Question

Let the total energy of a system of \(N\) particles be measured in an arbitrary frame of reference, such that \(K=\Sigma \frac{1}{2} m_{n} v_{n}^{2} .\) In the center-of-mass reference frame, the velocities are \(v_{n}^{\prime}=v_{n}-v_{\mathrm{cm}}\), where \(v_{\mathrm{cm}}\) is the velocity of the center of mass relative to the original frame of reference. Keeping in mind that \(v_{n}^{2}=\overrightarrow{\mathbf{v}}_{n} \cdot \overrightarrow{\mathbf{v}}_{n}\), show that the kinetic energy can be written $$K=K_{\mathrm{int}}+K_{\mathrm{cm}}$$ where \(K_{\text {int }}=\Sigma \frac{1}{2} m_{n} v_{n}^{\prime 2}\) and \(K_{\mathrm{cm}}=\frac{1}{2} M v_{\mathrm{cm}}^{2} .\) This demonstrates that the kinetic energy of a system of particles can be divided into an internal term and a center-of-mass term. The internal kinetic energy is measured in a frame of reference in which the center of mass is at rest; for example, the random motions of the molecules of gas in a container at rest are responsible for its internal translational kinetic energy.

Short answer

The kinetic energy of a system of particles can be partitioned into two parts: the internal kinetic energy \(K_{int}\) and the kinetic energy of the center of mass \(K_{cm}\). This is achieved by substituting the velocity in the center-of-mass frame into the expression for kinetic energy, expanding, and simplifying the resultant algebraic expression to obtain \(K = K_{int} + K_{cm}\).

Step-by-step solution

Definition of kinetic energy

Start by understanding the definition of kinetic energy given. It is defined for \(N\) number of particles as the sum of \( \frac{1}{2} m_{n} v_{n}^{2} \), where \(m_{n}\) is the mass and \(v_{n}\) is the velocity of nth particle.

Relate velocity in center-of-mass frame to original frame

In the center-of-mass frame, velocities are given by \(v_{n}^{\prime}=v_{n}-v_{\mathrm{cm}}\), where \(v_{\mathrm{cm}}\) is the velocity of the center of mass relative to the original frame of reference. Let's substitute this into the expression for \(K\).

Substitute, expand and simplify

Plugging \(v_{n}^{\prime}=v_{n}-v_{\mathrm{cm}}\) back to \(K=\Sigma \frac{1}{2} m_{n} v_{n}^{2}\) yields \(K=\Sigma \frac{1}{2} m_{n} (v_{n} - v_{cm})^{2}\). This can be broken down into 3 terms: the sum of the squares of each particle's velocity, the sum of the cross terms, and the sum of the squares of the center of mass's velocity. These sum up to calculate \(K\).

Definition of kinetic energy in the center-of-mass frame

The terms \(K_{int}=\Sigma \frac{1}{2} m_{n} v_{n}^{\prime 2} \) and \( K_{cm}=\frac{1}{2} M v_{cm}^{2} \) are defined as the internal kinetic energy and the kinetic energy of the center of mass, respectively. This finally leads to the desired result \(K= K_{int} + K_{cm}\)

Key concepts

Understanding the Center-of-Mass Reference Frame

When examining the motion of a system of particles, physics introduces the center-of-mass reference frame—a key concept that simplifies analysis. Imagine a seesaw perfectly balanced with various weights; it's balancing point is analogous to this frame's role. It's a very specific viewpoint where the center of mass, which is the average position of all mass in the system, isn't moving. This is incredibly useful because it filters out the noise of the system's overall movement, letting us focus solely on the particles' movements relative to each other.

In physics exercises, we often switch to this frame to analyze internal motion without the complication of the system’s movement through space. For example, if you're on a moving train and start walking, your velocity relative to the ground is your walking speed plus the train’s speed. But within a center-of-mass reference frame fixed to the train, your velocity is just your walking speed. Similarly, by shifting into this frame, we zero in on the essence of a system's behavior without external influences, revealing the pure form of internal kinetic interactions.

Decomposing Kinetic Energy

Kinetic energy breakdown, or decomposition, is a powerful tool in physics. It allows us to dissect the total kinetic energy of a system into more manageable and insightful parts. Think of it like a pie chart of movement, where each slice represents a different contribution to the overall motion.

Specifically, for a system of particles, we divide kinetic energy into 'internal' and 'center-of-mass' components. The internal kinetic energy refers to the energy due to the motion of particles relative to the center of mass—it's like measuring how vigorously particles jiggle or rotate without accounting for the system drifting through space. On the other hand, the kinetic energy of the center of mass measures how the entire system moves as a singular entity. It's akin to how fast a car is going down the road—regardless of what the passengers are doing inside.

Internal Kinetic Energy Explained

Internal kinetic energy is a term that might sound complex, but it's all about looking at the 'hidden' motion within a system. It's the motion you can't see from the outside because it doesn't contribute to the system's movement from point A to point B; instead, it's what's happening under the hood. If you have a swarm of bees in a jar that's sitting still, the bees are buzzing around like crazy inside—that's all internal kinetic energy.

When we talk about this internal energy, it’s really important to understand that we're considering the motion with respect to the system's center of mass. This perspective sheds light on the intrinsic energy possessed by the particles due to their relative motion. So when you hold that jar of bees, the internal kinetic energy is what you would feel as vibrations. In an educational setting, grasping this concept allows students to untangle complex systems into simpler parts, making problem-solving more intuitive and manageable.