# Kinematics derivation of conservation of angular momentum

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1. Jun 21, 2015

### kzf

1. The problem statement, all variables and given/known data
A welding robot consists of an arm (thin rod) that can rotate about the origin point O, and a welding tip, which can freely move along the rod, from the outermost point of the arm A all the way to the center point O. The design invokes two electric motors, one to turn the arm, and one to move the welding tip along the arm. The welding robot has moment of inertia IO about O and length L. The welding tip has mass m and radius of gyration kg.

a)
Derive and state the equations of motion for this welding robot.

Consistency check: Consider the situation where Motor 1 delivers Zero moment, and the welding tip is moving inwards. What does conservation of angular momentum say is going to happen? Is this reflected in your equations of motion?

Note: This is not the full problem. The rest of the problem involves optimization but I think that I can solve it with conservation of angular momentum. However, I cannot figure out how to state the equations of motion in such a way that conservation of angular momentum will be obeyed.

2. Relevant equations
Acceleration of rigid body in rotating reference frame:
$$a = \alpha \times \vec{r} + \vec{a}_{x' y'} + 2 \vec{\omega} \times \vec{v}_{x' y'} - \left|\vec{\omega}\right|^2 \vec{r}$$

Conservation of angular momentum
$$I_G \omega = const$$

3. The attempt at a solution
From the problem statement, it seems like I am to develop equations of motion without using conservation of angular momentum, but which will still ultimately agree with conservation of angular momentum.

The rotational motor has no torque, and the linear motor pulls inward in order to have the tip move inwards at constant velocity.

Attempt 1: Equations of motion.
I took the first equation governing the acceleration of a rigid body, and applied it to the arm. Let a' be the linear acceleration of A along the rod in the rotating frame, and v' be the linear velocity of A along the rod in the rotating frame.

$$\vec{a} = \alpha\left<-r\sin\theta, r\cos\theta\right> + a' \left<\cos\theta, \sin\theta\right> + 2\omega v' \left<-\sin\theta, \cos\theta\right> - \omega^2 \left<\cos\theta, \sin\theta\right>$$

where I have used <x, y> to denote xî + yĵ. Then,

$$\vec{a} = \left<(a' - \omega^2)\cos\theta - (\alpha r + 2 \omega v')\sin\theta, (a' - \omega^2)\sin\theta + (\alpha r + 2 \omega v')\cos\theta \right>$$

Now v' is some constant negative scalar (since the head is moving inwards), and a' is zero. At θ = 0, under the constraints of the problem, this reduces to

$$\vec{a} = \left< -\omega^2, \alpha r + 2 \omega v' \right>$$

but I cannot see if this is relevant or not.

Attempt 2: Working backwards

From the conservation of angular momentum equation,

$$(I_O + m k_g^2 + mr_f^2)r_f^2 \omega_f = (I_O + m k_g^2 + mr_i^2)r_i^2 \omega_i$$
$$\omega_f = \frac{(I_O + m k_g^2 + mr_i^2)r_i^2 \omega_i}{(I_O + m k_g^2 + mr_f^2)r_f^2}$$

But once again I fail to see its relevance.

Attempt 3: Forces

The force applied by the linear motor must be equal to the centripetal force,
$$\vec{F} = -\omega^2 \vec{r}m$$
but I instantly gave up on this.

Is there some crucial thing that I am missing, or am I misinterpreting the question?