# Help with writing the equation of motion for a rigid body assemblage

## Homework Statement

Problem figure

The rigid body assemblage shown in the figure above, has a force P(t) acting upon it (varying with time). Pick a generalized coordinate such that a single degree of freedom equation of motion can be written.

There is a black, massless, rigid bar. Call this Bar A. It is supported on its left end by a pin joint, A.

There is a red bar, with uniform mass per unit length, mbar. Call this Bar B. It is supported on its right end by pin joint B.

There are two dampers with damping ratio c. The damping force they exert is proportional to the velocity at their ends. Therefore, Fdamper = c*v. Call the damper on Bar A, Damper A, and the damper on Bar B, Damper B.

There is a spring, with spring constant k.

One may consider the displacements the system undergoes to be small enough so that small angle approximations may be applied: sin(theta) =~ theta.

## Homework Equations

For a simple single degree of freedom problem, the equation of motion can usually be written as: m*u''(t) + c*u'(t) + k*u(t) = p(t)

Where u(t) is the displacement relative to a given point, and u'(t) is the first derivative of u(t) with respect to time (velocity), and u''(t) is the second derivative of u(t) with respect to time (acceleration).

## The Attempt at a Solution

I will use D'Alembert's principle to formulate the equation of motion. D'Alembert's principle states that a dynamic system can be analyzed as a system in equilibrium if inertial forces are also considered to be acting on the body.

Probably one of the most important things to do here is pick a good generalized coordinate. There are several possible choices:

1) the angle with the horizontal at support A - call it angle alpha
2) the angle with the horizontal at support B - call it angle beta
3) The displacement relative to some points along the assemblage, u(t)

The option I went with was the displacement u(t) relative to the point on the black bar where the force P(t) acts. Call this point C.

Consider the up direction, and the counter-clockwise direction to be positive.

Therefore, for a given u(t):

Damper A, would exert a force of FdA = -u'(t)*c

Due to effect of the lever arm, a displacement u(t) at the location of the application of the force, would create a displacement 2*u(t).

Call the point at the end of the spring that connects to Bar B, point C. The spring will exert a force downwards of Fs = (2*u(t) - uC(t))*k. Where uC(t) is the displacement at point C.

Damber B, would a exert a force of FdB = -uC'(t)*c.

For the application of D'Alembert's principle, the mass M will exert an inertial force downwards of Fi = -m*uC''(t) - m*g, at the center of mass of the bar, where m = mbar*2*(L/6) = mbar*(L/3).

Where I get stuck is considering the inertial force due to the resistance to rotation of the bar. Should I figure out what force would be exerted at point C due to the rotation of Bar B around pin B, due to displacement uC(t)? Would that be MomentC = -((Io + m(L/3)^2)*u''C(t)/(L/3) + mg*(L/6)), where Io is the moment of inertia of the bar around its center of mass, and the angular acceleration is expressed as u''C(t)/(L/3) using small angles approximation? Would dividing MomentC by the length of Bar B then give me the equivalent force acting at the end of the bar?

Probably the biggest problem then staring me in the face is how I'd figure out uC(t) in terms of u(t). I suppose if everything above is correct, I would know the force at the end of Bar B, and due to equilibrium, I know this is the force is the force through the spring. Therefore, F(at the end of B) = k(2*u(t) - uC(t))...however, the LHS (left hand side) of that equation has uC''(t) terms...so I can't simply isolate for uC(t) in order to find what it is in relationship to u(t)...

That's the mental mess I need to untangle myself from right now. It's late at night, so hopefully some sleep will help me too, but some hints in the right direction would be greatly appreciated.

Related Introductory Physics Homework Help News on Phys.org
haruspex
$k(2L/3 \alpha - L/3 \beta - d) = (P - c \dot{\alpha})/2$
$= (c \dot{\beta} +I\ddot{\beta})/2$
From the last two, you can integrate to get an equation relating $∫P, \alpha, \beta, \dot{\beta}$. Using that you can eliminate α to get an equation relating $∫P, \beta, \dot{\beta}, \ddot{\beta}$.