A system of rigid bodies in its static equilibrium position is shown in Fig. 11.3. The homogeneous wheel B of radius a is free to rotate in a smooth bearing about Q, and the block of mass m is supported by a linear spring of stiffness k2 attached to an inextensible cable that wraps around the

Question

A system of rigid bodies in its static equilibrium position is shown in Fig. 11.3. The homogeneous wheel [latex]\mathscr{B}[/latex] of radius a is free to rotate in a smooth bearing about Q, and the block of mass m is supported by a linear spring of stiffness [latex]k_2[/latex] attached to an inextensible cable that wraps around the wheel. The other end of the cable is fastened to a linear spring of modulus [latex]k_1[/latex] fixed to the machine foundation . There is no slip between the cable and the wheel when the block is displaced vertically and released . Derive the equations of motion for the system.

Accepted Answer

Because the elastic spring response is linear, we may consider the motion about the prestretched static equilibrium position of the system . In consequence, gravity has no further influence on the motion. This system has two degrees of freedom characterized by the independent generalized coordinates [latex]\left(q_1, q_2 ight)=(x, \phi)[/latex] measured from the equilibrium state shown in Fig. 11.3. Due to the no slip and inextensibility constraints, the extension [latex] ilde{x}[/latex] the foundation spring is [latex] ilde{x}=a \phi[/latex]. The smooth bearing reaction force is workless, and the remaining forces that act on the system are conservative . Therefore, relative to the static equilibrium state, the Lagrangian function for the system is given by

[latex]L=T-V=\frac{1}{2} m \dot{x}^2 + \frac{1}{2} I \dot{\phi}^2 - \left[\frac{1}{2} k_1 a^2 \phi^2 + \frac{1}{2} k_2(x - a \phi)^2 ight][/latex]. (11.67a)

where I is the moment of inertia of the wheel about its principal axis at Q. Notice in passing that the total kinetic energy in (11.67a) has the form (11.24) in which [latex]\left[M_{j k} ight]=\operatorname{diag}[m, I][/latex]. With (11.66) in mind, we first determine

[latex]T=\frac{1}{2} M_{j k}\left(q_r ight) \dot{q}_j \dot{q}_k[/latex] (11.24)

[latex]\frac{d}{d t}\left(\frac{\partial L}{\partial \dot{q}_r} ight) - \frac{\partial L}{\partial q_r}=0, \quad r=1,2, \ldots, n[/latex] (11.66)

[latex]\frac{\partial L}{\partial \dot{x}}=m \dot{x}, \quad \frac{\partial L}{\partial x}=-k_2(x - a \phi)[/latex], (11.67b)

[latex]\frac{\partial L}{\partial \dot{\phi}}=I \dot{\phi}, \quad \frac{\partial L}{\partial \phi}=-k_1 a^2 \phi + a k_2(x - a \phi),[/latex] (11.67c)

and thereby obtain the following coupled pair of linear differential equations of motion for the system:

[latex]m \ddot{x} + k_2 x - k_2 a \phi=0, \quad I \ddot{\phi} + \left(k_1 + k_2 ight) a^2 \phi - k_2 a x=0 .[/latex] (11.67d)

The structure of these equations is similar to (11.40f) studied earlier.

[latex]m_1 \ddot{x}_1 + \left(k_1 + k_2 ight) x_1 - k_2 x_2=0, \quad m_2 \ddot{x}_2 + \left(k_1 + k_2 ight) x_2 - k_2 x_1=0 .[/latex] (11.40f)

Question 11.9: A system of rigid bodies in its static equilibrium position ...

Related Answered Questions

(i) Derive the equations for the uniaxial motion of the spring-mass system shown in Fig. 11.1. The supporting surface is smooth and all springs are linearly elastic and unstretched initially. (ii) Determine the motion of the system for the special symmetric case when m1 = m2 = m and k1 = k2 = k. ...

Verified Answer:

Verified Answer:

Verified Answer:

Verified Answer:

Verified Answer:

Derive the equation of motion for a particle P that falls from rest in a Stokes medium . ...

Verified Answer:

Apply Lagrange’s equations (11.15) to derive the equations of unconstrained motion of a particle in cylindrical coordinates. ...

Verified Answer:

The damped free vibration of a two degree of freedom system moving in the xy-plane has total kinetic energy T = 1/2m1x2 + 1/2m2y2, total elastic potential energy V = 1/2k1x2 + 1/2k2y2, and a total Stokes type dissipation described by the Rayleigh function D = 1/2(C1x2 + 2c12xy + c2y2). Derive ...

Verified Answer:

Verified Answer:

Derive the equations of motion for a particle of mass m moving in a plane under a central force F = -(μm/r²)er , where μ is a constant and cylindrical coordinates are used. ...

Verified Answer: