Fixed-Axis Rotation Kinetics

Hello! Welcome back to our course.

In the last lesson, we explored the kinetics of general planar motion, where a body can both translate and rotate. We learned to apply the three core equations of motion: two for forces ($\sum F_x,\sum F_y$) and one for moments ($\sum M_G$).

Today, we'll focus on a very common and important special case: fixed-axis rotation. This is where a rigid body is pinned or hinged about an axis that does not move. This type of motion is everywhere in engineering, from the rotor of a jet engine and the propeller of a turboprop aircraft to the flight control surfaces like ailerons and rudders moving on their hinges.

Our goal is to solve rigid body kinetics problems involving fixed-axis rotation. We will see how the general equations of motion are adapted for this scenario and learn a powerful shortcut that often makes solving these problems much more direct.

1. Recapping the Equations of Motion

Let's start with a quick review. From our previous lesson, we know the motion of any rigid body is governed by:

1. $\sum \vec{F}=m{\vec{a}}_G$ (The net force equals mass times the acceleration of the center of mass G)
2. $\sum M_G=I_G\alpha$ (The net moment about the center of mass G equals the moment of inertia about G times the angular acceleration)

When a body rotates about a fixed axis at a point O, the center of mass G moves in a circular path around O. This means its acceleration, ${\vec{a}}_G$, can be broken down into two familiar components:

• Normal acceleration ($a_{Gn}$): Points from G towards the center of rotation O. Its magnitude is $a_{Gn}=r_G{\omega }^2$, where $r_G$ is the distance from O to G.
• Tangential acceleration ($a_{Gt}$): Is tangent to the circular path. Its magnitude is $a_{Gt}=r_G\alpha$.

This suggests that it's often more convenient to use a normal-tangential (n-t) coordinate system instead of x-y. In this system, our force equations become:

• $\sum F_n=m{a}_{Gn}=m{r}_G{\omega }^2$
• $\sum F_t=m{a}_{Gt}=m{r}_G\alpha$

The video below gives a great overview of these equations and introduces the concept of the kinetic diagram for fixed-axis rotation, which visually represents the $m{\vec{a}}_G$ and $I_G\alpha$ terms.

2. The Most Direct Approach: Moments About the Fixed Axis

While using the three equations ($\sum F_n$, $\sum F_t$, $\sum M_G$) always works, it often requires solving a system of three simultaneous equations. Just as we found a shortcut using the Instantaneous Center (IC) for rolling motion, a similar powerful shortcut exists for fixed-axis rotation.

We can sum the moments directly about the fixed axis of rotation O:
$\sum M_O=I_O\alpha$

Why is this so useful?

1. Eliminates Unknowns: The reaction forces at the pin support (e.g., $O_x$ and $O_y$) pass through point O, meaning they create no moment about O. This removes them from the equation, often leaving you with a single equation for a single unknown, $\alpha$.
2. Direct Solution: It directly connects the external torques (from weight, applied forces, etc.) to the resulting angular acceleration.

There is one crucial detail: you must use $I_O$, the mass moment of inertia about the axis of rotation O, not $I_G$. As we saw in a previous lesson, we can find this using the parallel axis theorem:
$I_O=I_G+m{d}^2$
where $d$ is the distance between the center of mass G and the fixed axis O.

The resource below clearly explains the two main approaches: summing moments about G or summing moments about O.

3. A Complete Worked Example

Let's see how this is applied in a full problem. We will find the initial angular acceleration and the reaction forces at the pin for a plate released from rest.

Notice the strategy:

1. Draw FBD and Kinetic Diagram: This is always the first step.
2. Solve for $\alpha$: Use the moment equation $\sum M_A=\sum (M_k{)}_A$. The video refers to the right-hand side, the "kinetic moments," as $\sum (M_k{)}_A$. This is equivalent to our $\sum M_O=I_O\alpha$ equation. Since the object starts from rest ($\omega =0$), the normal acceleration is zero, which simplifies the kinetic diagram.
3. Solve for Reactions: With $\alpha$ known, use the force equations ($\sum F_x=m{a}_{Gx}$ and $\sum F_y=m{a}_{Gy}$) to find the pin reactions.

Test your understanding!

Consider the 25-kg flywheel shown below, which is released from rest in the position shown. Its radius of gyration is $k_G=0.18$ m. We want to find the initial angular acceleration $\alpha$ and the reaction forces at the pin O immediately after release.

Steps:

1. Calculate the mass moment of inertia about the center of mass, $I_G$. Remember, $I_G=m{k}_G^2$.
2. Use the most efficient moment equation to find the initial angular acceleration, $\alpha$.
3. Use the force equations in the n-t coordinate system to find the normal and tangential components of the pin reaction, $O_n$ and $O_t$.

Show answer

1. Moment of Inertia:
First, calculate $I_G$:
$I_G=m{k}_G^2=(25\text{ kg})(0.18\text{ m}{)}^2=0.81\text{ kg}\cdot {\text{m}}^2$

2. Find Angular Acceleration ($\alpha$):
The most efficient method is to sum moments about the fixed pin O.
The only external force creating a moment about O is the weight ($mg$). The lever arm is the horizontal distance from O to the line of action of the weight, which is 0.15 m.
$\sum M_O=W\cdot (0.15)=(25\cdot 9.81)\cdot (0.15)=36.79\text{ N}\cdot \text{m}$
Now, we set this equal to $I_O\alpha$. First, we need $I_O$ using the parallel axis theorem with $d=r_G=0.15$ m:
$I_O=I_G+m{d}^2=0.81+25\cdot (0.15{)}^2=0.81+0.5625=1.3725\text{ kg}\cdot {\text{m}}^2$
Now solve for $\alpha$:
$\sum M_O=I_O\alpha ⟹36.79=1.3725\alpha ⟹\alpha =\frac{36.79}{1.3725}=26.8{\text{ rad/s}}^2\text{ (clockwise)}$

3. Find Pin Reactions:
We use a normal-tangential coordinate system centered at G, with the tangential direction (t) pointing straight down and the normal direction (n) pointing from G to O.
Since the flywheel is released from rest, $\omega =0$.

• Normal acceleration: $a_{Gn}=r_G{\omega }^2=0$.
• Tangential acceleration: $a_{Gt}=r_G\alpha =(0.15\text{ m})(26.8{\text{ rad/s}}^2)=4.02{\text{ m/s}}^2$.

Now apply the force equations:

• $\sum F_n=O_n=m{a}_{Gn}=m(0)=0\text{ N}$
• $\sum F_t=W-O_t=m{a}_{Gt}⟹(25\cdot 9.81)-O_t=25\cdot (4.02)$
  $245.25-O_t=100.5⟹O_t=245.25-100.5=144.75\text{ N}$

The initial reaction at the pin is purely tangential (acting upwards) with a magnitude of 144.75 N.

4. Application to Connected Systems

The principles of fixed-axis rotation are also fundamental to analyzing systems where a rotating component is connected to other translating bodies, like a block and pulley system where the pulley has mass. In these cases, you analyze each body separately and connect their motions through the tension in the cord.

The following example demonstrates how to solve a problem with two blocks and a massive pulley. Notice how one moment equation for the pulley and one force equation for each block creates a solvable system.

Conclusion

You have now learned how to analyze the kinetics of rigid bodies undergoing fixed-axis rotation. This is a critical skill for understanding many mechanical systems you'll encounter in aerospace engineering and beyond.

Key Takeaways:

• Fixed-axis rotation problems are solved using the same fundamental principles as general planar motion, but we can adapt them for greater efficiency.
• The acceleration of the center of mass G has normal ($a_{Gn}=r_G{\omega }^2$) and tangential ($a_{Gt}=r_G\alpha$) components.
• The most powerful tool for these problems is often the moment equation about the fixed axis O: $\sum M_O=I_O\alpha$.
• This equation eliminates the unknown pin reactions and often allows for a direct solution for the angular acceleration $\alpha$.
• Once $\alpha$ is known, the force equations ($\sum F_n=m{a}_{Gn}$ and $\sum F_t=m{a}_{Gt}$) can be used to find the support reactions.

Next Lesson Preview:
In our next lesson, we will explore another powerful technique for solving kinetics problems: the work-energy method for rigid bodies. This approach is especially useful when you need to find changes in speed resulting from forces and moments acting over a distance or rotation, often providing a much quicker solution than using the equations of motion directly.