Kinetic energy of planar rigid body examples
In this post, I focus on the calculation of kinetic energy in systems that exhibit both translational and rotational dynamics. By analyzing practical examples, I aim to provide a clear understanding of the principles at play, making this accessible to graduate students in mathematics, mechanical engineering, and quantum physics.
Rolling Without Slipping
A wheel rolling without slipping is a classic example where translational and rotational motions are directly related. The no-slip condition ensures that the translational velocity v\_C is proportional to the angular velocity \omega through the wheel’s radius R:
v_C = \omega R
The total kinetic energy is composed of translational and rotational components, which combine to yield:
T = \frac{3}{4} m R^2 \omega^2
This result highlights how the geometry and mass distribution of the wheel influence its motion.
Unbalanced Rolling Wheel
For an unbalanced wheel, the center of mass does not align with the geometric center, introducing additional complexity. The velocity of the center of mass v_C is influenced by both the angular velocity and the offset r of the mass center:
v_C^2 = R^2\omega^2 + r^2\omega^2 - 2Rr\omega^2 \cos (\theta)
Including this in the kinetic energy calculation shows how asymmetry affects motion, requiring adjustments to the moment of inertia and velocity expressions.
Rotating Bar About a Fixed Axis
A slender bar rotating about a fixed axis provides another interesting scenario. The kinetic energy depends on the moment of inertia, which varies with the reference point. Calculating about the center yields:
I_C = \frac{1}{12} m L^2
While considering the endpoint gives:
I_A = \frac{1}{3} m L^2
Both approaches ultimately confirm the total kinetic energy is consistent, regardless of the reference point.
For more insights into this topic, you can find the details here.