Planar Rigid Body Impulse Momentum Relationship
The particle impulse momentum relationship was derived already for particles here and states that for a body of continuously distributed mass, the sum of forces acting on the body is related to the time rate of change of the linear momentum:
\sum \mathbf F = \frac{\mathrm d \mathbf P}{\mathrm d t}.
Similarly, for a rigid body of continuously distributed mass, the sum of moments about its center of mass is related to the time rate of change of angular momentum:
\sum \mathbf M_C = \frac{\mathrm d \mathbf L_C}{\mathrm d t}.
Integrating both sides over a time interval t_1 to t_2, we obtain the angular impulse and the change in angular momentum:
\int_{t_1}^{t_2} \sum \mathbf M_C \, \mathrm d t = \Delta \mathbf L_C = \mathbf L_{C_f} - \mathbf L_{C_i}.
For linear motion, integrating the forces gives the linear impulse and change in linear momentum:
\int_{t_1}^{t_2} \sum \mathbf F \, \mathrm d t = \Delta \mathbf P = m \mathbf v_{C_f} - m \mathbf v_{C_i}.
In 2D planar rigid body motion, the angular momentum about the center of mass can be expressed generally as:
\mathbf L_C = I_{xz}^C \omega \mathbf i + I_{yz}^C \omega \mathbf j + I_{zz}^C \omega \mathbf k.
For symmetric bodies, the products of inertia I_{xz}^C and I_{yz}^C vanish, reducing the angular momentum to:
\mathbf L_C = I_{zz}^C \omega \mathbf k.
Substituting this result into the angular impulse equation, we get:
\int_{t_1}^{t_2} \sum \mathbf M_C \, \mathrm d t = I_{zz}^C \omega_f \mathbf k - I_{zz}^C \omega_i \mathbf k.
Adding the equations for linear impulse and angular impulse gives us:
\mathbf{m} \mathbf{v}_{C_i} + I_{zz}^C \omega_i \mathbf{k} + \int_{t_1}^{t_2} \sum \mathbf{F} \, \mathrm{d}t + \int_{t_1}^{t_2} \sum \mathbf{M_C} \, \mathrm{d}t = \mathbf{m} \mathbf{v}_{C_f} + I_{zz}^C \omega_f \mathbf{k}
This equation expresses that the sum of initial linear and angular momentum, plus the impulses due to forces and moments over time, is equal to the sum of the final linear and angular momentum.
If there are no external forces or impulses acting on a system in a specific direction over a given time period, the linear and angular momentum in that direction are conserved. This means the conservation of linear momentum in the x-direction:
m \dot{x}_{C_i} = m \dot{x}_{C_f}
or the conservation of linear momentum in the y-direction:
m \dot{y}_{C_i} = m \dot{y}_{C_f}
If there are no external moments about the center of mass, angular momentum is conserved:
I_{zz}^C \omega_i = I_{zz}^C \omega_f
Skater
Consider a skater initially spinning with angular velocity \omega_1 about the z-axis. Assuming no external forces or moments act on the skater, angular momentum is conserved. Angular momentum is given by:
L = I \omega
where I is the mass moment of inertia about the axis of rotation, and \omega is the angular velocity. Conservation of angular momentum implies:
I_1 \omega_1 = I_2 \omega_2
Here, I_1 and I_2 represent the mass moments of inertia at positions 1 and 2, respectively, corresponding to the skater’s arms extended and pulled in.
The mass moment of inertia I about an axis is defined as:
I = \int \left( x^2 + y^2 \right) \, \mathrm{d}m
where x and y are the perpendicular distances from the axis of rotation to the infinitesimal mass element \mathrm{d}m. For the skater, I_1 > I_2 because, with arms extended, more mass is distributed farther from the axis of rotation.
Substituting into the conservation equation, we find the relationship between the angular velocities:
\omega_2 = \frac{I_1}{I_2} \omega_1
Since I_1 > I_2, it follows that \omega_2 > \omega_1, meaning the skater spins faster after pulling their arms in.
Next, we calculate the rotational kinetic energy T for positions 1 and 2. Rotational kinetic energy is expressed as:
T = \frac{1}{2} I \omega^2
For position 1 (initial condition), the rotational kinetic energy is:
T_1 = \frac{1}{2} I_1 \omega_1^2
For position 2 (final condition), substituting \omega_2 = \frac{I_1}{I_2} \omega_1 into the kinetic energy equation yields:
T_2 = \frac{1}{2} I_2 \omega_2^2 = \frac{1}{2} I_2 \left( \frac{I_1}{I_2} \omega_1 \right)^2 = \frac{1}{2} \frac{I_1^2}{I_2} \omega_1^2
Comparing T_1 and T_2, we see that:
T_2 = T_1 \frac{I_1}{I_2}
Since I_1 > I_2, the rotational kinetic energy increases in position 2 (T_2 > T_1).
This increase in kinetic energy, despite the conservation of angular momentum, is explained by the skater’s internal energy. As the skater pulls their arms in, they perform work on their body, decreasing their internal energy. This work converts internal energy into additional rotational kinetic energy. The overall system obeys the first law of thermodynamics, ensuring energy conservation; there is no violation of any laws of nature.
Center of percussion
We will solve a practical engineering problem to determine the center of percussion of a body. This concept helps identify the point of impact on an object, such as a bat or stick, that eliminates the reaction force at its pivot.
This point, commonly referred to as the “sweet spot”, ensures smooth motion without generating a transverse reaction force. By applying the impulse-momentum principle, we will analyze a simplified model of a stick hitting a ball to determine the optimal impact location that minimizes reaction forces at the pivot.
The goal is to eliminate the reaction force R_y at the hinge. To solve this problem we can use the principle of impulse-momentum, we begin with the initial momentum, impulse, and final momentum for the system.
Let’s start summing In the y-direction:
m \dot{y}_{C_i} + \int_{t_i}^{t_f} \left( R_y - F \right) \, \mathrm{d}t = m \dot{y}_{C_f}.
Summing the moment about the pivot point O, using the mass center C as the reference, it is possible to ignore m\dot x_C as it is not a quantity of interest in the problem:
I_C \omega_i + \frac{L}{2} m \dot{y}_{C_i} - \int_{t_i}^{t_f} d F \, \mathrm{d}t = I_C \omega_f + \frac{L}{2} m \dot{y}_{C_f}.
Substituting the moment of inertia for a bar about its center of mass, I_C = \frac{1}{12} m L^2, and expressing \dot{y}_C = \frac{L}{2} \omega, we rewrite the angular momentum equation as:
\frac{1}{12} m L^2 \omega_i + \frac{L}{2} \cdot m \cdot \frac{L}{2} \omega_i - \int_{t_i}^{t_f} d F \, \mathrm{d}t = \frac{1}{12} m L^2 \omega_f + \frac{L}{2} \cdot m \cdot \frac{L}{2} \omega_f
Simplifying:
\left( \frac{1}{12} m L^2 + \frac{1}{4} m L^2 \right) \omega_i - \int_{t_i}^{t_f} d F \, \mathrm{d}t = \left( \frac{1}{12} m L^2 + \frac{1}{4} m L^2 \right) \omega_f
Combining terms:
\frac{1}{3} m L^2 \omega_i - \int_{t_i}^{t_f} d F \, \mathrm{d}t = \frac{1}{3} m L^2 \omega_f
The condition for No Reaction Force is R_y = 0. Setting in in the linear momentum equation:
\int_{t_i}^{t_f} F \, \mathrm{d}t = m \dot{y}_{C_f} - m \dot{y}_{C_i}
Substitute \dot{y}_C = \frac{L}{2} \omega:
\int_{t_i}^{t_f} F \, \mathrm{d}t = m \frac{L}{2} (\omega_f - \omega_i)
The angular momentum equation becomes:
-\int_{t_i}^{t_f} d F \, \mathrm{d}t = \frac{1}{3} m L^2 (\omega_f - \omega_i)
Multiplying the linear impulse equation by d to match dimensions:
\int_{t_i}^{t_f} d F \, \mathrm{d}t = -\frac{m L d}{2} (\omega_f - \omega_i)
Combining these equations:
\frac{1}{3} m L^2 (\omega_f - \omega_i) = -\frac{m L d}{2} (\omega_f - \omega_i)
Factor out (\omega_f - \omega_i) and solve for d:
\frac{1}{3} m L^2 = \frac{m L d}{2}
d = \frac{2}{3} L
The center of percussion is located at d = \frac{2}{3} L from the pivot. At this point, the reaction force R_y at the hinge is eliminated, ensuring smooth motion.