14.1: Impulse-Momentum Equations for a Rigid Body
The concepts of impulse and momentum provide a third method of solving kinetics problems in dynamics. Generally this method is called the Impulse-Momentum Method , and it can be boiled down to the idea that the impulse exerted on a body over a given time will be equal to the change in that body's momentum. In a rigid body we will be concerned with not only linear impulse and momentum, but also angular impulse and momentum. The linear and angular impulse momentum equations are below, with the new term for angular impulse \((\vec{K})\) starting out the angular impulse momentum equation.
\[ \vec{J} = m \vec{v}_f - m \vec{v}_i \]
\[ \vec{K} = I_G \vec{\omega}_f - I_G \vec{\omega}_i \]
For two-dimensional problems, we can break the linear impulse equation down into two scalar components to solve. In the case of planar problems, we simply need to break all forces and velocities into \(x\) and \(y\) components - since all rotation will be about the \(z\)-axis, the angular impulse momentum equation remains a single equation. Notice, however, that we will need to take the mass moment of inertia about the center of mass of the body into account; similarly, we will use the velocity of the center of mass when discussing the velocity in the linear impulse momentum equations.
\[ J_x = m v_{f,x} - m v_{i,x} \]
\[ J_y = m v_{f,y} - m v_{i,y} \]
\[ K_z = I_G \omega_f - I_G \omega_i \]
Impulse:
As discussed with particles, a linear impulse in its most basic form is a force integrated over a time. For a force with a constant magnitude, we can find the magnitude of the impulse by multiplying the magnitude of the force by the time that force is exerted. If the force is not constant, we simply integrate the force function over the set time period. The direction of the impulse vector will be the direction of the force vector, and the units will be a force multiplied by a time (Newton-seconds or pound-seconds, for example).
\begin{align} \text{Constant magnitude force:} \quad &\, \vec{J} = \vec{F} * t \\ \text{Non-constant magnitude force:} \quad &\, \vec{J} = \int \vec{F}(t) \ dt \end{align}
An angular impulse is similar to a linear impulse, except it is the moment exerted over time instead of the force exerted over time.
\begin{align} \text{Constant magnitude moment:} \quad &\, \vec{K} = \vec{M} * t \\ \text{Non-constant magnitude moment:} \quad &\, \vec{K} = \int \vec{M}(t) \ dt \end{align}
This moment can come either in the form of a torque directly applied to a body, or an off-center force causing a linear and angular impulse at the same time. All moments should be taken about the center of mass of the body.
Momentum:
As discussed with particles, the linear momentum of a body will be equal to the mass of the body times its current velocity. Since velocity is a vector, the momentum will also be a vector, having both magnitude and a direction. Unlike the impulse, which happens over some set time, the momentum is captured as a snapshot of a specific instant in time (usually right before and after some impulse is exerted). The units for linear momentum will be mass times unit distance per unit time. This is usually kilogram-meters per second in metric, or slug-feet per second in English units.
Angular momentum, on the other hand, will be equal to the body's mass moment of inertia (about its center of mass) times its current angular velocity. It's important to note that while the mass moment of inertia remains constant if the body does not change shape, a shape that does change shape will likely have changes in mass moments of inertia along with changes in angular velocity.
Practice \(\PageIndex{1}\)
A miter saw has an operating speed of 1500 rpm. The blade and motor armature have a combined weight of 3 lbs and a radius of gyration of 1 inch.
- What is the time required for the bearing friction alone (torque=0.015 inch-lbs) to stop the blade?
- What is the torque a braking system would need to apply to stop the blade in just 0.25 seconds?
- Solution (not yet available):
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Not yet available.
Practice \(\PageIndex{2}\)
A bowling ball is modeled as a 7-kilogram uniform sphere, 300 mm in diameter. The ball is released with an initial velocity of 6 m/s on a horizontal wooden floor (\(\mu_k\) = 0.1) with zero angular velocity.
- How long does it take before the ball begins to roll without slipping?
- What is the linear velocity of the ball at this time?
- Solution (not yet available):
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Not yet available.