Deriving equations of rotational motion for connected rods

Question

I'm working on a problem involving deriving equations of rotational motion for a system of connected rods (see the picture), and I'm a bit confused about a few things to do with rotational motion. I won't go into detail about the forces involved or their precise nature - it's more something I don't understand about rotational motion. I know how to derive local and global momentum balances by taking forces acting on each point, applying Newton's second law and also considering the motion of the centre of mass of the whole system, but I am getting incredibly confused with doing the same for the rotational motion.

If the moments of inertia about each of the first three blue dots are $I_1$, $I_2$, $I_3$, would we have the following (using conservation of angular momentum, i.e. equivalent to Newton's Second Law for rotational motion):

• $\frac{\mathrm{d}}{\mathrm{d}t}(I_1 \dot{\alpha})=$ net anticlockwise moment on rod 1 about the first point,
• $\frac{\mathrm{d}}{\mathrm{d}t}(I_2 \dot{\beta})=$ net anticlockwise moment on rod 2 about the second point,
• $\frac{\mathrm{d}}{\mathrm{d}t}(I_3 \dot{\gamma})=$ net anticlockwise moment on rod 3 about the third point?

Also, how would we write down the equation for the overall moment balance? One of my supervisors suggested that the global moment balance can be obtained by considering

$$\frac{\mathrm{d}}{\mathrm{d}t}(I_1 \dot{\alpha})=\sum\textrm{(anticlockwise moments over system as a whole)} $$

since $\alpha$ could be describing the orientation of the whole body as the rods are connected, but I'm not so sure about this. Surely it doesn't make sense for the LHS of the above to equal two different things?

Also, a few of questions:

1. Is it usually true that adding all the angular momentum equations for each rod (constituent parts of a system) would result in the overall angular momentum balance equation, just like you can close a system of N2L equations by adding them all together and recovering an equation for the whole system?
2. Would you still be able to write down an equation of angular momentum of the first rod involving $\alpha$ if moments on the rod are taken to be about a different point on it? I know the moment of inertia would be different, but what about the angle?
3. Is $\alpha$ an appropriate choice of angle for describing the orientation of the whole system here?

Answer 1

What you need to do is a bunch of free body diagrams to establish a relationship for each pin force as a function of the next body pin force and motion of the targeted body.

Consider the force/moment balance of body (1) as it receives the equal and opposite pin forces from body (2).

1. Definitions - Here we track the joint (pin) location with $\boldsymbol{r}_1 = (x_1,y_1)$ and the orientation of the body relative to horizontal with the angle $\theta_1$. The relative location of the center of mass with respect to the pin is $\boldsymbol{c}_1 = (c_x, c_y)$ and the relative location of the next pin $\boldsymbol{b}_2 = (b_x, b_y)$.

2. Degrees of Freedom - Each body contributes three degrees of freedom that track the location of the pin and the orientation of the body, ${q}_1 = ({x}_1, {y}_1, {\theta}_1)$ as well as the derivatives $\dot{q}_1 = (\dot{x}_1, \dot{y}_1, \dot{\theta}_1)$ and $\ddot{q}_1 = (\ddot{x}_1, \ddot{y}_1, \ddot{\theta}_1)$.

3. Constraints - The DOF of each pin relates to the DOF of the previous pin due to the rigid links between them.

   • Positions $$\begin{aligned} {x}_2 & = {x}_1 + b_x \\ {y}_2 & = {y}_1 + b_y \end{aligned}$$
   • Velocities $$\begin{aligned} \dot{x}_2 & = \dot{x}_1 - b_y \dot{\theta}_1 \\ \dot{y}_2 & = \dot{y}_1 + b_x \dot{\theta}_1 \end{aligned}$$
   • Accelerations $$\begin{aligned} \ddot{x}_2 & = \ddot{x}_1 - b_y \ddot{\theta}_1 - b_x \dot{\theta}_1^2 \\ \ddot{y}_2 & = \ddot{y}_1 + b_x \ddot{\theta}_1 - b_y \dot{\theta}_1^2 \end{aligned}$$

4. Kinematics - The velocity of the center of mass is $\boldsymbol{v}_C = (\dot{x}_1 - c_y \dot{\theta}_1, \dot{y}_1 + c_x \dot{\theta}_1)$. More importantly the acceleration of the center of mass is $\boldsymbol{a}_C = (\ddot{x}_1 - c_y \ddot{\theta}_1 - c_x \dot{\theta}_1^2, \ddot{y}_1 + c_x \ddot{\theta}_1 - c_x \dot{\theta}_1^2)$

5. Inertial Loads - Using D'Alembert's principle convert the accelerations into equal and opposite forces and torques and then consider the problem like a statics problem. At each pin the inertial loads are $$\begin{aligned} Fx_{\rm acc} &= m_1 (\ddot{x}_1 - c_y \ddot{\theta}_1 - c_x \dot{\theta}_1^2) \\ Fy_{\rm acc} &= m_1 (\ddot{y}_1 + c_x \ddot{\theta}_1 - c_x \dot{\theta}_1^2) \\ \tau_{\rm acc} &= I_1 \ddot{\theta} + c_x Fy_{\rm acc} - c_y Fx_{\rm acc} \end{aligned}$$

6. Force Balance at the pin - Add up all the loads and equipollent moments at the pin to write down the equations of motion for each body.

   $$ \left. \begin{aligned} Fx_1 - Fx_{\rm acc} - Fx_2 & = 0 \\ Fy_1 - Fy_{\rm acc} - Fy_2 & = 0 \\ \tau_1 - \tau_{\rm acc} - \tau_2 - b_x Fy_2 + b_x Fx_2 & = 0 \end{aligned}\; \right\} $$

   So start from the last body and move down the chain, taking the loads at the next pin $(Fx_2, Fy_2, \tau_2)$ and calculating the loads at this pin $(Fx_1, Fy_1, \tau_1)$. Since the last body doesn't have a next pin, the forces are directly a result of motion only.

   $$ \boxed{ \begin{aligned} Fx_1 & = Fx_2 + m_1 \left( \ddot{x}_1 - c_y \ddot{\theta}_1 - c_x \dot{\theta}_1^2 \right) \\ Fy_1 & = Fy_2 + m_1 \left( \ddot{y}_1 + c_x \ddot{\theta}_1 - c_y \dot{\theta}_1^2 \right) \\ \tau_1 & = \tau_2 -b_y Fx_2 + b_x Fy_2 + m_1 \left( -c_y \ddot{x}_1 + c_x \ddot{y}_1 + (c_x^2+c_y^2) \ddot{\theta}_1\right) + I_1 \ddot{\theta}_1 \end{aligned} }$$

7. Solution - There are $2n$ constraint equations (in acceleration form) and $3n$ force balance equations (where $n$ is the number of bodies) for a total of $5n$ equations. From the total of $3n$ DOF and $3n$ pin loads (and torques), $n$ must be specified in order to solve the system. For example, each pin produces zero torque $(\tau_1 =0, \tau_2 = 0, \ldots, \tau_n=0)$, or $n$ equations relating the rotations of each body, like $\ddot{\theta}_2 = \ddot{\theta}_1 + \text{(known value)}$.

8. Extra Credit - You can do something similar to the recursion of forces with linear and angular momentum $(px_1,py_1,L_1)$ defined at each pin

   $$ \begin{aligned} px_1 & = px_2 + m_1 \left( \dot{x}_1 - c_y \dot{\theta}_1 \right) \\ py_1 & = py_2 + m_1 \left( \dot{y}_1 + c_x \dot{\theta}_1 \right) \\ L_1 & = L_2 -b_y px_2 + b_x py_2 + m_1 \left( -c_y \dot{x}_1 + c_x \dot{y}_1 + (c_x^2+c_y^2) \dot{\theta}_1\right) + I_1 \dot{\theta}_1 \end{aligned} $$

   and this will help you establish how is the linear and angular momentum of each body relate to each other body, and to the total values. But when the system is connected to the ground on the first pin, the total (sum) momenta is not conserved because it is not an isolated system. There can be momentum exchange with the ground. So if you start by the conservation of momentum you will get an incorrect result for constrained systems (like a robotic arm).