Rigid Body Impulse-Momentum

Hello! Welcome to the final lesson in our module on the planar kinetics of rigid bodies.

In the last lesson, we used the principle of work and energy to relate forces, displacement, and changes in speed. This scalar method is incredibly effective when you're analyzing motion over a specific distance.

Today, we'll explore the principle of impulse and momentum. This is a powerful vector method that connects forces, the time over which they act, and the resulting change in velocity. It's the go-to tool for analyzing situations involving short-duration forces, impacts, and collisions—scenarios common in aerospace, such as landing gear touchdowns or payload deployment mechanisms.

Your learning outcome for this lesson is to apply the principles of linear and angular impulse-momentum to rigid bodies.

1. Contrasting Work-Energy with Impulse-Momentum

Before we dive in, let's clarify when you would choose one method over the other:

Feature	Work-Energy Principle	Impulse-Momentum Principle
Core Idea	Force applied over a distance	Force applied over a time interval
Equation Type	Scalar ($T_1+\sum U_{1-2}=T_2$)	Vector (${\vec{L}}_1+{\overrightarrow{\text{Imp}}}_{1-2}={\vec{L}}_2$)
Key Variables	Force, displacement, velocity	Force, time, velocity
Best For...	Problems involving springs, gravity, and finding changes in speed after moving a certain distance.	Problems involving impacts, collisions, and finding velocity changes after a force acts for a specific duration.

2. The Principle of Impulse and Momentum

The core idea is that the initial momentum of a body, plus the impulse applied to it, equals its final momentum. For rigid bodies, we have two separate but related principles: one for linear motion and one for angular motion.

This diagram provides a great visual summary of the principle. It shows that by integrating the equations of motion ($F=ma$ and $M=I\alpha$) with respect to time, we arrive at the impulse-momentum equations.

The diagram illustrates a crucial problem-solving tool: the impulse-momentum diagram. We draw the body's momentum at time $t_1$, add the impulses applied between $t_1$ and $t_2$, and set it equal to the momentum at time $t_2$.

Let's break down the components.

2.1 Linear Impulse and Momentum

• Linear Momentum ($\vec{L}$): This is the momentum of the body's center of mass, $G$. It's a vector quantity defined as:
  $\vec{L}=m{\vec{v}}_G$
• Linear Impulse ($\overrightarrow{\text{Imp}}$): This is the effect of a force $\vec{F}$ applied over a time interval from $t_1$ to $t_2$. It's the integral of the force with respect to time:
  $\overrightarrow{\text{Imp}}={\int }_{t_1}^{t_2}\vec{F}dt$
  If the force is constant, this simplifies to $\overrightarrow{\text{Imp}}=\vec{F}(t_2-t_1)$.

The Principle of Linear Impulse-Momentum combines these:
$m({\vec{v}}_G{)}_1+\sum {\int }_{t_1}^{t_2}\vec{F}dt=m({\vec{v}}_G{)}_2$

Since this is a vector equation, we can break it into x and y components for 2D problems.

2.2 Angular Impulse and Momentum

• Angular Momentum ($H$): This is the "moment" of the linear momentum and also accounts for the body's rotation. For planar motion, its magnitude is most often calculated about the center of mass $G$:
  $H_G=I_G\omega$
• Angular Impulse ($\overrightarrow{\text{Ang Imp}}$): This is the effect of a moment $\vec{M}$ applied over a time interval.
  $\text{Ang Imp}={\int }_{t_1}^{t_2}Mdt$
  If the moment is constant, this simplifies to $\text{Ang Imp}=M(t_2-t_1)$. An angular impulse can be created by a pure couple moment or by an off-center force ($M=Fd$).

The Principle of Angular Impulse-Momentum (about the center of mass G) is:
$I_G{\omega }_1+\sum {\int }_{t_1}^{t_2}{M}_{G}dt=I_G{\omega }_2$

This clear and concise web page summarizes these fundamental equations.

3. A Complete Overview and Worked Examples

The best way to see how these principles work together is through guided examples. The following video first reviews the definitions we just discussed and then solves three different problems, showing how to apply the equations in practice.

4. Conservation of Momentum

A very important special case of the impulse-momentum principle occurs when there is no external impulse acting on a system over the time interval.

• Conservation of Linear Momentum: If $\sum \int Fdt=0$, then $m({\vec{v}}_G{)}_1=m({\vec{v}}_G{)}_2$. The linear momentum is conserved.
• Conservation of Angular Momentum: If $\sum \int Mdt=0$, then $(H{)}_1=(H{)}_2$. The angular momentum is conserved.

A classic example is a figure skater pulling their arms in. By reducing their mass moment of inertia ($I$), their angular velocity ($\omega$) must increase to keep the angular momentum ($H=I\omega$) constant. This principle is fundamental in controlling the attitude (orientation) of satellites and spacecraft using reaction wheels or thrusters.

The following text provides a good explanation of this concept.

5. Test Your Understanding

Let's apply these principles to an aerospace problem.

A 750 kg helicopter body (excluding the main rotor) has a mass moment of inertia $I_G=1300\text{ kg}\cdot {\text{m}}^2$ about its center of mass $G$. The main rotor suddenly speeds up, causing the helicopter body to start rotating counter-clockwise at ${\omega }_1=0.5\text{ rad/s}$.

To counteract this, the pilot uses the tail rotor, which provides a horizontal thrust $T$. The tail rotor is located 6 m from the helicopter's center of mass.

If the tail rotor provides a constant thrust of $T=450$ N, how long must this thrust be applied to stop the helicopter body's rotation (i.e., to make ${\omega }_2=0$)?

Diagram: A top-down view of a helicopter. The body has an initial angular velocity ω1. A thrust T from the tail rotor is applied at a distance d from the center of mass G.

Problem-Solving Steps:

1. Identify which principle to use: linear or angular impulse-momentum?
2. Write down the chosen equation: Initial Momentum + Impulse = Final Momentum.
3. Calculate the initial angular momentum, $I_G{\omega }_1$.
4. Determine the angular impulse caused by the tail rotor thrust. Remember that an off-center force creates a moment. Is the impulse positive or negative relative to the initial rotation?
5. What is the final angular momentum?
6. Substitute the values into the equation and solve for the unknown time, $\Delta t$.

Test your understanding!

Let's walk through the solution.

1. Principle: Since we're dealing with rotation and a force applied over time to stop it, the principle of angular impulse-momentum is the correct choice. We will consider counter-clockwise (CCW) as the positive direction.

2. Equation:
   $I_G{\omega }_1+\sum {\int }_{t_1}^{t_2}{M}_{G}dt=I_G{\omega }_2$

3. Initial Angular Momentum: The body is rotating CCW, which we defined as positive.
   $(H_G{)}_1=I_G{\omega }_1=(1300\text{ kg}\cdot {\text{m}}^2)(+0.5\text{ rad/s})=+650\text{ kg}\cdot {\text{m}}^2/\text{s}$

4. Angular Impulse: The tail rotor thrust $T$ creates a moment about $G$. The force is 450 N and the moment arm is 6 m. This moment will cause a clockwise (negative) rotation to counteract the initial CCW spin.
   $M_G=-(T\times d)=-(450\text{ N}\times 6\text{ m})=-2700\text{ N}\cdot \text{m}$
   Since the thrust is constant, the angular impulse is:
   $\int M_{G}dt=M_G\Delta t=-2700\Delta t$

5. Final Angular Momentum: The goal is to stop the rotation, so ${\omega }_2=0$.
   $(H_G{)}_2=I_G{\omega }_2=0$

6. Solve for $\Delta t$:
   $650+(-2700\Delta t)=0$
   $2700\Delta t=650$
   $\Delta t=\frac{650}{2700}\approx 0.241\text{ seconds}$

So, the tail rotor must provide thrust for about 0.24 seconds to stop the unwanted rotation. This shows just how responsive these systems need to be!

Conclusion

You have now completed the core topics of planar rigid body kinetics! You have three powerful methods at your disposal:

1. Equations of Motion ($F=ma,M=I\alpha$): For finding forces and accelerations at a single instant.
2. Work and Energy: For relating forces, distance, and speed (scalar).
3. Impulse and Momentum: For relating forces, time, and velocity (vector), especially useful for impacts.

Key Takeaways from this Lesson:

• The principle of impulse-momentum relates the initial and final states of motion of a body when forces and moments are applied over a time interval.
• The principle is expressed in two vector equations: one for linear motion ($m{\vec{v}}_{G1}+\text{Impulse}=m{\vec{v}}_{G2}$) and one for angular motion ($H_1+\text{Ang Impulse}=H_2$).
• When the net external impulse (linear or angular) on a system is zero, the corresponding momentum is conserved. This is a crucial concept for analyzing collisions and control systems.

Next Lesson Preview:
We will now move on to a new topic: Mechanical Vibrations. In the next module, you will learn how to derive the equation of motion for a simple oscillating system (a spring-mass system). This is the foundation for understanding how structures and machines, from aircraft wings to engine components, respond to dynamic loads, and it's a critical area of study in mechanical and aerospace engineering.