Two particle s of equal mass m are attached to the ends of a massless rigid rod of length ℓ initially oriented parallel to the y-axis of a frame ψ = {O;ik} and at rest on a smooth horizontal surface. An instantaneous impulsive normal force P = Pi acts on the particle closer to O. Determine the

Question

Two particle s of equal mass m are attached to the ends of a massless rigid rod of length [latex]\ell[/latex] initially oriented parallel to the y-axis of a frame [latex]\psi=\left\{O ; \mathbf{i}_k\right\}[/latex] and at rest on a smooth horizontal surface. An instantaneous impulsive normal force P = Pi acts on the particle closer to O. Determine the subsequent instantaneous generalized velocities of the system, find the instantaneous increase of the total energy of the system due to the impulse, and describe the subsequent motion for all time.

Accepted Answer

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

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

[latex]T=\frac{1}{2}(2 m)\left(\dot{x}^{* 2} + \dot{y}^{* 2} ight) + \frac{1}{2}\left(\frac{m \ell^2}{2} ight) \dot{ heta}^2 .[/latex] (11.59a)

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

[latex]\left.\frac{\partial T}{\partial \dot{q}_k} ight|_{t_0} ^t=\Delta p_k[/latex] (11.56)

[latex]\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{ heta}[/latex] (11.59b)

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

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

[latex]\delta \mathscr{W}^*=\mathscr{Q}_1^* \delta x^* + \mathscr{Q}_2^* \delta y^* + \mathscr{Q}_3^* \delta heta=P\left(\delta x^* + \frac{1}{2} \ell \delta heta ight),[/latex] (11.59c)

for all arbitrary virtual displacements. Therefore,

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

We thus see that [latex]\mathscr{Q}_1^*[/latex] is the instantaneous impulsive force while [latex]\mathscr{Q}_3^*[/latex] 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 [latex]\left(\dot{x}_i^*, \dot{y}_i^*, \dot{ heta}_i ight)[/latex]:

[latex]\mathscr{D}_k^*=\Delta p_k^* ;[/latex] (11.58)

[latex]\dot{x}_i^*=\frac{P}{2 m}, \quad \dot{y}_i^*=0, \quad \dot{ heta}_i=\frac{P}{m \ell} .[/latex] (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): [latex]T_i=P^2 / 2 m[/latex].

The values (11.59e) of the instantaneous translational velocity [latex]\dot{\mathbf{x}}^*=\dot{x}_i^* \mathbf{1}[/latex] of the center of mass and the instantaneous angular velocity [latex]\boldsymbol{\omega}=\dot{ heta}_i \mathbf{k}[/latex] 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 [latex]\mathbf{v}^*=P / 2 m \mathbf{i}[/latex], its initial value. Moreover, there are no applied torques, so the moment of momentum for the system about the center of mass is constant: [latex]h_C=m \ell^2 \dot{ heta} / 2 \mathbf{k}[/latex], and hence the angular velocity is constant, [latex]\omega=P / m \ell \mathbf{k}[/latex]. 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:

[latex]p_e\left(\dot{q}_r, q_r, t ight)=\frac{\partial L\left(\dot{q}_r, q_r, t ight)}{\partial \dot{q}_e}=\gamma_e, ext { a constant. }[/latex] (11.47)

[latex]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{ heta}=\gamma_3,[/latex] (11.59f)

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

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

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