quad p_2=2 m \dot{y}^*=\gamma_2, \quad p_3=\frac{m \ell^2}{2} \dot{\theta}=\gamma_3,                             (11.59f)

where the constants \gamma_k are determined by the initial values (11.59e). We thus find   \gamma_1=P, \gamma_2=0, \gamma_3=P \ell / 2,   and hence,  \mathbf{v}^*=P / 2 m \mathbf{i}  and   \boldsymbol{\omega}=P / m \ell \mathbf{k}  for all t.